text:
\section{Introduction}

Throughout the text let $p$ denote a fix prime.
Let $G$ be a finite group and $P$ a Sylow $p$-subgroup of $G$.
We denote by $H^\bullet (G,\mathbb{F}_p)$ the mod $p$ cohomology algebra. It is well known that the restriction map in cohomology
\begin{equation}
\label{eq:resisom}
\xymatrix{
H^\bullet (G,\mathbb{F}_p)\ar@{^{(}->}[r]&H^\bullet (P,\mathbb{F}_p)
}
\end{equation}
is injective (see \cite[Proposition 4.2.2]{ev:cgr}). Suppose that $G$ is $p$-nilpotent, i.e., $P$ has a normal complement $N$ in $G$. In this situation
the composition
\begin{equation}
\label{eq:composition}
\xymatrix{
P\ar[r] & G\ar[r] & G/N,
}
\end{equation}
is an isomorphism. Therefore the composition
\begin{equation}
\xymatrix{
H^\bullet (G/N,\mathbb{F}_p)\ar[0,2]^{inf^G_{G/N}} & & H^\bullet (G,\mathbb{F}_p)\ar[0,2]^{res^G_P} & & H^\bullet (P,\mathbb{F}_p)
}
\end{equation}
is also an isomorphism. This together with \eqref{eq:resisom} implies that, if
$G$ is $p$-nilpotent, then the
restriction map in cohomology
$\res^G_P :H^\bullet (G,\mathbb{F}_p)\longrightarrow H^\bullet (P,\mathbb{F}_p)$ is an isomorphism.
The following result of M. Atiyah shows that the converse is also true.

\begin{teo}[Atiyah]
If $\res^G_P :H^i (G,\mathbb{F}_p)\longrightarrow H^i (P,\mathbb{F}_p)$ are isomorphisms for all $i$ big enough,
then $G$ is $p$-nilpotent. In particular $G$ is $p$-nilpotent if and only if
$\res^G_P :H^\bullet (G,\mathbb{F}_p)\longrightarrow H^\bullet (P,\mathbb{F}_p)$ is an isomorphism.
\end{teo}

\begin{proof}
A proof of this can be found in the introduction of \cite{qu:pnil}.
\end{proof}

Atiyah $p$-nilpotency criterion uses the cohomology in high dimension.
Another cohomological criterion for $p$-nilpotency using cohomology in dimension $1$
was provided by J. Tate (\cite{tat:nil}).

\begin{teo}[Tate]
If $\res^G_P:H^1 (G,\mathbb{F}_p)\longrightarrow H^1 (P,\mathbb{F}_p)$ is an isomorphism, then $G$ is $p$-nilpotent.
\end{teo}

\begin{proof}
See  \cite{tat:nil}.
\end{proof}

D. Quillen generalized Atiyah's $p$-nilpotency criterion for odd primes (\cite{qu:pnil}).

\begin{teo}[Quillen]
\label{teo:quipnil}
Let $p$ be an odd prime. Then $G$ is $p$-nilpotent if and only if $\res^G_P:H^\bullet (G,\mathbb{F}_p)\longrightarrow H^\bullet  (P,\mathbb{F}_p)$ is an $F$-isomorphism.
\end{teo}

\begin{proof}
See \cite{qu:pnil}.
\end{proof}

Atiyah's $p$-nilpotency criterion can be reinterpreted in terms of $p$-fusion.
We recall that  a subgroup $H$ of $G$ \textit{controls $p$-fusion in $G$} if
\begin{itemize}
\item[(a)] $H$ contains a Sylow $p$-subgroup of $G$ and
\item[(b)] for any subgroup $A$ of $G$ and for any $g\in G$ such that $A, A^g\leq H$,
there exists $x\in H$ such that for all $a\in A$, $a^g=a^x$.
\end{itemize}
By a result of G. Mislin \cite{mis:modp}, a subgroup $H$ of $G$ controls $p$-fusion in $G$
if and only if $\res^G_P:H^\bullet (G,\mathbb{F}_p)\longrightarrow H^\bullet  (H,\mathbb{F}_p)$ is an isomorphism.
Using Mislin's result Atiyah's $p$-nilpotency criterion follows from Frobenious $p$-nilpotency criterion.

