text:
\section{Introduction}
The power graph of finite groups has been studied in \cite{Power finite}, \cite{power semi}, \cite{power combi}, \cite{Power finiteII}.
In this work, we denote the group of integers mod $n$ under addition by $\mathbb{Z}_n$, or $\mathbb{Z} / n\mathbb{Z}$.  These notations are used interchangeably in this article.  It can be shown that any cyclic group of order $n$ is isomorphic to $\mathbb{Z}_n$.

A $\textit{graph}$ is a pair $\Gamma=(V(\Gamma),E(\Gamma))$ where $V(\Gamma)$ is a non-empty set, called the \textit{vertex set} whose elements are called \textit{vertices}, and $E(\Gamma)$ is a (possibly empty) set consisting of sets of pairs of elements of $V(\Gamma)$ called the $\textit{edge set}$, whose elements are called $\textit{edges}.$  If $e\in E(\Gamma)$ is an edge and $v\in e$, then $e$ is said to be $\textit{incident }$to $v$.  The \textit{degree} of a vertex $v$ is the number of edges incident to $v$.  If $\{v_1,v_2\}\in E(\Gamma)$, then $v_1$ and $v_2$ are said to be adjacent. Where there is no ambiguity, $V(\Gamma)$ is sometimes denoted as just $V$, and similarly $E(\Gamma)$ as E.  A $\textit{directed graph}$ or digraph is a pair $\Gamma=(V(\Gamma),E(\Gamma))$ where $V(\Gamma)$ is a non-empty vertex set, and $E$ is a (possible empty) set consisting of ordered pairs of elements in $V(\Gamma)$ called the edge set.

Let $G$ be a group.  The \textit{power graph} of $G$, denoted by $\mathfrak{g}(G)$ is defined as the graph with vertex set consisting of the elements of $G$, and edge set $E=\{\{x,y\}\mid x\neq y $ and $\langle x\rangle \leq \langle y\rangle$ or $\langle y\rangle \leq \langle x\rangle \}$. The \textit{directed power graph} of $G$, denoted $\vec{\mathfrak{g}}(G)$ is defined as the graph with vertex set consisting of the elements of $G$, and edge set $E=\{(x,y)\mid x\neq y $ and $\langle y\rangle \leq \langle x\rangle$\}.

\section{Some Properties of Power Graphs of Groups}
A graph $\Gamma$ is called $\textit{connected}$ if for any two vertices $u$ and $v$ in $\Gamma$, there is a path joining $u$ and $v$.  In a connected graph $\Gamma$, the distance between $u$ and $v$, denoted $d_{\Gamma}(u,v)$, can be defined as the length of the shortest path joining $u$ and $v$.  The $\textit{eccentricity}$ of a vertex $u$ is the maximum distance between $u$ and any other vertex in $\Gamma$.  The $\textit{radius}$ of $\Gamma$ is the minimum eccentricity of a vertex in $\Gamma$, and the $\textit{diameter}$ of $\Gamma$ is the maximum eccentricity of a vertex in $\Gamma$.  A central vertex in $\Gamma$ is a vertex with eccentricity equal to the radius of $\Gamma$.  The $\textit{center}$ of $\Gamma$ is the set of central vertices in $\Gamma$.

\begin{proposition}
It is well known that the Power graphs of finite groups are connected.
\end{proposition}
\begin{proof}
Let $G$ be a group with power graph $\mathfrak{g}(G)$ and identity $e$.  Let $g$ be an arbitrary element of $G$.  Since $G$ is a group, $\langle e\rangle \leq \langle g\rangle$, so in the power graph of a group there is an edge between the identity and every other group element.
\end{proof}

Note that the distance between the identity and any other vertex is $1$, in the power graph of a group, giving the following corollary.

\begin{Corollary}
The radius of the power graph of a group is $1$, and the center of the power graph of a group of order $n$ is the set of vertices with degree $n-1$.
\end{Corollary}

The $\textit{center}$ of a group $G$ is the set of elements of $G$ which commute with all other elements of $G$.  The following proposition gives a relation between the center of a power graph of a group and the center of the group.

\begin{proposition}
Let $G$ be a group of order $n$ with power graph $\mathfrak{g}(G)$. The vertices in the center of the power graph $\mathfrak{g}(G)$ are in the center of the group $G$.
\end{proposition}
\begin{proof}
Let $g\in G$ and suppose $deg_{\mathfrak{g}(G)}(g)=n-1$.  Then for any other $h\in G$, either $\langle h\rangle \leq \langle g\rangle$, or $\langle g\rangle \leq \langle h\rangle$, either way $g$ and $h$ commute.  Hence, $g$ is in the center of $G$.
\end{proof}

Since power graphs of groups are connected and the identity is adjacent to every other element, there is a path of length at most $2$ between any two vertices of a power graph of a group, so the diameter of the power graph of a group is $2$, unless the power graph is complete, in which case it is $1$.

\begin{proposition}
Let $G$ be a non trivial Abelian group of order $n$.  The center of $\mathfrak{g}(G)$ has cardinality
\begin{enumerate}
    \item $\phi(n)+1$, if $G$ is cyclic and $n\neq p^k$ for any prime $p$ and positive integer $k$.
    \item $n$, if $G$ is cyclic and $n=p^k$ for some prime $p$ and positive integer $k$.
    \item $1$, if $G$ is non-cyclic.
\end{enumerate}
\end{proposition}
\begin{proof}$\phantom e$
\begin{enumerate}
\item[(1):] First let $G$ be a cyclic group whose order is $n$, and suppose $n$ is not a power of a prime.  Note, all the $\phi(n)$ many generators of $G$ and the identity are in the center of $\mathfrak{g}(G)$, so the center has size at least $\phi(n)+1$.  Since $n$ is not a power of a prime, $n$ has at least two distinct prime factors. Consider any $b\in G$, where $b$ is neither a generator of $G$, nor the identity of $G$. So, if order of $b$ is $m$, then $m<n$. Hence, by converse of Lagrange's Theorem over finite Abelian group, there exists $c\in G$, where order of $c$ is $r$ and $r$ does not divide $m$, $m$ does not divide $r$. So, there is no edge between $b$ and $c$. Since $b$ is an arbitrary non-generator of $G$, which is not the identity of $G$ either, the cardinality result  follows.

\item[(2):] Now suppose $G$ is cyclic and it's order is $n=p^k$ for some prime $p$ and positive integer $k$. Then, the graph being complete, by \cite{power semi}, the center has cardinality $n$.

\item[(3):] Let, $G$ be not cyclic. Then, the result follows by the proposition \ref{prop 8} in the next section.
\end{enumerate}
\end{proof}
\section{Some results relating Composition series in power graphs}

     The Jordan-H\"{o}lder theorem asserts that any two composition series of a given group have the same length, and isomorphic factor groups up to permutation.  The length of a composition series of a group $G$ is called its \textit{composition length} and denoted as $\ell$(G). The composition length $\ell(G)$ is also the maximum length a normal series of $G$ can have.  Then a group $G$ can be partitioned in the following way, let $X_i$ be the set of all elements of $G$ which generate a cyclic subgroup with composition length $i$. Then $G=\bigcup_i X_i$, and $X_i\cap X_j=\varnothing$ for any $i\neq j$.

    \begin{Lemma}
    The composition length of an Abelian group is the sum of the exponents of the order of the group, when written as a product of prime numbers.
    \end{Lemma}
    \begin{proof}
    An Abelian group is simple only if its order is prime.  Then if $\lvert G\rvert=p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}$, then each $G_{i}/G_{i+1}$ must have prime order, that is $\lvert G_{i}\rvert=p_i\lvert G_{i+1}\rvert$.  Then each $G_{i-1}$ has an order whose exponents sum to one less than those in the order of $G_i$, so there must be $k_1+k_2+\cdots+k_n$ inclusions in a maximal length normal series of an Abelian group.
    \end{proof}
    \begin{proposition}
    If $X_i$ is non-empty, then $X_{i-1}$ is non-empty.
    \end{proposition}
    \begin{proof}
    Let $x\in X_i$ and $\langle x\rangle\supset \langle x_{i-1}\rangle\supset\cdots\langle x_2\rangle \supset \{e\}$ be a composition series.  Then there is a normal series of length $i-1$ from $\langle x_{i-1}\rangle$. Suppose there were a normal series of length $i$ or greater from $\langle x_{i-1}\rangle$, then $\langle x\rangle \supset \langle x_{i-1}\rangle \supset\cdots\supset \{e\}$ is a normal series from $\langle x\rangle$ of length greater than $i$, contradicting the assumption that $x\in X_i$.
    \end{proof}
    \begin{Corollary}
    Removing subgroups from the beginning of a composition series of length greater than zero from a cyclic subgroup leaves a composition series.
    \end{Corollary}
    \begin{proposition}
    If some vertex in $X_i$ is adjacent to all other vertices in $X_i$, then $X_i$ is a clique in $\mathfrak{g}(G)$.
    \end{proposition}
    \begin{proof}

    If $x_1,x_2\in X_i$ and they are adjacent in $\mathfrak{g}(G)$, then either $\langle x_1\rangle \leq \langle x_2\rangle$ or $\langle x_2\rangle \leq \langle x_1\rangle$.  Without loss of generality suppose $\langle x_2\rangle$ is a proper subgroup of $\langle x_1\rangle$.  Then there exists a composition series $\langle x_2\rangle\supset \langle x_3\rangle\supset\cdots\supset\{e\}$ of length $i$ and $\langle x_1\rangle\supset \langle x_2\rangle\supset\cdots\supset\{e\}$ is a normal series of length $i+1$, contradicting the assumption that $x_1\in X_i$, so it must be the case that $\langle x_2\rangle=\langle x_1\rangle$.  Then, if any element of $X_i$ is adjacent to all other elements of $X_i$ in $\mathfrak{g}(G)$, then all elements of $X_i$ generate the same subgroup and form a clique in $\mathfrak{g}(G).$
    \end{proof}
    \begin{proposition}
    If $X_i$ is a clique in $\mathfrak{g}(G)$, then $X_{i+1}$ is a clique in $\mathfrak{g}(G)$.
    \end{proposition}
    \begin{proof}
    Suppose $X_i$ is a clique in $\mathfrak{g}(G)$ and $X_{i+1}$ is not a clique in $\mathfrak{g}(G)$, and let $x\in X_i$.  Then there exist two distinct elements in $X_{i+1}$, say $y$ and $z$, with orders $p\lvert x\rvert$ and $q\lvert x\rvert$ respectively where $p$ and $q$ are prime and $p\neq q$.  Write $\lvert x\rvert=p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}$, with $\sum_{t=1}^n k_t=i$.  Then $\langle y\rangle$ has a subgroup of order $pp_1^{k_1-1}p_2^{k_2}\cdots p_n^{k_n}$, and $\langle z\rangle$ has a subgroup of order $qp_1^{k_1-1}p_2^{k_2}\cdots p_n^{k_n}$.  Both of these subgroups are in $X_i$, but since $p\neq q$ at least one of the subgroups is not equal to $\langle x\rangle$, contradicting the assumption that $X_i$ is a clique in $\mathfrak{g}(G)$.
    \end{proof}
    \begin{proposition}
    If $X_i$ is a clique in $\mathfrak{g}(G)$ and $x\in X_i$ and $\langle x\rangle \neq p^k$ for some prime $p$ and positive integer $k$, then $X_{i+1}$ is empty.
    \end{proposition}
    \begin{proof}
    Suppose $x\in X_i$, $\langle x\rangle \neq p^k$, so $\lvert x\rvert$ has at least two distinct prime factors, say $p$ and $q$.  Suppose there exists a $y\in X_{i+1}$, then $pq \mid \lvert y\rvert$, but then $\lvert y\rvert=pqm$ for some $m$ (possible equal to $p$ or $q$), and in that case $\langle y\rangle$ has subgroups of orders $pm$ and $qm$, both in $X_i$, contradicting the assumption that $X_i$ is a clique in $\mathfrak{g}(G)$.
    \end{proof}
    \begin{proposition}\label{prop 8}    If $G$ is a finite Abelian group and not cyclic, then $max_iX_i$ is not a clique in $\mathfrak{g}(G)$.
    \end{proposition}
    \begin{proof}
    Since $G$ is a finite non-cyclic Abelian group, $G\cong \mathbb{Z}_{p_1}^{k_1}\times\mathbb{Z}_{p_2}^{k_2}\times\cdots\times\mathbb{Z}_{p_n}^{k_n}$, where not all of the $p_{i}'s$ are distinct.  Then, $G\cong \mathbb{Z}_m\times\mathbb{Z}_{p_i}^{k_i}\times\cdots\times\mathbb{Z}_{p_t}^{k_t}$, where $p_i^{k_i}$ $\mid$ $m,\cdots$ $p_t^{k_t}$ $\mid$ $m$, and $m=q_1^{l_1}q_2^{l_2}\cdots q_r^{l_r}$ where $q_1,q_2,\cdots,q_r$ are relatively prime.  Notice that $g=(1,0,\cdots,0)$ and $g'=(1,1,0,\cdots,0)$ both have the maximum possible order in $G$, that is $m$. Also, notice that $\langle g\rangle \neq \langle g'\rangle$, so $g$ and $g'$ are both in $max_iX_i$ but not adjacent to each other, so $max_iX_i$ is not a clique in $\mathfrak{g}(G)$.
    \end{proof}
    Since $max_iX_i$ is not a clique in $\mathfrak{g}(G)$, then no $X_i$ is a clique in $\mathfrak{g}(G)$.  Since $G=\bigcup_iX_i$, no non-identity vertex in $\mathfrak{g}(G)$ has degree $n-1$.
\section{Perfect Graphs}
A \textit{clique} in a graph $\Gamma$ is a subset $C$ of the vertices of $\Gamma$, such that any two vertices in $C$ are adjacent. The size of the largest clique in $\Gamma$ is called the \textit{clique number} of $\Gamma$, and is denoted $\omega(\Gamma)$.