Mislin's type of result can also be provided
for the concept of $F$-isomorphism.  In order to do this we introduce the following concept. Let $\ca{C}$ be a class of finite $p$-groups. We say that a subgroup $H$ of $G$ \textit{controls fusion of $\ca{C}$-groups in $G$} if
\begin{itemize}
\item[(a)] Any $\ca{C}$-subgroup of $G$ is conjugated to a subgroup of $H$ and
\item[(b)] for any $\ca{C}$-subgroup $A$ of $G$ and for any $g\in G$ such that $A, A^g\leq H$, there exists $x\in H$ such that for all $a\in A$, $a^g=a^x$.
\end{itemize}
The condition (b) can be rewritten as
\begin{itemize}
\item[(b$^\prime$)] if $A$ is a $\ca{C}$-subgroup of $H$ and $g\in G$ satisfies that $A^g\leq H$, then $g\in C_G(A).H$.
\end{itemize}

Theorem A bellow, which will be proved in Section \ref{s:cofus}, follows
naturally from Quillen's work on cohomology (see \cite{qu:strat} and \cite{qu:pnil}). Note that the
``if'' was proved in \cite{GSW} and it is a direct consequence of Quillen's stratification (\cite{qu:strat}).
The converse follows from a careful reading of \cite[Section 2]{qu:pnil}.

\begin{teoa}
Let $G$ be a finite group and $H$ a subgroup of $G$. Then $\res^G_H:H^\bullet (G,\mathbb{F}_p)\longrightarrow  H^\bullet (H,\mathbb{F}_p)$
is an $F$-isomorphims if and only if $H$ controls fusion of elementary abelian $p$-subgroups of $G$.
\end{teoa}

In Section \ref{s:pnilcriterion} we will prove the following $p$-nilpotency criterion that can be seen as a generalization of Quillen $p$-nilpotency criterion (Theorem \ref{teo:quipnil} above) to the prime $p=2$.

\begin{teob}
Let $G$ be a finite group and $P$ a Sylow $p$-subgroup of $G$. Then  the following two conditions are equivalent
\begin{enumerate}
\item $G$ is $p$-nilpotent.
\item $P$ controls fusion of cyclic subgroups of order $p$ in case $p$ is odd, and cyclic subgroups of order $2$ and $4$ in case $p=2$.
\end{enumerate}
\end{teob}

Note that Theorem A and Theorem B imply Quillen's $p$-nilpotency criterion. We will finish this short note by giving two applications of Theorem B. The first application will consist on reproving a result of H-W. Henn and S. Priddy  that implies that "most" finite groups are $p$-nilpotent (see \cite{hp:pcen}). The second application is a generalization to the prime $p=2$ of the following fact: if all elements of order $p$ of a finite group $G$ are in some upper center of $G$ and $p$ is an odd prime, then $G$ is $p$-nilpotent (see \cite{th:pcenp} and \cite{GSW}). For the prime $p=2$ we will show that if all elements of order $2$ and $4$ are in some upper center of $G$, then $G$ is $2$-nilpotent.

We would like to end this introduction with an example of Quillen \cite{qu:pnil} where
the necessity of considering cyclic groups of order $2$ and $4$ for the case $p=2$ in Theorem B
is illustrated.

\begin{eje}
Consider $Q=\{ 1,-1,i,-i,j,-j,k,-k\}$ the quaternion group and $\alpha$ an automorphism of order $3$ that permutes $i$, $j$ and $k$. Let $G$ be the semidirect product between $Q$ and $\langle \alpha\rangle$ given by the action of $\alpha$ in $Q$. $A=\{ 1,-1\}$ is the only subgroup of exponent $2$ in $G$. Clearly $Q$ controls fusion of cyclic subgroup of order $2$. However $G$ is not $2$-nilpotent.
\end{eje}

\section{Cohomology and fusion}
\label{s:cofus}
The aim of this section is to sketch the proof of Theorem A.
In subsections \ref{ss:strat}, \ref{ss:ea} and \ref{ss:spec}
we will recall Quillen work in the mod $p$ cohomology algebra
of a finite group. This will be used in subsection \ref{ss:cf} to
prove Theorem A.

For a finite group $G$ the {\it mod $p$ cohomology algebra}
\begin{equation}
\label{eq:bas-11}
H^\bullet(G)=H^\bullet(G,\mathbb{F}_p)
\end{equation}
is a finitely generated, connected, anti-commutative, $\mathbb{N}_0$-graded $\mathbb{F}_p$-algebra.