A \textit{vertex coloring} of a graph is an assignment of labels (called colors) to the vertices of a graph $\Gamma$ such that no two adjacent vertices share the same color.  The smallest number of colors which can be used in a coloring of $\Gamma$ is called the \textit{chromatic number} of $\Gamma$, and is denoted $\chi(\Gamma)$.

A graph $\Gamma$ is called \textit{perfect} if for every induced subgraph $\Gamma_i$ of $\Gamma$, $\omega(\Gamma_i)=\chi(\Gamma_i).$ It was conjectured by Berge in 1961, and was proved by Chudnovsky et. al. in \cite{PerfectGraphs} that graphs are perfect if and only if they contain no holes or anti-holes of odd length greater than $3$. A \textit{hole} in a graph is a cycle such that no two vertices in the cycle are joined by an edge which does not itself belong to the cycle.  An anti-hole is the edge complement of a hole.

\begin{Theorem}[Strong Perfect Graph Theorem]
    A graph is perfect if and only if it contains no odd holes or odd anti-holes of length greater than 3.
\end{Theorem}

Let $G$ be a group with power graph $\mathfrak{g}(G)$.  Then $\mathfrak{g}(G)$ can contain no holes of odd length.  To prove this a few short lemmas are used.
\begin{Lemma}[Path]
Let $G$ be a group with directed power graph $\vec{\mathfrak{g}}(G)$.  If a path exists between two vertices then they are adjacent.
    \begin{proof}
        Denote the start of the path as vertex $a$ and label the vertices along the path as $a_1,a_2,\cdots a_m$.  Then $a_1=a^n,a_2={a_1}^{n_1},a_3={a_2}^{n_2}\cdots a_m={a_{m-1}}^{n_{m-1}}$.  Then $a_m=a^{nn_1n_2\cdots n_{m-1}}$, so there is an edge from $a$ to $a_m$.
    \end{proof}
\end{Lemma}
\begin{Lemma}[Strong Path]
Let $G$ be a group with directed power graph $\vec{\mathfrak{g}}(G)$, and let $\vec{\mathfrak{g}}(G)$ contain a directed path of length $n$, then the vertices making up the directed path form a clique of size $n$ in $\mathfrak{g}(G)$.
\end{Lemma}
\begin{proof}
 The proof follows from the previous (Path) Lemma.
\end{proof}

The strong path lemma shows that whenever there is a path of length $n$ in $\vec{\mathfrak{g}}(G)$  there is a clique of size $n$ consisting of the same vertices in $\mathfrak{g}(G)$.  Here it will be shown that the converse is true as well, that is, whenever there is a clique of size $n$ in $\mathfrak{g}(G)$, then those vertices are traversable by a path in $\vec{\mathfrak{g}}(G)$.  Path-clique equivalence in power graphs of finite groups is a consequence of R\'{e}dei's theorem~\cite{Redei}.

As per our need, later in our work, we reproduced the proof of the following well known theorem here one more time.

\begin{Theorem}[R\'{e}dei's Theorem]
    Every orientation of a complete graph contains a directed Hamiltonian path.
\end{Theorem}

\begin{Theorem}[Path-clique Equivalence]\label{theo 3}
Let $G$ be a group with directed power graph $\vec{\mathfrak{g}}(G)$ and undirected power graph $\mathfrak{g}(G)$.  Then whenever there is a path in $\vec{\mathfrak{g}}(G)$, its constituent vertices form a clique in $\mathfrak{g}(G)$, and whenever there is a clique in $\mathfrak{g}(G)$, its vertices are traversable by a path in $\vec{\mathfrak{g}}(G)$.
\end{Theorem}
\begin{proof}
The proof of the first direction is given above, here we show that a clique in $\mathfrak{g}(G)$ is traversable by a path in $\vec{\mathfrak{g}}(G)$.  Let $\mathfrak{g}(G)$ contain a clique of size $\alpha$.  The proof is by induction on $\alpha$.  If $\alpha=1$, then the clique is traversable by the path consisting of only the single vertex in the clique.

Suppose the result holds for $0<\alpha \leq k$, and let there exist in $\mathfrak{g}(G)$ a clique of size $k+1$.  By the induction hypothesis, a clique of size $k$ is traversable by a directed path, so at most one vertex is excluded from the longest path through the clique. Proceed along that directed path through these vertices and label them in the order they are encountered, $v_1, v_2, v_3,\cdots,v_k$. If for any $v_i$ in the clique both $(v_i,v_{k+1})$ and $(v_{k+1},v_i)$ are edges in $\vec{\mathfrak{g}}(G)$, then the sequence can be modified from $(\cdots v_i,v_{i+1}\cdots)$ to $(\cdots v_i, v_{k+1}, v_{i+1}\cdots)$ adding $v_{k+1}$ to the path.  Suppose no two-sided edges exist in the clique. If $(v_{k+1}, v_1)$ is an edge in $\vec{\mathfrak{g}}(G)$, then $v_{k+1}$ can be added to the beginning of the existing path.  If $(v_k,v_{k+1}) $ is an edge in $\vec{\mathfrak{g}}(G)$, then $v_{k+1}$ can be added to the end of the existing path. Suppose neither of these are edges in $\vec{\mathfrak{g}}(G)$.  If $(v_{k+1}, v_2)$ is an edge is $\vec{\mathfrak{g}}(G)$, then $v_{k+1}$ can be inserted in the path between $v_1$ and $v_2$.  Then, if $(v_{k+1}, v_2)$ is not an edge in $\vec{\mathfrak{g}}(G)$, $(v_2,v_{k+1})$ must be an edge in $\vec{\mathfrak{g}}(G)$.  Proceeding in this way we get the desired directed path in $\vec{\mathfrak{g}}(G)$.
\end{proof}

\begin{figure}[h]
\centering
\caption{An illustration of Theorem \ref{theo 3}.  Only relevant edges from the new vertex are shown in each case.}
\begin{subfigure}[b]{0.45\textwidth}
\centering

    \begin{tikzpicture}
            \node[circle,fill] (A) at (0,0) {};
            \node[circle,fill] (B) at (-1,-1) {};
            \node[circle,fill] (C) at (1,-1) [label={[yshift=-0.9cm]$v_{i+1}$}] {};
            \node[circle,fill] (D) at (-1,1) {};
            \node[circle,fill] (E) at (1,1) [label={$v_i$}] {};
            \node[circle,fill] (F) at (2,0) [label={[yshift=-0.48cm, xshift=0.62cm]$v_{k+1}$}] {};
            \Edge(A)(B);
            \Edge(A)(C);
            \Edge(A)(D);
            \Edge(A)(E);
            \Edge(B)(C);
            \Edge(B)(D);
            \Edge[style={bend left}](B)(E);
            \Edge[style={bend right}](C)(D);
            \Edge[style={<-}](C)(E);
            \Edge(D)(E);
            \Edge[style={<->}](E)(F)
            \Edge[style={->, dashed}, color=red](F)(C)
    \end{tikzpicture}
    \caption{An edge in both directions between $v_{k+1}$ and any other vertex implies $v_{k+1}$ can be added to the path}
\end{subfigure}
\qquad
\begin{subfigure}[b]{0.45\textwidth}
\centering

    \begin{tikzpicture}
            \node[circle,fill] (A) at (0,0) [label={[yshift=-0.9cm]$v_{3}$}] {};
            \node[circle,fill] (B) at (-1,-1) [label={[yshift=-0.9cm]$v_{1}$}] {};
            \node[circle,fill] (C) at (1,-1) [label={[yshift=-0.9cm]$v_{5}$}] {};
            \node[circle,fill] (D) at (-1,1) [label={$v_2$}] {};
            \node[circle,fill] (E) at (1,1) [label={$v_4$}] {};
            \node[circle,fill] (F) at (2,0) [label={[yshift=-0.48cm, xshift=0.62cm]$v_{6}$}] {};
            \Edge[style={->}](A)(B);
            \Edge[style={->}](A)(C);
            \Edge[style={<-}](A)(D);
            \Edge[style={->}](A)(E);
            \Edge[style={->}](B)(C);
            \Edge[style={->}](B)(D);
            \Edge[style={->,bend left}](B)(E);
            \Edge[style={->,bend right}](C)(D);
            \Edge[style={<-}](C)(E);
            \Edge[style={<-}](D)(E);
            \Edge[style={<-, dashed}, color=red](F)(C)
    \end{tikzpicture}
    \caption{If there is an edge from $v_k$ to $v_{k+1}$, then $v_{k+1}$ can be added to the end of the path.  Here $k=5$.}
\end{subfigure}
\linebreak
\begin{subfigure}[b]{0.45\textwidth}
\centering

    \begin{tikzpicture}
            \node[circle,fill] (A) at (0,0) [label={[yshift=-0.9cm]$v_{3}$}] {};
            \node[circle,fill] (B) at (-1,-1) [label={[yshift=-0.9cm]$v_{1}$}] {};
            \node[circle,fill] (C) at (1,-1) [label={[yshift=-0.9cm]$v_{5}$}] {};
            \node[circle,fill] (D) at (-1,1) [label={$v_2$}] {};
            \node[circle,fill] (E) at (1,1) [label={$v_4$}] {};
            \node[circle,fill] (F) at (2,0) [label={[yshift=-0.48cm, xshift=0.62cm]$v_{6}$}] {};
            \Edge[style={->}](A)(B);
            \Edge[style={->}](A)(C);
            \Edge[style={<-}](A)(D);
            \Edge[style={->}](A)(E);
            \Edge[style={->}](B)(C);
            \Edge[style={->}](B)(D);
            \Edge[style={->,bend left}](B)(E);
            \Edge[style={->,bend right}](C)(D);
            \Edge[style={<-}](C)(E);
            \Edge[style={<-}](D)(E);

            \draw[<-, dashed, color=red] (B) to[out=-30,in=-90,swap] (F);
    \end{tikzpicture}
    \caption{If there is an edge from $v_{k+1}$ to $v_{1}$, then $v_{k+1}$ can be added to beginning of the the path.  Here $k=5$.}
\end{subfigure}
\qquad
\begin{subfigure}[b]{0.45\textwidth}
\centering

    \begin{tikzpicture}
            \node[circle,fill] (A) at (0,0) [label={[yshift=-0.9cm]$v_{i}$}] {};
            \node[circle,fill] (B) at (-1,-1) {};
            \node[circle,fill] (C) at (1,-1) [label={[yshift=-0.9cm]$v_{i+1}$}] {};
            \node[circle,fill] (D) at (-1,1) {};
            \node[circle,fill] (E) at (1,1) {};
            \node[circle,fill] (F) at (2,0) [label={[yshift=-0.48cm, xshift=0.62cm]$v_{k+1}$}] {};
            \Edge[style={->}](A)(B);
            \Edge[style={->}](A)(C);
            \Edge[style={<-}](A)(D);
            \Edge[style={->}](A)(E);
            \Edge[style={->}](B)(C);
            \Edge[style={->}](B)(D);
            \Edge[style={->,bend left}](B)(E);
            \Edge[style={->,bend right}](C)(D);
            \Edge[style={<-}](C)(E);
            \Edge[style={<-}](D)(E);
            \Edge[style={->, dashed}, color=red](A)(F);
            \Edge[style={->, dashed}, color=red](F)(C);
    \end{tikzpicture}
    \caption{If none of the other cases hold, then there must exist a pair of vertices in between which $v_{k+1}$ can be placed.}
\end{subfigure}
\end{figure}

\setcounter{figure}{2}

It is well known that
Power graphs of groups are perfect, which has been proved by using
 Theorem~\cite{PerfectGraphs}.