Let $\alpha_\bullet\colon A_\bullet\to B_\bullet$ be a homomorphism of
finitely generated, connected, anti-commutative, $\mathbb{N}_0$-graded $\mathbb{F}_p$-algebras.
Then $\alpha_\bullet$ is called an {\it $F$-isomorphism} if
$\kernel(\alpha_\bullet)$ is nilpotent, and for all $b\in B_n$ there exists
$k\geq 0$ such that $b^{p^k}\in\image(\alpha_\bullet)$.

\subsection{Quillen's stratification}
\label{ss:strat}
Let $G$ be a finite group. Let $\ca{E}_G$ denote the category
whose objects are the elementary abelian $p$-subgroups of $G$ and whose morphisms are given
by conjugation, i.e., for $E, E^\prime\in \text{ob}(\ca{E}_G)$ one has
\begin{equation}
\label{eq:bas-4}
\mor_G(E,E^\prime)=\{\,i_g\colon E\to E^\prime\mid g\in G,\ g\,E\,g^{-1}\leq E^\prime\,\},
\end{equation}
where $i_g(e)= g\, e\, g^{-1}$, $e\in E$. Then
\begin{equation}
\label{eq:bas-5}
H^\bullet(\ca{E}_G)=\textstyle{\varprojlim_{\ca{E}_G} H^\bullet(E)}
\end{equation}
is a finitely generated, connected, anti-commutative, $\mathbb{N}_0$-graded $\mathbb{F}_p$-algebra.
Moreover, the restriction maps $\res^G_E$ yield a map
\begin{equation}
\label{eq:bas-6}
q_G=\textstyle{\prod_{E\in ob(\ca{E}_G)}\rst^G_E}\colon H^\bullet(G)\longrightarrow
H^\bullet(\ca{E}_G).
\end{equation}
The following result is  known as Quillen stratification.

\begin{teo}[Quillen]
\label{thm:strat}
Let $G$ be a finite group. Then $q_G\colon H^\bullet(G)\to H^\bullet(\ca{E}_G)$
is an $F$-isomorphism.
\end{teo}

\begin{proof}
See \cite[Cor. 5.6.4]{ben:coh2} or \cite{qu:strat}.
\end{proof}

\subsection{Cohomology of elementary abelian $p$-groups}
\label{ss:ea}
One can easily deduce the cohomology
of an elementary abelian $p$-group from the cohomology of the cyclic group of exponent $p$
and the Kunneth formula.

\begin{lem}
Let $A$ be an elementary abelian $p$-group. Then
\begin{equation}
H^\bullet (A,\mathbb{F}_p)\cong \begin{cases}
\Lambda (A^*) \otimes S (\beta (A^* )) & \text{if $p$ is odd} \\
S (A^*) & \text{if $p=2$},
\end{cases}
\end{equation}
where $\Lambda$ denotes the exterior algebra functor, $S$ the symmetric algebra functor,
$A^*=\Hom (A,\mathbb{F}_p)=H^1(A,\mathbb{F}_p)$ and $\beta$ the Bockstein homomorphism from $H^1(A,\mathbb{F}_p)$ to
$H^2(A,\mathbb{F}_p)$.
\end{lem}

\begin{proof}
See \cite[Chap. 3 Section 5]{ev:cgr}.
\end{proof}

From the previous lemma one can easily deduces that

\begin{equation}
\label{eq:elemabelian}
H^\bullet (A,\mathbb{F}_p)/\sqrt{0}\cong
S (A^*).
\end{equation}

\subsection{The spectrum of  $H(G)$}
\label{ss:spec}
Let $G$ be a finite group. Following  Quillen (\cite{qu:pnil}) we define
\begin{equation}
H(G)=\begin{cases}
\oplus_{i\geq 0} H^{2i}(G,\mathbb{F}_p) & \text{if $p$ is odd} \\
\oplus_{i\geq 0} H^{i}(G,\mathbb{F}_p)  & \text{if $p=2$}.
\end{cases}
\end{equation}
$H(G)$ is a graded commutative ring.  For an elementary abelian $p$-subgroup $A$ of $G$,
denote by $\eu{g}_A$
the ideal of $H (G)$ consisting of elements $u$ such that $u|_A$ is nilpotent. From
\eqref{eq:elemabelian}, $\res^G_A:H(G)\longrightarrow H (A)$ induces a
monomorphism
\begin{equation}
\label{eq:spec-1}
\xymatrix{
H (G)/\eu{g}_A\ar@{^{(}->}[r]&S (A^*).
}
\end{equation}
In particular, the ideal  $\eu{g}_A$ is a prime ideal of $H(G)$. Furthermore,