\section{The Cyclic Subgroup Graph}

Let $G$ be a group and define the relation $\sim$ on $G$ by $x\sim y$ if $\langle x\rangle=\langle y\rangle.$  Define the graph $\vec{C}(G)$ by $V(\vec{C}(G))= G / \mathord{\sim}$ and $(A,B)\in E(\vec{C}(G))$ if there exists elements $b\in B$ and $a\in A$ such that $\langle b\rangle \leq \langle a\rangle$ and $\langle b \rangle \neq \langle a\rangle.$  Also define a weight function $w:V(\vec{C}(G))\to \mathbb{N}$ by $w(A)=\lvert A\rvert$.  Then $\vec{C}(G)$ is a directed acyclic graph with a similar structure to the directed power graph $\vec{\mathfrak{g}}(G).$ \begin{proposition}The weight of the path with the largest weight in $\vec{C}(G)$ is the length of the longest path in $\vec{\mathfrak{g}}(G).$
\end{proposition}
\begin{proof}
The vertices in $\vec{C}(G)$ with only out-edges represent generators of the maximal cyclic subgroups of $G$.  As in the proof of Theorem 3 above any longest path in $G$ must start with these vertices.  If this maximal cyclic subgroup has order $n$ then a vertex adjacent to it represents generators of a cyclic subgroup of order $\frac{n}{d}$ where $d$ is a divisor of $n$, and if the subgroup is maximal $d$ will be a prime divisor of $n$. Then the sum of the weights of the vertices in a path from a maximal cyclic subgroup of $G$ of order $n$ through all of its maximal subgroups of maximum order to the trivial subgroup will be given by $\Psi(n)$, the length of the longest path in $\vec{\mathfrak{g}}(G)$.
\end{proof}
When $G$ is cyclic there is no ambiguity in naming vertices in $\vec{C}(G)$ by their corresponding isomorphic group $\mathbb{Z}_n$, for example
\begin{figure}[H]
    \caption{$\vec{C}(\mathbb{Z}_{18})$ with vertex weights shown in parentheses.}
\begin{tikzpicture}
        \node[shape=circle, draw=black] (A) at (0,0) {$\mathbb{Z}_{6}(2)$};
        \node[shape=circle, draw=black] (B) at (1.5,1) {$\mathbb{Z}_{18}(6)$};
        \node[shape=circle, draw=black] (C) at (3,0) {$\mathbb{Z}_{9}(6)$};
        \node[shape=circle, draw=black] (D) at (3,-2) {$\mathbb{Z}_{3}(2)$};
        \node[shape=circle, draw=black] (E) at (1.5,-3) {$\{e\}(1)$};
        \node[shape=circle, draw=black] (F) at (0,-2) {$\mathbb{Z}_{2}(1)$};
        \path [->] (B) edge node[left] {} (A);
        \path [->] (B) edge node[left] {} (C);
        \path [->] (B) edge node[left] {} (D);
        \path [->] (B) edge node[left] {} (E);
        \path [->] (B) edge node[left] {} (F);
        \path [->] (A) edge node[left] {} (D);
        \path [->] (A) edge node[left] {} (E);
        \path [->] (A) edge node[left] {} (F);
        \path [->] (F) edge node[left] {} (E);
        \path [->] (D) edge node[left] {} (E);
        \path [->] (C) edge node[left] {} (D);
        \path [->] (C) edge node[left] {} (E);

    \end{tikzpicture}
\end{figure}
The non-oriented graph $C(G)$ also contains a lot of the information in the power graph in a smaller form.  For example $d_{\mathfrak{g}(G)}(u,v)=d_{C(G)}([u]_{\sim},[v]_{\sim})$ as long as $[u]_{\sim}\neq [v]_{\sim}$ and the independence number $\alpha(\mathfrak{g}(G))=\alpha(C(G))$.

\begin{proposition}
The cyclic subgroup graph is isomorphic to an induced subgraph of the power graph.
\end{proposition}
\begin{proof}
Define $f:V(C(G))\to(V\mathfrak{g}(G))$ by $f([x]_{\sim})=x^{'}$, for a fixed choice of the corresponding equivalence class representative $x^{'}$. The result follows by the definition of cyclic subgroup graph.
\end{proof}

\begin{proposition}
$d_{\mathfrak{g}(G)}(u,v)=d_{C(G)}([u]_{\sim},[v]_{\sim})$ as long as $[u]_{\sim}\neq [v]_{\sim}$.
\end{proposition}
\begin{proof}
let $G$ be a group and let $u,v\in G$ with $d_{\mathfrak{g}(G)}(u,v)=k$.
Let $P=\{u, u_1,u_2,\cdots,u_k, v\}$ denote the shortest path from $u$ to $v$ in $\mathfrak{g}(G)$.  It must be the case that for each $u_i,u_j\in P$, $[u_i]_{\sim}\neq[u_j]_{\sim}$ otherwise deleting the $u_i$ or $u_j$ we get a shorter path in $\mathfrak{g}(G)$ which is a contradiction.  Then $\{[u]_{\sim},[u_1]_{\sim},\cdots,[u_k]_{\sim}\}$ is a path from $[u]_{\sim}$ to $[v]_{\sim}$ in $C(G)$.  Suppose there is a shorter path from $[u]_{\sim}$ to $[v]_{\sim}$ in $C(G)$, and denote that path $P_1=\{[u]_{\sim},[y_1]_{\sim},\cdots[y_j]_{\sim},[v]_{\sim}\}$ where $j<k$.  Then $\{u,y_1,\cdots, y_j,v\}$ is a path in $\mathfrak{g}(G)$ which is shorter than $P$, a contradiction.  Then $\{[u]_{\sim},[u_1]_{\sim},\cdots,[u_k]_{\sim}\}$ is the shortest path from $[u]_{\sim}$ to $[v]_{\sim}$ in $C(G)$ meaning $d_{C(G)}([u]_{\sim},[v]_{\sim})=k=d_{\mathfrak{g}(G)}(u,v)$.
\end{proof}
\begin{proposition}
Let $G$ be a group.  Elements $\{g_1,g_2,\cdots,g_k\}$ form an independent set in $\mathfrak{g}(G)$ if and only if $\{[g_1]_{\sim},[g_2]_{\sim},\cdots,[g_k]_{\sim}\}$ is an independent set in $C(G)$.
\end{proposition}
\begin{proof}
Suppose $I=\{g_1,g_2,\cdots,g_k\}$ forms an independent set in $\mathfrak{g}(G)$.  Then for each $g_i,g_j\in I$, $\langle g_i\rangle \nleq \langle g_j\rangle$ and $\langle g_j\rangle \nleq \langle g_i\rangle$. So, $\{[g_i]_{\sim},[g_j]_{\sim}\}\notin E(C(G))$,  giving an independent set of size $k$ in $C(G)$.

Now suppose $\{[g_1]_{\sim},[g_2]_{\sim},\cdots,[g_k]_{\sim}\}$ is an independent set in $C(G)$. Then  $\{g_1,g_2,\cdots,g_k\}$ is an independent set in $\mathfrak{g}(G)$.
\end{proof}
\begin{Corollary}
$\alpha(\mathfrak{g}(G))=\alpha(C(G))$.
\end{Corollary}
\begin{proposition}
$\mathfrak{g}(G)$ is complete if and only if $C(G)$ is complete.
\end{proposition}
\begin{proof}
Suppose $\mathfrak{g}(G)$ is complete and let $[u]_{\sim}$ and $[v]_{\sim}$ be arbitrary vertices in $C(G)$ Since $u\in [u]_{\sim}$ and $v\in [v]_{\sim}$ and $\{u,v\}\in E(\mathfrak{g}(G))$, $\langle u\rangle \leq \langle v\rangle$ or $\langle v\rangle \leq \langle u\rangle$, so   $\{[u]_{\sim},[v]_{\sim}\}\in E(C(G))$.  Since $[u]_{\sim}$ and $[v]_{\sim}$ were arbitrary, $C(G)$ must be complete.

Now suppose $\mathfrak{g}(G)$ is not complete, then there exist vertices $u$ and $v$ such that $\{u,v\}\notin E(\mathfrak{g}(G))$.  Then $\{u,v\}$ is an independent set in $\mathfrak{g}(G)$ and consequently $\{[u]_{\sim},[v]_{\sim}\}$ is an independent set in $C(G)$.  Then there exist elements $[u]_{\sim}$ and $[v]_{\sim}$ in $C(G)$ such that $\{[u]_{\sim},[v]_{\sim}\}\notin E(C(G))$ so $C(G)$ is not complete.
\end{proof}
\begin{proposition}
Let $u_1,u_2,\cdots u_{2k}$ be vertices in $\mathfrak{g}(G)$, then $u_1, \cdots u_{2k}$ is a hole in $\mathfrak{g}(G)$ if and only if $[u_1]_{\sim},[u_2]_{\sim},\cdots,[u_{2k}]_{\sim}$ is a hole in $C(G)$.
\end{proposition}
\begin{proof}
Let $P=\{u_1,u_2,\cdots,u_{2k}\}$ be a hole in $\mathfrak{g}(G)$, then each $u_i\in P$ has exactly two neighbors in $P$, call them $u_{i+1}$ and $u_{i-1}$.  Then either $\langle u_i\rangle \leq \langle u_{i+1}\rangle$ and $\langle u_i\rangle \leq \langle u_{i-1}\rangle$ or $\langle u_{i+1}\rangle \leq \langle u_{i}\rangle$ and $\langle u_{i-1}\rangle \leq \langle u_{i}\rangle$.  Either way, the following are true, $\langle u_i\rangle \neq \langle u_{i+1}\rangle$ and $\langle u_i\rangle \neq \langle u_{i-1}\rangle$, so $[u_i]_{\sim}\neq [u_{i+1}]_{\sim}$ and $[u_i]_{\sim}\neq [u_{i-1}]_{\sim}$,     so $[u_{i+1}]_{\sim}$ and $[u_{i-1}]_{\sim}$ are neighbors of $[u_i]_{\sim}$ in $C(G)$. There is no chord in $C(G)$ inside that hole in $\mathfrak{g}(G)$, for if so, then there will be a chord inside the hole in $\mathfrak{g}(G)$ giving a contradiction. So $[u_1]_{\sim},[u_2]_{\sim},\cdots,[u_{2k}]$ forms a hole in $C(G)$.The other direction also follows via a similar argument.
\end{proof}
\begin{Corollary}
$C(G)$ are perfect graphs.
\end{Corollary}
\begin{proposition}
$\mathfrak{g}(G)$ is claw free if and only if $C(G)$ is so.
	\end{proposition}
	\begin{proposition}
		$\mathfrak{g}(Z_n)$ is chordal if and only if $C(Z_n)$ is chordal.
		\begin{proof}
			The result follows since a hole exists in $C(Z_n)$ if and only if a hole exists in $\mathfrak{g}(Z_n).$
		\end{proof}
		\end{proposition}
	\begin{proposition}
		A vertex $x\in \mathfrak{g}(Z_n)$ is simplicial if and only if  $[x]_{\sim}\in C(Z_n)$ is simplicial.
		\begin{proof}
			The result follows as induced subgraph of a complete graph is complete.
		\end{proof}
	\end{proposition}
\begin{proposition}
If $C(G)$ is Hamiltonian, then $\mathfrak{g}(G)$ is also Hamiltonian.
\end{proposition}

\begin{proof}
Suppose $C(G)$ is Hamiltonian, and let $\{[u_1]_{\sim},[u_2]_{\sim},\cdots,[u_2]_{\sim}$, $[u_1]_{\sim}$ be a Hamiltonian cycle in $C(G)$.  $[u_i]_{\sim}$ is a clique in $\mathfrak{g}(G)$ since for $g_1,g_2\in [u_i]_{\sim}$, $\langle g_1\rangle =\langle g_2\rangle$, so $\langle g_1\rangle \leq \langle g_2\rangle$, so $\{g_1,g_2\}\in E(\mathfrak{g}(G))$, so denote $[u_i]_{\sim}$ by $\{u_{1i},u_{2i},\cdots,u_{k_ii}\}$, then $\{u_{11},u_{21},u_{31},\cdots,u_{k_11},u_{12},u_{22},$\\
$\cdots, u_{k_22},\cdots,u_{1n},u_{2n},\cdots u_{k_nn},u_{11}\}$ is a Hamiltonian cycle in $\mathfrak{g}(G).$
\end{proof}
Singh and Devi showed in \cite{CyclicSubgroupGraph} that the cyclic subgroup graph of cyclic groups of non-prime order is Hamiltonian in.  Power graphs of groups of prime order are complete, and therefore Hamiltonian for all orders except for $2$, so we note the following corollary by using R\'{e}dei's theorem,~\cite{Redei}.
\begin{Corollary}
Let $G\cong$ $\mathbb{Z}_n$ be a cyclic group with $n\neq 2$, then $\mathfrak{g}(G)$ is Hamiltonian.
\end{Corollary}
Since the cyclic subgroup subgraph is isomorphic to an induced subgraph of the power graph, if $\mathfrak{g}(G)$  is planner, then so is $C(G)$. For example, consider $G= \mathbb{Z}_9$. Here $\mathfrak{g}(G)$ being a complete graph with $9$ vertices it is not planner. $C(G)$ being a triangle is planner.

\section{Chordless Cycles}

It has now been shown that power graphs contain no holes of odd-length. Here it will be shown that for arbitrary even integer $n$, there exists a finite group whose power graph contains a hole of length $n$.

\begin{proposition}
Let $n$ be an even integer, then for even $n>4$, the power graph of a cyclic group will contain a hole of length $n$, if the order of the group has $\frac{n}{2}$ distinct prime factors.  The power graph of the group will contain a hole of length $4$, if the order of the group has at least two prime factors of multiplicity two or more.
\end{proposition}

\begin{proof}
First consider the case that $n=4$.  Then in the group $\mathbb{Z}_{p^2q^2}$ there is a subgraph consisting of the vertices $p,pq^2,q,$ and $p^2q$.  This subgraph will be a hole of length $4$.

Now consider the case that $n\geq 6$, and take primes $p_1,p_2,\cdots p_{\frac{n}{2}}$. Then the group contains a hole of length $n$ namely subgraph consisting of vertices $$p_1-p_1p_2- p_2-p_2p_3-p_3\cdots  p_{\frac{n}{2}} \cdots p_{\frac{n}{2}}p_{1}-p_1$$
Then this subgraph is a hole of length $n$ in $\mathfrak{g}(\mathbb{Z}_{p_1p_2...p_{\frac{n}{2}}})$.
\end{proof}
Note: Following the proof of the theorem, it is possible to create many holes of even length permuting the positions of the primes and allowing various exponent of them.

\begin{proposition}
If the power graph of a finite cyclic group $G$ contains a hole of length $n$, then $\lvert G \rvert$ has at least $\frac{n}{2} $ distinct prime factors.
\end{proposition}

\begin{proof}
Suppose $\mathfrak{g}(\mathbb{Z}_m)$ contains a hole of length $n$. This cycle in the corresponding directed power graph consists of $\frac{n}{2}$ vertices with only out-edges and $\frac{n}{2}$ vertices with only in-edges.  Each vertex with out-edges is non-adjacent to each other vertex with out-edges, so certainly if $x, y$ are group elements represented by vertices in the hole with out-edges, then $\lvert x\rvert $ does not divide $\lvert y \rvert$, and $\lvert y \rvert$ does not divide $\lvert x\rvert $.  Then the order of each of these $\frac{n}{2}$ vertices with out-edges has a prime factor which is not shared by the other $\frac{n}{2}$ vertices with out-edges.  By Lagrange's theorem, the order of each element must divide the order of the group, so the order of the group must contain at least $\frac{n}{2}$ prime factors.
\end{proof}

It has been shown that for an arbitrary even integer $n$, a finite group can be found whose power graph contains a hole of length $n$.  The proof relied on the fact that there were $\frac{n}{2}$ primes dividing the order of the group so that the multiples of the primes were a subset of their multiples in $\mathbb{Z}$.  Then $\mathfrak{g}(\mathbb{Z})$ contains holes of any even length.  In fact since there are infinitely many prime numbers, there will be an infinite number of holes of any even length in $\mathfrak{g}(\mathbb{Z})$.

A necessary and sufficient condition for the existence of a hole of length $n$ in finite cyclic groups has been given above.  This is not a necessary condition for the existence of holes of length $n$ in general Abelian groups.  Consider the Abelian group $\mathbb{Z}_{12}\times \mathbb{Z}_{12}$.  This group has order $144=2^4\cdot 3^2$.   consider the subgraph of $\mathbb{Z}_{12}\times \mathbb{Z}_{12}$ consisting of the elements $(1,0),(2,0),(1,6),(3,6),(1,2),(2,4),(1,8),(3,0)$. These elements (in order around the cycle) form a hole of length $8$.

\begin{proposition}
Let $G$ be an Abelian group whose order is $p^k$ for some prime $p$ and positive integer $k$.  Then $\mathfrak{g}(G)$ can contain no hole.
\end{proposition}
\begin{proof}
The result follows as the power graph is complete in this case.
\end{proof}

\begin{Theorem}
An element whose order is a power of a prime cannot be a vertex with out-edges in a hole of even length.
\end{Theorem}
\begin{proof}
	Suppose, that is not the case. Then, $\vec{\mathfrak{g}}(G)$ contains a hole with an element $x$ of order $p^n$ as a vertex with out-edges in the hole. Let $y$ and $z$ be  the elements adjacent to $x$.  Then $y$ and $z$ are of orders $p^l$ and $p^m$ respectively, where $l,m<n$.   $l\neq m$. Without loss of generality, suppose $l<m$, then $\langle y \rangle \subset \langle z \rangle$ as each are proper subgroups of $\langle x \rangle$ which is a primary cyclic group. So there is a chord from $y$ to $z$ giving a contradiction.
\end{proof}

\begin{proposition}
Power graphs of groups can contain no anti-holes of length greater than $4$.
\end{proposition}
\begin{proof}
Let $G$ be a group and suppose that $\mathfrak{g}(G)$ contains an anti-hole of length $n$ greater than $4$. Claim: All vertices in the anti hole have in-degree zero or out-degree zero in the corresponding directed power graph. Proof of the claim:
First note that, if we consider $n=4$, then the claim follows clearly. Arbitrarily choose a vertex $d$ in the anti hole. There must be a vertex $s_1$ in the antihole adjacent to $d$. Without loss of generality, let the source of the corresponding edge be $s_1$ and the destination vertex $d$. Now choose an edge between a vertex $s_2$, which is non-adjacent to vertex $s_1$, and adjacent to $d$.  Such an edge must exist as we assume $n>4$, since no two vertices can share the same pair of non-adjacent vertices.  The direction of this edge must be from $s_2$ to $d$ since if it were from $d$ to $s_2$ then a directed path would exist between $s_1$ and $s_2$, which by the path lemma would make them adjacent. Repeat the process for a vertex non-adjacent to one of either $s_1$ or $s_2$ and adjacent to $d$ again. The stopping point of this process is when all $n-3$ vertices adjacent to vertex $d$ have been selected. Following this procedure we see that $d$ has out-degree zero. Hence the claim follows.
Now power graph of a group being perfect, here $n>4$ means $n\geq 6$, as it can't have any anti-hole of odd length. So, for any arbitrary vertex $d$ in the antihole, degree of $d=n-3\geq 3$. So, it is possible to choose two vertices $v_1$ and $v_2$ in the neighborhood of $d$ that are adjacent to each other. Without loss of generality, if we assume $d$ has in degree zero, then as there is at least one directed edge between $v_1$ and $v_2$, we get a contradiction to the above claim. Hence the result follows.
  \end{proof}

\section{Completeness}
Here an alternative proof of the well known result regarding completeness of power graphs of cyclic groups of prime-power order in terms of the strong path lemma is presented.
\begin{proposition}
The power graph of a cyclic group of order $p^n$ where $p$ is prime and $n$ is a non-negative integer is complete.
\end{proposition}
\begin{proof}
The proof is by induction on $n$. By the Strong path lemma it suffices to show that there exists a directed path through all vertices in the power graph. When $n=0$ the graph consists of one vertex and the result follows.

Suppose a directed path exists through all vertices in the power graphs of cyclic groups of order $p^k$ for $0\leq k < n$.   Let $G\cong \mathbb{Z}_{p^k}$. Let $x,y$ denote two arbitrary generators of $G$, then in $\vec{\mathfrak{g}}(G)$ there is an edge from $x$ to $y$ and there is an edge from $y$ to $x$, as both elements are members of $G$ and therefore generated by each other. Then, there is a directed path between all generators of $G$.  This path can be extended towards a generator of a subgroup of order $p^{k-1}$, which contains a directed path through all of it's vertices by the induction hypothesis.  As every subgroup of $G$ divides the order of $G$ by Lagrange's theorem, every subgroup of $G$ is properly contained inside the subgroup of order $p^{k-1}$, so there exists a directed path through the entire vertex set of $\mathfrak{g}(G)$.

By the strong path lemma, there is a clique of size $p^n$ in the cyclic group of order $p^n$, so the graph is complete.
\end{proof}

\section{Chromatic Number of Power Graphs of Cyclic Groups}
\begin{proposition}
    Let $G$ be the cyclic group of order $n$ and let $\mathfrak{g}(G)$ be its power graph.  Let $H$ denote the set of non-generators of $G$. Let $\chi(\Gamma)$ denote the chromatic number of a graph $\Gamma$. Then $\chi(\mathfrak{g}(G))=\phi(n)+\chi(\mathfrak{H})$ where $\mathfrak{H}$ is the subgraph of $G$ consisting of vertices representing non-generators and the edges between them and $\phi$ is Euler's totient function.
    \begin{proof}
        Since elements in $G\setminus H$ generate $G$, for any $g\in G\setminus H$ and $x\in G$, $x\in\langle g \rangle$, so vertex $g$ is adjacent to all elements in $\mathfrak{g}(G)$.  Then no color used in a coloring of the portion of $\mathfrak{g}(G)$ consisting of elements of $G\setminus H$ can be used in a coloring of $\mathfrak{H}$.  Additionally that portion of the graph is internally complete as well so the subgraph of $\mathfrak{g}(G)$ consisting of elements of $G\setminus H$ is isomorphic to $K_{\phi(n)}$ and has chromatic number $\phi(n)$.  Since every color used in the coloring of $\mathfrak{H}$ is distinct from every color in the subgraph of $\mathfrak{g}(G)$ consisting of elements of $G\setminus H$, and $\mathfrak{H}$ has chromatic number $\chi(\mathfrak{H})$, $\chi(\mathfrak{g}(G))$ is at most $\phi(n)+\chi(\mathfrak{H})$.  Also since the colors used in the colorings of the two subgraphs of $\mathfrak{g}(G)$ are distinct, $\chi(\mathfrak{g}(G))$ cannot be less than $\phi(n)+\chi(\mathfrak{H})$, so $\chi(\mathfrak{g}(G))=\phi(n)+\chi(\mathfrak{H})$.
    \end{proof}
\end{proposition}
\begin{Corollary}
    Let $G\cong \mathbb{Z}_n$ then $\phi(n)<\chi(\mathfrak{g}(G))\leq n$
\end{Corollary}

The result above gives a lower bound for the chromatic number of a power graph of a cyclic group by identifying a clique of a known size in every cyclic group.  By the strong path lemma, the existence of a directed path of length $k$ implies that the vertices along the path also make up a clique of length $k$.  If $m$ is the length of the longest directed path in a group $G\cong \mathbb{Z}_n$, then the chromatic number of the cyclic group of order $n$ is at least as large as the length of that path.
\begin{proposition}
Let $G$ be a group and let $m$ be the length of a directed path in $\vec{\mathfrak{g}}(G)$.  Then $m\leq \chi(\mathfrak{g}(G)) \leq n$.
\end{proposition}