\begin{teo}[Quillen]
\label{teo:conjugacion}
Let $A,A^\prime \subset G$ be elementary abelian subgroups of $G$. Then $\eu{g}_A\subseteq \eu{g}_{A^\prime}$
if and only if $A^\prime$ is conjugated to a subgroup of $A$. In particular $\eu{g}_A= \eu{g}_{A^\prime}$
if and only if $A$ and $A^\prime$ are conjugated in $G$.
\end{teo}

\begin{proof}
See \cite[Theorem 2.7]{qu:pnil}.
\end{proof}

Let us consider the extension of quotient fields associated to the monomorphism in \eqref{eq:spec-1},
\begin{equation}
\xymatrix{
k(\eu{g}_A)\ar@{^{(}->}[r]&k(A).
}
\end{equation}

We have that

\begin{teo}[Quillen]
\label{teo:extension}
The extension $k(A)/k(\eu{g}_A)$ is a normal extension and
\begin{equation}
 \Aut (k(A)/k(\eu{g}_A))\cong N_G(A)/C_G(A).
\end{equation}
\end{teo}

\begin{proof}
See \cite[Theorem 2.10]{qu:pnil}.
\end{proof}

\subsection{F-isomorphisms and fusion}
\label{ss:cf}
The following lemma is a standard result in commutative algebra.

\begin{lem}
\label{lem:fisomspec}
Let $A$ and $B$ be commutative $\mathbb{F}_p$-algebras and $f\colon A\longrightarrow B$ an $F$-isomorphism. Then $f^*: \Spec (B)\longrightarrow \Spec (A)$ is a homeomorphism.
\end{lem}

\begin{proof}
Since the kernel of $f$ is
nilpotent, then for any radical ideal $\eu{a}$ of $A$ one has that $f^{-1} (\sqrt{f(\eu{a})})=\eu{a}$. Since
for any $x\in B$ there exits $y\in A$ and $n\geq 0$ such that $f(y)=x^{p^n}$, then for any radical ideal $\eu{b}$ of $B$
one has that $\sqrt{f (f^{-1}(\eu{b}))}=\eu{b}$. Therefore
\begin{eqnarray}
\xymatrix{\eu{a}\ar[r]& \sqrt{f(\eu{a})}} \\
\xymatrix{\eu{b}\ar[r]& f^{-1}(\eu{b})}
\end{eqnarray}
is a bijection between the radical ideals of $A$ and the radical ideals of $B$. In particular
$f^*$ is an isomorphism of varieties.
\end{proof}

We are now ready to prove Theorem A.

\begin{proof}[Proof of Theorem A]
Suppose first that $H$ controls fusion of elementary abelian $p$-subgroups of $G$. Then the embedding functor
\begin{equation}
\label{eq:fisom-1}
\xymatrix{j_{H,G}:\ca{E}_H \ar[r] & \ca{E}_G}
\end{equation}
is an equivalence of categories. Therefore
\begin{equation}
\label{eq:fisom-2}
\xymatrix{H^\bullet (j_{H,G}):H^\bullet(\ca{E}_G) \ar[r] & H^\bullet (\ca{E}_H)}
\end{equation}
is an isomorphism. Consider the commutative diagram
\begin{equation}
\label{dia:fusF}
\xymatrix{
H^\bullet(G)\ar[d]_{\res^G_H}\ar[r]^{q_G}& H^\bullet(\ca{C}_G)\ar[d]^{H^\bullet(j_{H,G})}\\
H^\bullet(H)\ar[r]^{q_H}& H^\bullet(\ca{C}_H).
}
\end{equation}
By Theorem \ref{thm:strat} and equation \eqref{eq:fisom-2} it follows that $\res^G_H$ is an $F$-isomorphism.