In a cyclic group $G$ of order $n$, a path $m$ of length longer than $\phi(n)$ can be constructed as follows.  Follow the path of length $\phi(n)$ through the generators of of $G$.  After visiting the last generator, follow the path to a generator of a proper subgroup of $G$, which must exist as a path exists from a generator of $G$ to every element of $G$. As any two generators of any group have edges from each vertex to the other,  each of the generators of this subgroup can be added to the path.  This process can than be continued for a subgroup of this subgroup and so on until the the identity element is added to the path.  In fact, the longest path through any cyclic subgroup will be of this form, a descending chain of generators of proper subgroups.

\begin{Theorem}
Let $\mathbb{Z} / n\mathbb{Z}$ be the cyclic group of order $n$, and let $S_k$ be the set of generators of $\mathbb{Z} / k\mathbb{Z}$.  The longest path through $\mathbb{Z} / n\mathbb{Z}$ will be of the form \\$s_{n1},s_{n2},...s_{n\phi(n)}, d\cdot s_{\frac{n}{d}1}, d\cdot s_{\frac{n}{d}2}, d\cdot s_{\frac{n}{d}\phi(\frac{n}{d})}, ...(dd_1...d_j)\cdot s_{\frac{n}{dd_1d_2...d_j}}\phi(\frac{n}{dd_1d_2...d_j})$ where $d, d_1, d_2...d_j$ are prime divisors of $n$ with $d\leq d_1 \leq d_2 \leq ... \leq d_j$. and $s_{ix}\in S_i$.
\end{Theorem}
\begin{proof}
Suppose a path longer than the one given exists. Such a path necessarily begins with the set of elements of order $n$, that is the elements of $S_n$, since if it did not these elements could simply be added to the beginning of the path to make a new longer path.  After every element of order $n$ is added to the path an element $g_1$ from a proper subgroup of $\mathbb{Z} / n\mathbb{Z}$ can be added to the path, but this element will have order $\frac{n}{d}$ where $d$ is a divisor of $n$.  In fact this element will generate a subgroup $\langle g_1 \rangle$ of order $\frac{n}{d}$ which is isomorphic to $\mathbb{Z} / \frac{n}{d}\mathbb{Z}$ with the mapping given by $z\in \mathbb{Z} / \frac{n}{d}\mathbb{Z} \mapsto d\cdot g\in \langle g_1 \rangle$.  Then for each of the $\phi(\frac{n}{d})$ generators $z\in z\in \mathbb{Z} / \frac{n}{d}\mathbb{Z}$, the corresponding element $d\cdot z \in \langle g_1 \rangle$must be added to the path, since if it were not added to the path a new path could be constructed with these elements following (or preceding) $g_1$ which is longer. The same process can be repeated from $\langle g_1 \rangle$, add an element from a subgroup of $\langle g_1 \rangle$ and all of the other generators of the same subgroup.  In this way generators of a descending chain of subgroups are added to the path, terminating with the generator of the trivial group, the identity element.

Since the longest path must be in the form of generators of a chain of subgroups, it remains to be shown that by always choosing the largest possible proper subgroup whose generators to add to the path, the path size is maximized.  That is, by traversing the subgroups of $\mathbb{Z} / n\mathbb{Z}$ in the order \\$\mathbb{Z} / n\mathbb{Z}\to d\mathbb{Z} / n\mathbb{Z}\to d\cdot d_1\mathbb{Z} / n\mathbb{Z}\to...\to (d\cdot d_1\cdot ... \cdot d_j)\mathbb{Z} / n\mathbb{Z}$ where $d\leq d_1 \leq ... \leq d_j$, the number of elements added to the path is as large as possible.  To see this, some properties of Euler's totient function are examined.  First observe that increasing any single prime factor in a number will increase the totient of that number, that is if $n=p_{11}p_{12}...p_{1k}$ where $p_1 \leq p_2 \leq... \leq p_k$ and $m=p_{21}p_{22}...p_{2k}$ where $p_{1x}=p_{2x}$ for all $x$ except one, and at that one index $p_{2x}>p_{1x}$, then $\phi(m)\geq \phi(n)$.  It is known that the totient function is multiplicative over relatively prime arguments, that is $\phi(ab)=\phi(a)\phi(b)$ if $\gcd(a,b)=1$, so we can write $\phi(n)$ and $\phi(m)$ as $\Pi_{x\in X} \phi(p_{1x}^{k_x})$ and $\Pi_{y\in Y} \phi(p_{2y}^{k_y})$ respectively where $X$ is the set of prime factors of $n$ and $Y$ is the set of prime factors of $m$.  Then both $\phi(n)$ and $\phi(m)$ can be divided by the prime factors and multiplicities for which they agree, leaving only the prime factors which differ between $n$ and $m$.  Then observe that $\phi(p^k) < \phi(q^k)$ if $p < q$ are distinct primes.  Also $\phi(p^{k})\leq \phi(p^{k-1})q$ if $p< q$ since $p^k-p^{k-1} \leq (p^{k-1}-p^{k-2})(q-1)$.  Then it is clear that if $n$ is the order of a cyclic group the subgroups must be traversed in the order above to make the number of elements in the path as large as possible.
\end{proof}
Here the largest path through $\vec{\mathfrak{g}}(\mathbb{Z} / n\mathbb{Z})$ has been constructed, which will have a length equal to the size of the largest clique in $\mathfrak{g}(\mathbb{Z} / n\mathbb{Z})$, by the path-clique equivalence theorem.  Since power graphs are perfect, the size of the largest clique in a power graph is also equal to its chromatic number.  Then the following result is true for power graphs of cyclic groups

\begin{Corollary}
Let $G\cong \mathbb{Z} / n\mathbb{Z}$ be a group with power graph $\mathfrak{g}(G)$.  Also let $d_1\leq d_2\leq... \leq d_k$ be (not necessarily distinct) prime divisors of $n$ then $\chi(\mathfrak{g}(G))=\phi(n)+\phi(\frac{n}{d_1})+\phi(\frac{n}{d_1d_2})+...+\phi(\frac{n}{d_1d_2...d_k})$
\end{Corollary}

Here the proof of Theorem \ref{theo 3} did not depend on the fact that $G$ was a cyclic group in any way, except that the path constructed through the elements of $G$ could always be started with an element of order $n$, where $n$ is the order of $G$.  In a general group, there are no elements with order equal to the order of the group, so there are many possibilities for where to begin the longest path.  Let $g_1, g_2,...g_n$ be elements of a group $G$, and define $\Psi(n)=\phi(n)+\phi(\frac{n}{d_1})+\phi(\frac{n}{d_1d_2})+...+\phi(\frac{n}{d_1d_2...d_k})$, where $d_1, d_2, ...d_k$ are prime divisors of $n$, then $\chi(G)=\max_{i}\{\Psi(g_i)\}$

\section{Chordallity of Power Graph of cyclic groups}

		\begin{proposition}
			Consider $n\neq p^{m}$ for some prime $p$ and a positive integer $m$. If a vertex $k \in \mathfrak g(\mathbb Z_{n})$ is simplicial then $gcd(k,n)\neq 1$.
			\begin{proof}
				Let $k \in \mathfrak g(\mathbb Z_{n})$ be simplicial with $gcd(k,n)= 1$. Then $k$ generates $\mathbb Z_{n}$. So, $k$ being adjacent to every $x\in \mathfrak g(\mathbb Z_{n})$, $\mathfrak g(\mathbb Z_{n})$ turns to be complete, which is a contradiction as $n\neq p^m$.
			\end{proof}
		\end{proposition}

		Converse of the above result is not true in general. For example: Consider $Z_{12}, gcd(6,12)\neq 1$. Though, $6$ is not a simplicial vertex in $\mathfrak g(\mathbb Z_{12})$ as because $2,3$ are adjacent to $6$, but they are not adjacent to each other.

		\begin{proposition}\label{prop 27}
			$\mathfrak g(\mathbb Z_{n})$ is chordal if and only if $n=p^{m}$ for some prime $p$ and positive integer $m$ or $n=p^{m}q$, for two distinct primes $p,q$ and positive integer $m$.
		\end{proposition}
		\begin{proof}
			If $n=p^{m}$, then $\mathfrak g(\mathbb Z_{n})$ being complete is chordal. If $n=p^{m}q$, then subgroup diagram of the group $\mathbb Z_{n}$ is chordal. And hence, $C(Z_n)$ is also so, as the subgroup diagram is a subgraph of   $C(Z_n)$. Thus, $\mathfrak g(\mathbb Z_{n})$ is chordal.
			Conversely, if $n\neq p^{m}, p^{m}q$, then $n$ has at least two distinct prime factors $p,q$ with powers $m,n$ where both $m,n$ are positive integers bigger than or equal to $2$. In that case, $p-pq^{2}-q-p^{2}q-p$ is a chordless cycle in $\mathfrak g(\mathbb Z_{n})$  giving the graph as  non-chordal.
		\end{proof}

		\begin{proposition}
			A vertex in $C(Z_{n})$ other than $[n]_\sim$ and $[1]_\sim$ is simplicial iff it has only one parent and one child.

			\begin{proof}
				Let $x$ be a vertex in $C(Z_{n})$ other than $[n]_\sim$ and $[1]_\sim$. If it has more than one parent, then any two parents are not adjacent. Similarly, if it has more than one child then any two children are not adjacent to each other.\\
				Conversely let $x$ be a simplicial vertex. Then, it can neither have more one parent, nor more than one child.
			\end{proof}
		\end{proposition}
		Note that, even if $\mathfrak g(\mathbb Z_{n})$ is not chordal for $n= p^{m}q^{r}$ where $m,r \geq 2$ and $p,q$ are distinct primes, they have simplicial vertices namely $p^{m}, q^{r}$, as because they are simplicial in the corresponding $C(Z_n)$.

		Thus, we can now state the following result:
		\begin{Theorem}
		If $n = {\prod_{i=1}^{k} p_{i}^{\alpha_{i}}}$,  then whenever $k\geq 3, \mathfrak g(\mathbb Z_{n})$ does not have any simplicial vertex.
		\end{Theorem}
		\begin{proof}
			$n$ and $1$ are not simplicial. Other wise, $\mathfrak g(\mathbb Z_{n})$ will be complete, giving more than one prime factor of $n$ which contradicts the hypothesis of the theorem. On the other hand, if we consider any vertex in $\mathfrak g(\mathbb Z_{n})$ namely $ m={\prod_{i=1}^{k} p_{i}^{\beta_{i}}}$,

			  then the equivalence classes of  $p_{j}p_{i},p_{j}p_{k}$ are two parents of that of  $p_{j}$ in $C(\mathbb Z_{n})$. So, the vertex is not simplicial in $C(\mathbb Z_{n})$ and hence is not simplicial in $\mathfrak g(\mathbb Z_{n})$.\\
			 \end{proof}

\section{Power graph of $U_n$ and $Q_n$}
For any positive integer $n$, let $\mathbb U_{n}$ be the multiplicative group of integers modulo $n$ and let $ \mathbb Q_n$ be it's subgroup of quadratic residue modulo $n$. Then, we have the following results.

\begin{proposition}
	$\mathfrak g(\mathbb Q_{n})$ is not planner whenever,

\begin{enumerate}
		\item [i.] $n=p$ or $2p$, where $p$ is a prime bigger than $37$.

		\item[ii.] $n=p^{m}$ or $2p^{m}$, where $p\geq 7, m\geq 2$ or else, $p=3$ or $5$ and $m\geq 2$.

		\end{enumerate}
\end{proposition}
\begin{proof}$\phantom e$
\begin{enumerate}
		\item [i.]	 For $n=p$ or $2p$, the cardinality of $\mathbb Q_{n}$ is $\mu =\frac {\phi(n)}{2}= \frac{p-1}{2}$ by Lemma 7.3 of \cite{book}. In that case $\mathbb Q_{n}$ is cyclic as $\mathbb U_{n}$ is so. Hence by lemma 4.7 in \cite{power semi}, $\mathfrak g(\mathbb Q_{n})$ is not planner if $\phi{(\frac{p-1}{2})}>7$, that is if $\frac{p-1}{2}>18$, that is if $p>37$.

\item[ii.] Now let $n=p^{m}$, where $p$ is an odd prime, then by lemma 4.7 in \cite{power semi}, $\mathfrak g(\mathbb Q_{n})$ is not planner if $\frac {\phi(n)}{2}= \frac{p^{m}}{2}= \frac {p^{m-1}(p-1)}{2}= p^{m-2}(p-1){\phi(\frac{p-1}{2})}>7$. Hence, the result follows.
\end{enumerate}
\end{proof}
\begin{proposition}
	$\mathfrak g(\mathbb Q_{n})$ is planner if $n$ divides $240$.
\end{proposition}
\begin{proof} If $n$ divides $240$, then $\mathfrak g(\mathbb U_{n})$ is planner by theorem 4.10 in \cite{power semi}. Hence by proposition 4.5 in \cite {power semi}, $\mathfrak g(\mathbb Q_{n})$ is also planner.
\end{proof}
\begin{proposition}
	Let $n= p^{m}$ or $n=2p^{m}$, where $p$ is a Fermat prime. Then, $\mathfrak g(\mathbb U_{n})$ is chordal iff $m\leq 2$ or $p=F_{0}=3$.
	\end{proposition}
	\begin{proof}
		Let $n= p^2$ or $n=2p^2$ where $p$ is a Fermat prime. Then, by Corollary 6.14 in \cite{book}, $\mathbb U_{n}$ is isomorphic to $\mathbb Z_{p}\times \mathbb Z_{p-1}= \mathbb Z_{p}\times \mathbb Z_{2^{m}}=\mathbb Z_{2^{m}p}$, for some positive integer $m$ as $p$ is a Fermat Prime. Hence the graph is chordal by proposition  \ref{prop 27}. In a similar way, for $n=p$ or $n=2p$ where $p$ is a Fermat prime, $\mathbb U_{n}$ is isomorphic to $\mathbb Z_{p-1}=\mathbb Z_{2^{m}}$ for some positive integer $m$ and hence is chordal by proposition \ref{prop 27}. Now let $p=F_{0}=3$. Then, if $n=p^{m}$ or $n=2p^{m}, U_{n}$ is isomorphic to $\mathbb Z_{p^{m-1}}\times\mathbb Z_{2}=\mathbb Z_{2p^{m-1}}$, as $p$ is odd. Hence the graph is chordal.\\
		Conversely, let $n\neq p,2p,p^{2},2p^{2}$ where $p\neq 3$ and let $n\neq p^{m}, 2p^{m}$ where $p=F_{0}$ and $m$ is any positive integer. So, here if $p>F_{0}, m\geq 3$, then as in \cite{book}, $\mathbb U_{n}$ is isomorphic to $Z_{p^{m-1}}\times\mathbb Z_{2^{i}}, i\geq 2$. Thus the power graph is not chordal by proposition~\ref{prop 27}.
			\end{proof}
			\begin{Corollary}
				For a positive integer $n$, if $p^{m}$ or $2p^{m}$ divides $n$, where $p$ is a Fermat prime bigger than $3$ and $m\geq 3$, then $\mathfrak g(\mathbb U_{n})$ is not chordal.
			\end{Corollary}
			\begin{proposition}\label{prop 32}
				Let, $n$ be an odd integer which is not square free and no Fermat prime be a factor of $n$. Then, $\mathfrak g(\mathbb U_{n})$ is not Chordal.
			\end{proposition}
			\begin{proof}
				As $n$ is not square free, there is an odd prime $p$, which is not a Fermat prime and $p^{m}$ divides $n$ for some $m\geq 2$. In that case, as in Corollary 6.14 of \cite{book}, $\mathbb Z_{p^{m-1}(p-1)}$ appears inside the direct product decomposition of $\mathbb U_{n}$, where $p-1$ is not a power of $2$ since $p$ is not a Fermat prime. So, $p-1$ being it has an odd prime factor $q$ where $p\neq q$, thus $p^{m-1}(p-1)$ contains at least $3$ distinct odd primes and hence the power graph is not chordal.
			\end{proof}
			\begin{proposition}\label{prop 33}
				Let $n$ be an even integer. Then, if
\begin{enumerate}
\item [i.] $2^{f}\parallel n$ (that means $f$ is the largest integer so that $2^{f}$ divides $n$), where $f\geq 4$ and if there are at least two distinct primes $p$ and $q$ and positive integers $\alpha, \beta$, where neither of $p,q$ are Fermat's prime, then $\mathfrak g(\mathbb U_{n})$ is not chordal.
\item[ii.] If $2^{f}\parallel n, f\geq 1$ and $p^{\alpha}, q^{\beta} \parallel n$, where $\alpha,\beta \geq 2$ and neither of $p,q$ are Fermat's prime, then
					 $\mathfrak g(\mathbb U_{n})$ is not chordal.
				\end{enumerate}
			\end{proposition}
\begin{proof}
	The proof follows by Corollary 6.14 in \cite{book} and by our proposition \ref{prop 27}.
\end{proof}
We have analogous result regarding chordality for $\mathfrak g(\mathbb Q_{n})$.

\begin{proposition}
Let $n= p^{m}$ or $n=2p^{m}$, where $p$ is a Fermat prime. Then, $\mathfrak g(\mathbb Q_{n})$ is chordal iff $m\leq 2$ or $p=F_{0}=3$ or $p=F_{1}=5$.
\end{proposition}
\begin{proof}
	The proof follows by using Lemma 7.1 and Corollary 6.14 in \cite{book} as in that case, $\mathbb Q_{n} =\mathbb Z_{\frac{\phi(p^{m})}{2}}= Z_{p^{m-1}(\frac{p-1}{2})}$. Hence, by similar method as in the proof of proposition \ref{prop 32}, the result follows.
\end{proof}
\begin{Corollary}
	For a positive integer $n$, if $p^{m}$ or $2p^{m}$ divides $n$, where $p$ is a Fermat prime bigger than $3, 5$ and $m\geq 3$, then $\mathfrak g(\mathbb Q_{n})$ is not chordal.
\end{Corollary}
\begin{proposition}
	Let, $n$ be an odd integer which is not square free and no Fermat prime be a factor of $n$. Then, $\mathfrak g(\mathbb Q_{n})$ is not Chordal.
\end{proposition}
\begin{proof}
	The proof follows as in proposition \ref{prop 32} and Lemma 7.1 in \cite{book}.
\end{proof}

	\begin{proposition}
		Let $n$ be an even integer. Then, if \begin{enumerate}
			\item[i.] $2^{f}\parallel n$  where $f\geq 5$ and if there are at least two distinct primes $p$ and $q$ and positive integers $\alpha, \beta$, where neither of $p,q$ are Fermat's prime, then $\mathfrak g(\mathbb Q_{n})$ is not chordal.
			\item[ii.] If $2^{f}\parallel n, f\geq 2$ and $p^{\alpha}, q^{\beta} \parallel n$, where $\alpha,\beta \geq 2$ and neither of $p,q$ are Fermat's prime, then
			$\mathfrak g(\mathbb Q_{n})$ is not chordal.
		\end{enumerate}

	\end{proposition}
	\begin{proof}
		The proof follows as in proposition \ref{prop 33}.
	\end{proof}
	Finally, we conclude with a question: Precisely, for what positive integers $n,\mathfrak g(\mathbb U_{n})$ and $\mathfrak g(\mathbb Q_{n})$ are chordal?

\textit{Acknowledgment:\/} Authors acknowledge Dr. M. K. Sen for his helpful suggestions and advice towards this work.

redpajama_set_name: RedPajamaArXiv


28 Mai This entry was written by MICHAEL LEVY, posted on 28 mai 2011 at 10 h 48 min, filed under Art, Photographie and tagged ANISH KAPOOR, GRAND PALAIS, MONUMENTA. Bookmark the permalink. Follow any comments here with the RSS feed for this post. Both comments and trackbacks are currently closed.


text:
tv News Al Jazeera January 31, 2022 12:00pm-12:31pm AST
ah ah attack and counter attack the you a response off to who the rebels also another miss off and threatening operation deep inside its territory. ah, i'm can vanelle, this is elegy they're alive from go home also coming up a shout from space north korea's nations miss all launch puts areas like the u. s. territory of guam within striking distance. china 0 cove at 19 policy is on the pressure of the beijing rewards, more infections among athletes,
just days before the winter olympics and the lost tribes of self in chile, descendants seek recognition for, for gumption. people ah, the united arab emirates says it's intercepted and destroyed a missile fod by who the rebels of yemen is the 3rd salt attack. in recent weeks, the target was unclear, but the defense ministry says debris fell outside of populated areas and that nobody was hurt. the u. a has launched a strike in yemen in response, but that's not comparing the who these war that planning more attacks. lot of law doesn't know how not critical installation to not be targeted by a number of listed missiles. also, the image of dubai was targeted by number of drones. the operation by the grace of god hit its targets with high precision. we reiterate that the enemy state of the
united arab emirates will remain an unsafe country so long as they continuously belligerence against our beloved human people. we renew our warning to ex patry. it's residence in foreign companies to stay away from critical installations. where tween shall the last attack the u. e. z general authority of civil aviation ation rather said air movement in the country is going as usual and offline to operating normally. and there is no impact on flights and airports as a result of the ballistic missile launched by what back, holding the who, the terrorist militia in the country. while it all off top joins us now from santa mom and was, should we make this warning from the who? the yes, as you, as you mentioned that we are witnessing this kind of tag and counter attack by both sides of the houses from unsigned and also the saudi lit quality. and from the other side, the spoke spokesperson,
your head says that the will continue their attacks and the name of emeralds, especially as long as this is what he described, the aggression to continue. and also that the, the, what he described as a tightening blockade is, is not lifted. so when witnessing this kind of cycle of violence, the spokesperson says it, during his pressure that he sent a number of ballistic missiles which are called god, the sword missiles. he also said that the have, those was sent, are launched a tax with guns. but to from the other side, we are witnessing that the meds have only as the have described the intercepted one missile on that that has fallen in and populated area and has no damage or has not imposed any kind of threat to the civilians there. so we are witnessing this kind of cycle violence, the promise of more attacks until the say the aggression is
put to an end. we goodness. also in the early hours of today that the saudi, the quality carried out a number of tags on how these positions on sites here. and so now we have bombed the area, the old area of the tv and radio, the also bomb, the different regions in the, in the capital sun, the, i've had the military maintenance met at the cam, which has been heads and several times previous periods as so this kind of cycle of violence is not going to solve the problem between the warring sides. many people believe that the and go, she shows will be the, the real answer for what is going on here. i'm home and i'll come there from funnel . thank you. north korea's relief photos is proof that conducted its most powerful missiles,
hast and full use. it's the latest and a flurry of missile testing activity, making it the 7th and january record in a single month. the photos published by thank media show the hall on 12 miss all taking off on sunday the release also for each of these images showing the korean peninsula from space. the media says they were taken by a camera installed on the missiles warhead. it's believe the rocker reached a height of 2000 kilometers. it was on 12 miss all of the nuclear capable weapon. and according to south korea and japan, sunday past travelled around 800 kilometers. but its maximum range is much further than mass at 4 and a half 1000 kilometers. that's enough to reach the u. s. territory of glum a target young house threatened to hit before washington urging north korea to engage in direct talks. we are open to having diplomatic discussions. we've offered this over and over to to the do p r k and they've not
accepted it, but we're absolutely open to a diplomatic engagement without pre conditions. our goal is to, to, in the threatening actions that the dpr k is taking against their neighbors. john provides is the head of asia pacific use with jane's defense intelligence here, because i was young maybe trying to achieve by ramping up launches long career. spend a lot of time and investment in the past 2 or 3 years. certainly during cove, it in developing some of its missile capabilities. and that, that is a line very closely with a 5 year weapons planned that north career announced at the beginning of 2021. with the aim to develop capabilities that will enable the north korea to fight a modern, a modern conflict. so, so these developments are in line with those capabilities that it onto outline,
wanted to achieve then. and again, it's consistent with what we saw in 2017 leading up to 2017 to when north korea did test nuclear weapons and did test icbm into inter continental myself. but would that will provide a very clear risk? and then in 2018, of course, what north korea it was, it's suspended a lot of these developments in response to a requirement, a perceived requirement to get close to, to move closer to the us in meetings between kim and donald trump. but that now that kind of effort has changed tactic in north career and they're now looking to put pressure on the us administration to go back to the table and to see if they can ease some of the sanctions. i think that's the real kind of strategy behind these, these escalating missile tests that we're seeing. certainly since the beginning of january are just a few days to go before the winter. olympic games opened in beijing and ethics have been arriving. our district covered 19 protocols. 37 new cases have been reported
among olympic personnel. team members, coaches, support staff and media adjoining and the big bubble to keep them separate from the public. some ethics have already had to pull out of the games after testing positive. let's speak now to katrina. you who's in beijing, katrina so the bubble has still been breached once the latest. that's right. the slight, china's best efforts covered 19 cases are continuing to mount within the winter olympics bubble. now, according to the official statistics we've had about 7000 people arrive in china and recent days to participate in the games. and already at least 176 people have tested positive for corvette 19. now the authorities say that they have this under control, that there is no outbreak spreading within the bubble because they're managing to catch most of these cases when they arrive at the airport. but china does have very
rigorous measures in place to try to keep tabs on all of those cases. not only do athletes and, and other participants have to record the temperatures in the health status, 14 days and the lead up to the flight to china. once they arrive, they also have to test at the airport than every day while they're here and aging. they also have to complete a corporate $900.00 tests. and what that has created is a very tense environment within this closed loop system. athletes in particular and the team members are terrified of catching coven 19. because if you do catch covered 19, if you test positive, you will be immediately quarantine. and you have to test negative at least twice before you will be allowed to return to your normal routine. so that means there is a very big chart that you completely miss out on competing in your event, and that will be 4 years of preparation wasted. so we've already said that a swedish skier and a civilian snowboarder has been among those who have tested positive covered 1900
katrina, even outside the bubble. of course, authorities are continuing to try and contain an outbreak of both delta and only call china's worst outbreak has occurred just in the weeks and months leading up to the olympics. since december, we've had about 4000 cases. norris at the latest vega was 58 noted on sunday and as a new hot spot here in china, the southern city of hunger where they didn't fight about 2 dozen cases on the weekend. now what's most alarming for authorities that cases continue to rise in the capital city? beijing we had on sunday 3 new cases noted, and that doesn't sound like much compared to international standards, but for china and china, there were covered in 1900 strategy. that's 3 cases. true many. so we've had a number of residential neighborhoods invading under locked out at least 2000000 people, forced to undergo a mass testing twice this past week in the city. and today on monday it's the
beginning of the lunar new year festival. so traditionally, we should see hundreds of millions of people traveling from the city to the countryside to celebrate with a family book because a very strict corporate restrictions ahead of the opening of the winter games. many people are not able to do that, so they've been separated for the family for this big celebration this week. i katrina you there in beijing. thank you. a cabinet minister in hong kong, a step down author, attending a birthday party despite a government call for people to avoid large gatherings. casper 3 was the home affairs minister and arising figure in the probate ging political party, the da b. the celebration has become a source of embarrassment for hong kong, chief executive carry lamb among the $200.00 guests that the party, we're more than a dozen top politicians, including the cities, police, immigration, and anti corruption chiefs in full gum guy. his conduct has brought the hong kong government into disrepute and he and his action has seacoast a negative
a perception by the general public in relation to the conduct of principal officials a we have high expectations and requirements and these officials a shoot of serve the highest standards helped us no conduct and integrity at all times, and they should promote and support these principles by leadership and example. still ahead on al jazeera, how many it said it could not be done. portugal socialist prime minister has won an outright majority and parliamentary elections on the world's biggest music streaming platform. spotify takes action, offer a growing dispute about coven 19 misinformation? ah,
hey, they're good to see you. australia's kimberly region has seen some flooding. so broom over the past 24 hours is scooped up more than a months worth of rain and it's still coming out on tuesday right through the interior in time for that se we've had high humidity that's been sparking some pretty big storms and we've got a southerly porsha there, here. so what, what it does so the temperature is on wednesday, melbourne. 20 degrees hobart just 18 in by the way, the darker the red here, the higher the temperature. hey, off to new zealand. wet weather is creeping closer. but for now, on tuesday, just some clouds drifting around while intense got a high of 24 of to southeast asia, concentrated bands of rain over sumatra, west java, east java. and next stop. we're going to the philippines, the ne monsoon given us, what weather are from low are great down to devour on tuesday. not the major, some showers, but something to be on the lookout for some driving rain for that southeast corner of china. it's putting a cap on the temperature. so hong kong,
just 15 degrees for you on tuesday. and now we'll and is weather report in japan. we're in between those waters between korea and japan is shield of snow. so that's going to spread snow over the hills of honju, but tokyo all to while sunshine for you on tuesday, 10 degrees sanskrit season. ah, the corona virus has been indiscriminate in selecting its victims. it's devastating effects of plague, every corner of the globe, transcending class creed and color. but in britain, a just proportionally high percentage of the fallen have been black or brown skins . the big picture traces the economic disparities and institutional racism that is seen united kingdom platelets, citizens, britons, true colors, part one on al jazeera blue.
the algebra, her reminder about told stories this hour given to the rebels have only you either planning more strikes of launching missile and join the tax on monday morning. he says the attack was intercepted and that launch retaliatory action in yemen. north korea is released photos. the proof that conducted at post office miss all passed in 2017. the 7th launch in january, a record and a single month. us calling for direct talks. 37 new cases of corona virus have been reported among the personnel, rubbing and beijing for the winter game. off of being kept apart from the public will be tested daily. candidate defense minister is in ukraine for talks,
announced canadian troops. there were moving to the west inside of the country. canada is temporarily withdrawing non essential employees from its embassy and also providing military aid and training to ukraine. russia foreign minister has accused nato of tried to pull the ukraine into the alliance. because there are well known shooting for which as it turns out that the nato defense line continues to move eastwards. now it has come close to ukraine, which is they also want to drag this country in tomato. although it is clear to everyone that ukraine is not ready and will not make any contribution to strength, may nato security. and politicians in the us are debating what to do about russia, which is more than $100000.00 troops stationed on board with ukraine. one senators, composing a potential compromise. petticoat came reports from washington, d. c. a most of shoes, stores tanks for artillery, more than 100000 russian troops. the u. s. military says, this is now
a strong enough force to take over the country of ukraine. on monday, the un security council will put the issue front and center with the us ambassador saying she will try to make russia feel isolated in the world. our voices are unified and calling for the russians to explain themselves. we're, we're going to go in the room prepared to listen to them, but we're not going to be distracted by their propaganda. and the ukranian ambassador to the us is hoping to rally the international community as well. with a warning, if ukraine will be further attacked by russia, of course, they will not stop in india after ukraine, but with russia able to veto any action at the un. this is likely to be more theatre than a solution. but one prominent us senator has a new suggestion saying the u. s. can't give into russia's demands that ukraine will not join nato, but ukraine can,
but it's his decision to make if he decides that the future membership, if there's a b, one in nato, for ukraine. and the question of russian occupation of ukraine are 2 things to put on the table. i think we may move toward a solution to this. this week, the u. s. senate is expected to take up a bill that would spell out severe consequences if russia acts, which would include sanctions, which would include more military assistance, which would include helping them to fight back against the cyber attacks that the russians will continue to use against ukraine. and also to help on the disinformation campaign, rushes actively involved and to try to stabilize the zaleski government. but the biden administration wants to hold off on sanctions as a deter it, hoping that the threat of freezing russia out of the international banking system is enough to stop any military action. which russia says has no intention of taking in care of on sunday, a show of gratitude. ukrainians take to the streets to send a message that they appreciate the support even as many hope they won't need it.
patty calhane al jazeera washington. portugal is prime minister has won a 3rd consecutive term and office tony cost is the governing socialist security parliamentary parliamentary rather majority. in sunday staff election. he found the thing to govern for all pulled together the economy struggles to bounce back from the pandemic. i'll just those atom rainy reports from lisbon. the stunning mandate prime minister in 20th cost. as ruling socialists have won a majority of people in portugal, snap election. it's a feat, few could have imagined coming into the election pre election pulling, had them nearly, even with the 2nd place center, right. social democrats. the election was triggered when other left his parties voted with wrangling parties to kill the 2022 budget. that put cost his rule at risk until he pulled off sundays victory. the socialists will have the votes now to
pass the budget firm of the thought. it was a victory of humility, of confidence, and for stability. firstly, i want to say thank you to all the portuguese for the example away. the electoral campaign took place for the civic way that these elections took in such a difficult moment for all of us. his main rival, really rio, the leader of the center, right? social democrats acknowledging defeat, and implying he may step down his party leader bank while he's on the phone with the socialist party governing period of 4 years. i really don't see how i can be useful in this framework. the party will decide, but i sincerely don't see how i can be useful in this context for 4 more years. the the far right wing, party, shaker, or enough office celebrating they received the 3rd highest number of votes. the 1st such showing for far right when party since the rival of democracy in 1974. a sign
of the traction populism hold. after 2 years have pandemic pain. one of the problems that we have, and this might explain a little bit why this rendering party is rising, is as part of was not growing as the other comes in to be a union and it's getting backwards. now it's the 7th poorest country in the $27.00 this due to be in union dawning protective gear poll workers even received, covert positive voters who left isolation to cast their ballots. and what many saw as the most important election in decades? the election was defined by the pandemic with voters calling on parliament to get the economy going again into affectively manage the handling of billions of dollars from the e. u. recovery and resilient fund. adam rainy out to 0. lisbon gunman and pakistan have killed a christian priest in the northwest and city of push. our father william serrato
was shot multiple times. another priest was injured. no one has claimed responsibility for the attack. members of the christian community have previously been targeted and shallow. come all harder, has more from the some about the security for their growth bog. it's on. i don't know height, their state of alert off the targeting of the senior christian bridge and the video for gun down by 2 people writing a motor bike. there was a gun badger recently, and fish were also in ridge, a number of islamic stairs, quarter sawn fighter. and we're we're got that they are now trying to target insight buckets on it. read the christian faith. killing, of course, has fog wave of broad day by the minority christian community and the capital of the high book. while province dick glowed the missionary schools and protest your day. that the forces of god saying that the real ap rand, appropriate great day of britain was in our young badges recently with
a number of islamic quarters on fighters in that video for the military. d for dog doing it more district and just on to boost the morale of georgia. adoption of their men were killed in an attack which appears to have come from across the border budget on god says send a message to the dollar bond that they want more action against the group. you're doing one story. chinese naval vessels carrying disaster release supplies of departed photography in the south pacific. the 1400 ton shipment includes food, medical supplies, tractors, and temporary housing pong. it is still recovering from a volcanic eruption into nami, that strong. earlier this month. a delivery and distribution has been slowed by the countries strict pandemic restrictions. so, so, so unless so do, we was short on time preparing for the misha was, and the voyage is long. the loading is heavy and the requirements for the epidemic
prevention are high with we fully anticipated various complex situations and carefully planned the operations to make sure that the relief supplies are safely and quickly delivered to tonga for decades to lands. having taught the original inhabitants of patagonia were all but extinct, but descendants of 2 indigenous groups have come forward to demand recognition. and that in america editor, lucy newman, met them into deaf wiggle. the dealer, the full archipelago, near the tip of cape horn, is known as the end of the world. for 7000 years, it was inhabited by the janish, nomadic people who walked barefoot, and you seal and whatnot. gaskins to keep warm. 135 kilometers north lived the south. none a people so tall that the 1st european explorers named the region patagonia, meaning land of bypass or long legs. what can happen to day replicas of the self
financing, full spiritual body paint have become fashionable as souvenirs of the people who are presumed extinct except they aren't. now self, nuns are finally coming out to demand recognition. is it but on the fact that it's a southerly speaker for history though we often heard from our grandchildren out of shame, is painful to recognise though we come from a family that was practically enslaved and exterminated in the 20th century, is hard. lydia gonzales, scalded on tells a similar story about her people, the janish, her 93 year old mother. christina is the last who still speaks their language fluently. i had a good backyard. nickina you're gone. is ledia an indigenous delegate to chili's constitutional convention shocked. many. when she addressed the chamber k marchina, but on my main focus, imagine not doing grandparents because outsiders annihilated them. imagine being taught at school that your language and your pupil had disappeared. imagine that
you direct ancestors were taken across the ocean to far away lands to be exhibited of exotic animal muscle which drama g. as the documentary humans zill illustrates, she wasn't exaggerating, abusive. lydia says they're still in estimated 200 yet, annie's. and in the last census, more than 1100 people said they identified as said none until a 100 years ago. they lived here from hunting and fishing. and by the coast there was also plenty to eat. you can still feel lots of muscles and other presentations here and historic document to demonstrate that they had an if and the segments could feast on, for example, reached wales like this one. so it wasn't hunger that wiped out their numbers, but rather the arrival of the foreigners, the europeans that brought with them unknown diseases like to procure lotus and influenza. many the survivors were taken from their parents to religious
institutions or to brussels or hunted down in cathy, the f. one might not look as if the hunters were paid for bringing back a hand abreast of penis, a head or an air in today. the sultan language exists on paper, but no one speaks it. while lydia's mother is recording the phonetic sound of the young man language, both indigenous nations are pushing for chiles, new constitution to recognize, support, and promote their linguistic and cultural heritage. a keep or from us, we are all important and deserve recognition. no matter how small a number. because i insisted we didn't choose to be so few. they're determined, she says, to stop justifying their existence and start working towards reviving their nearly last heritage to see and human al jazeera. do you have another foil, chilly? thousands of protest isn't chilly, have gathered to voice their anger against undocumented migrants mostly from venezuela. they pay them for rise and violence in the lane, city of k,
and the demanding more security. the international organization for migration estimates, nearly 1700000 migrants are in chile. us based music streaming site specify has announced new measures to combat coven 19 misinformation. it follows a decision by several high profile musicians to remove that music from the platform in protest against a podcast, who's been accused of spreading falsehoods about vaccines. julian wolf has the story. signing with its highest paid podcast personality, spotify has chosen to keep joe rogan show on the platform as it attempts to tackle the growing backlash over his controversial views on cobra. 19 the company has announced plans to add an advisory to any podcast episode. discussing the pandemic, directing listeners to the latest information from trusted sources in health and science. but despite an open letter penned by hundreds of health professionals
calling on spotify to moderate misinformation, and a growing number of celebrities raising concern, including the duke and duchess of sussex, who have a 25000000 dollar content deal with the platform. some say it's an issue that isn't solved in silencing one voice. there's a misinformation challenge tied to joe rogan in particular, but there's also a broader mas information problem, i think, in this entire ecosystem. that is a lot more challenging simply because it is so decentralized and there isn't really one platform. there isn't really one player involved who can take the lead and moderating. it's very decentralized. you know, you basically could be playing effectively whack of all the fall out began when rock star neil young accused rogan of spreading misinformation while offering spotify an old maiden. remove his work or the comedians, the singer joanie mitchell soon followed,
announcing her solidarity with young and while the streaming service says it has removed more than 20000 podcast episodes. that cover covert 19 since the start of the pandemic. it fell short of ditching one of its biggest stars after it's exclusive 2020 licensing deal that netted rogan, more than a $100000000.00. gillian wolf, al jazeera. oh, this is al jazeera. these are the top stories. gibbons, who the rebels have warned the you are ye that the planning more strikes off to launching miss ireland drawing a tax on monday morning the you a, he says the attack was intercepted and that it launched. retaliatory action in yemen, balboa darvin, maria de la ha critical installations and abu dhabi were targeted by a number of plastic mussel.
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Sound familiar? Yeah, that's how the routine usually goes. And it is a routine for many of us. Our patterns of behavior are so ingrained in us that we rarely pause to reflect; let alone seek an active and positive way towards change. It escalates to a habit, and one that is clearly not working for us (sometimes under the guise of multitasking).
So how do we achieve excellence?
One way is through reading, studying, and then applying the great wisdom that can be gleaned from the ever-famed self-improvement books that we all know and love. Yeah, I say that tongue-in-cheek a little bit, because, let's face it, many of us don't want to be seen in that section of the bookstore. But let's be honest for a moment: bombastic claims and marketing tactics aside, there is much we can benefit from in that section of the bookstore. Believe me, I wasn't always a "believer," but once I started reading a number of these books, it really started to click. And my life started to change, and still is. It's not all cheese and fluff my friends (but you do have to choose wisely, admittedly).
One such wise choice is Brian Tracy's Eat That Frog!. It's a short read, just under 120 pages in all. But it's an impactful read. The 21 ways that he shares are real game-changers, if you read with an eye towards improvement and an intention to make a change. I have benefited greatly from this book, and I highly recommend that you pick up your own copy today.
It all starts with the basic premise that your "frog" is your "biggest, most important task, the one you are most likely to procrastinate on if you don't do something about it." And what if you have two frogs? Eat the ugliest one first, he says.
You see, if you tackle the most unappealing task first, everything from then out out is easier. Once that major thing is out of the way, you will gain momentum and will find greater ease in the other tasks that you have to complete. Just be careful not to celebrate too much after "eating" your biggest and ugliest "frog" though, it could just cause you to slack off on the other tasks! It really is about balance.
The key to reaching high levels of performance and productivity is to develop the lifelong habit of tackling your major task first thing each morning.
~ Brian Tracy, Eat That Frog!
And the book just cascades on from there: set the table, plan every day in advance, practice creative procrastination (this part was interesting!), leverage your special talents (and we all have them), maximize your personal power (sounds like some classic Tony Robbins there), get out of the technological time sinks (iPhone addicts out there!, ummm, me), and much more. He really does cover all the major bases.
So let me close with a question for you all.. What is procrastination really about, and what do you do to overcome it? Share your thoughts about that, and about this post, in the comments below. Let's chat!
Christopher J. Moore is an internationally renowned educator, public speaker, author, and entrepreneur. He is the Founder & CEO of Moore Creative Ideas, the publishing platform for Reflect On This, and many other sites to come. Chris first started Reflect On This in 2005 as a way to express his personal reflections on life and living. After completely revamping the site, it has now grown to what you see today.


Anyway, speaking of dabbling, remember that free pattern that nearly blinded me when I was in my local fabric store a couple weeks ago? The apple core quilt? Yeah, it's not done. I stink at binding. I really have no idea what I'm doing. (It's fun that way!) And, I swear, I didn't even read the pattern card that said experience was necessary to do a project like that. I just saw that as I started the horrible binding process. It all makes sense now. Not the smartest pick to get my feet wet in the quilt world.
I think, deep down, I was hoping that it would be so challenging that it would turn me off and I would/could forget about quilting. Of course, the opposite happened. This wonky, puckered, over-sized dishrag has literally rocked my nerdy little world! I have caught myself thinking about how I'm going to tackle this or that when I'm driving in the car or brushing my teeth. It would be an embarrassment to an experienced quilter, but to me it's the gateway to another intoxicating realm of sewing excitement.
I'm not sure that I'll do another apple core anytime soon (but I won't rule it out either!), but I have already had Matt rig up my Minnesota A to do free-motion quilting. Thank goodness he's a fan of the mechanical machines that make my heart go pitter-patter; otherwise, I don't think he'd be able to stand me!
My 1910(?) Minnesota, Model A.
And, yes, sewing friends, I have been using my scanner. A lot. A whole lot. Actually, I have three permanent markers; and two of them need to be replaced. I've been doing a lot of drawing…and I'm very excited for what's to come!
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I think it turned out well, but I feel your pain on binding. I have always been too afraid to tackle that. Maybe one day.
I would love to, just don't expect miracles. I love to try new things.


text:
Tom Brady shrugs off thumb injury to throw two touchdowns in Bucs win over Eagles
Having become the NFL's all-time leader in passing yards earlier this month, it was fitting that Tom Brady, quite literally, stuck out like a sore thumb last night.
The Tampa Bay Buccaneers quarterback threw two touchdown passes and added another 297 yards to his tally in a 28-22 victory over the Philadelphia Eagles, despite having his thumb taped.
Brady was seen icing the injury on the sidelines during Sunday's crushing win over the Miami Dolphins, but insisted it had caused him no major issues following last night's win.
"I feel great," Brady said afterwards. "Awesome."
Both of the 44-year-old's touchdown passes came before the break -- to Antonio Brown and O.J Howard -- firing the Buccaneers into a 21-7 lead at halftime.
The Eagles threatened a late comeback as quarterback Jalen Hurts rushed for two TDs, but the Bucs survived the scare to make it three wins in a row for the defending Super Bowl champions and move to 5-1 on the season.
Having lost three-time All-Pro cornerback Richard Sherman to a hamstring injury during the first quarter, adding to a proliferating injury list, Brady and the Bucs will relish the break before their week seven game against the Chicago Bears on October 24.
"It was a good win," Brady said.
"It's tough to win on the road. Obviously, we didn't play our best, so we've got to get back to work. We've got a good little break here -- use it and get back to work."
Brady heaped praise on his running back, Leonard Fournette, who ran for 81 yards, caught six passes for 46 yards and scored two touchdowns.
"He's a great back," Brady said. "Big, tough, catches it, runs, blocks, does everything for us -- so great to have him in there.
"When he's rolling, it's tough to stop us," Brady added.
Fournette, who signed for the Bucs last season, has started the season in strong form. Having already rushed for 332 yards in the first six weeks of the season, Fournette is already just 35 yards short of matching his rushing tally from his 13 games last season.
PREV South African cricketer Quinton de Kock apologizes for refusing to take the knee, insists he is 'not a racist'
Trump participates in 'Tomahawk chop' at World Series game NEXT


A little more about who we are and what we do.
Sunderland Guild of Help is based in the city of Sunderland and historically the Mayor of Sunderland has assumed the honorary role of President, whilst the Vice President position is held by a local businessman, who is a Deputy Lieutenant of the county.
Sunderland Guild of Help is still predominantly occupied with meeting the needs of those adversely affected by poverty or illness, referral usually coming from professionals employed by the statutory agencies operating in those areas.
To ensure that funds are directed where most needed and to confirm the integrity of all applications, the administration of the Charity is undertaken by part time staff, who direct referrals to the Chairman of Trustees for decision. In cases where referrals are ambiguous, or require further investigation, a Welfare Subcommittee of senior trustees meets when required to consider such applications. The charity is able to provide grants in excess of 90% of all applications received.
The Guild operates through its premises at Regus House, Doxford International Business Park in Sunderland, but for many years were based at Toward Road, near the Winter Gardens, with a number of other charities.
For more information please click the sub-menu links from the main menu above.
Since its foundation in 1907, The Sunderland Guild of Help has been involved in meeting the needs of the poor and needy of Sunderland at their point of need.
All donations of however much you can spare are greatly appreciated, and all money raised goes towards the work of the Sunderland Guild of Help. For more information please contact our team.
At this present time we can only accept applications from those in need through an appointed Social Worker, and you must be living within the city boundary. For more info please click the title link above.
From time to time we like to let our supporters know about the work of the Guild and good causes we support. Please sign-up if you'd like to be involved.


text:
Q: zoom in dynamically to fit all the marker React-leaflet I'm using react-leaflet. to show the map in my react app. I'm also showing marker on the map too. The problem is the zoom level is not appropriate because sometimes the marker might be quite near to each other and sometimes they will be very far apart. Is there any way to set the zoom level depending on the distance of the markers so that the user can see all the markers at once?
Here is my code

<Map center={center} maxZoom={9} zoom={5}>
  <MarkerClusterGroup showCoverageOnHover={false}>
    {
      markers.map(({fillColor, position, id}) =>
         <CircleMarker fillColor={fillColor} color={darken(0.1, fillColor)} radius={10} fillOpacity={1} key={id} center={position} onClick={this.onClick} />
    }
  </MarkerClusterGroup>
</Map>

P.S: My react-leaflet version is 2.4.0

A: Assuming MarkerClusterGroup is a component from react-leaflet-markercluster package, the following example demonstartes how to auto-zoom to cover visible markers:
function CustomLayer(props) {
  const groupRef = useRef(null);
  const { markers } = props;
  const mapContext = useLeaflet();
  const { map} = mapContext; //get map instance

  useEffect(() => {
    const group = groupRef.current.leafletElement; //get leaflet.markercluster instance
    map.fitBounds(group.getBounds());  //zoom to cover visible markers
  }, []);

  return (
    <MarkerClusterGroup ref={groupRef} showCoverageOnHover={false}>
      {markers.map(({ fillColor, position, id }) => (
        <CircleMarker
          fillColor={fillColor}
          radius={10}
          fillOpacity={1}
          key={id}
          center={position}
        />
      ))}
    </MarkerClusterGroup>
  );
}

Usage
function MapExample(props) {
  const { markers, center } = props;
  return (
    <Map center={center} maxZoom={9} zoom={5}>
      <TileLayer
        url="https://{s}.tile.openstreetmap.org/{z}/{x}/{y}.png"
        attribution='© OpenStreetMap contributors'
      />
      <CustomLayer markers={markers} />
    </Map>
  );
}

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