\smallskip

Suppose now that $\res^G_H\colon H^\bullet (G,\mathbb{F}_p)\longrightarrow H^\bullet (H,\mathbb{F}_p)$ is an
$F$-isomorphism. Then $\res^G_H$ induces an $F$-isomorphism $f\colon H (G)\longrightarrow H (H)$. Consider $A$ and $A^\prime$ two elementary abelian $p$-subgroups of $H$.

\bigskip

\noindent \textit{Subclaim 1:} If $A$ and $A^\prime$ are conjugated in $G$, then they are conjugated in $H$.

\begin{proof}[Subproof]
By Lemma \ref{lem:fisomspec}, $f^*\colon \Spec (H (H))\longrightarrow
\Spec (H (G))$ provides a bijection between the prime ideals of $H (H)$ and the prime ideals
of $H (G)$. Furthermore, if $A$ is an elementary abelian $p$-subgroup of
$H$, then $\eu{g}_A=f^*(\eu{h}_A)$. By Theorem \ref{teo:conjugacion}, if $A$ and $A^\prime$ are conjugated in $G$, then $\eu{g}_A=\eu{g}_{A^\prime}$. In particular $f^* (\eu{h}_A)=\eu{g}_A=\eu{g}_{A^\prime}=f^*(\eu{h}_{A^\prime})$. Therefore
$\eu{h}_A=\eu{h}_{A^\prime}$. Hence,
by Theorem \ref{teo:conjugacion}, $A$ and $A^\prime$ are conjugated in $H$.
\end{proof}

\noindent \textit{Subclaim 2:} $N_G(A)=C_G(A)N_H(A)$.

\begin{proof}[Subproof] Since $k(\eu{h}_A)$ is a purely inseparable extension of $k (\eu{g}_A)$, then
\begin{equation}
 \Aut (k(A)/k(\eu{h}_A))\cong \Aut (k (A)/k (\eu{g}_A)).
\end{equation}
Therefore, by Theorem \ref{teo:extension},
$N_H(A)/C_H(A)\cong N_G(A)/C_G(A)$.
\end{proof}

\noindent \textit{Subclaim 3:} $H$ controls fusion of elementary abelian $p$-subgroups of $G$.

\bigskip

\noindent \textit{Subproof:} Let $A$ be an elementary abelian $p$-subgroup of $H$ and $g\in G$ such that $A^g\leq H$.
Then, by Subclaim 1 there exists $h\in H$ such that $A^g=A^h$. In particular, by Subclaim 2,
$gh^{-1}\in N_G(A)=C_G(A).N_H(A)$. Therefore $g\in C_G(A).H$
\end{proof}

\section{A $p$-nilpotency criterion}

\label{s:pnilcriterion}

In this section we will prove our main result Theorem B. To ease the notation we denote by $\Cp$ the class of cyclic groups of order $p$ in case $p$ is odd and cyclic groups of order $2$ and $4$ in case $p=2$. Put $\bp=p$ if $p$ is odd and $\bp=4$ in case $p=2$.

\begin{teo}
\label{main}
Let $G$ be a finite group and $P$ a Sylow $p$-subgroup of $G$. Then  the following two conditions are equivalent
\begin{enumerate}
\item $G$ is $p$-nilpotent.
\item $P$ controls fusion of $\Cp$-groups.
\end{enumerate}
\end{teo}

\begin{proof}
It is clear that if $G$ is $p$-nilpotent, then $P$ controls fusion
of $\Cp$-groups.

Let us show the converse. Using Frobenius
$p$-nilpotency criterion it is enough to prove that for any subgroup $B$ of $P$ and for any
$p^\prime$-element $g\in N_G(B)$, then $g$ centralices $B$. The subgroup $B$
is contained in  $Z_l(P)$ for some $l\geq 1$ where $Z_l(P)$ denotes the $l$-upper center of $P$.
We will show by induction on $l$ that $g\in C_G(B)$.
Suppose first that $B\leq Z(P)$ and consider $a\in B$ such that
$a^{\bp}=1$. Since $P$ controls
fusion of $\Cp$-groups, there exists $x\in P$ such that $a^g=a^x$ and since $a\in Z (P)$, then $a^x=a$.
Hence we have that $g$ centralices all elements of order $p$ ($2$ and $4$ in case $p=2$) in $B$.
Thus, by \cite[Chap. V Lemma 5.12]{hupp1:grp},  $g$ centralices $B$.

For the general case, consider  $B\leq Z_l(P)$ and suppose the assumption to be true for any subgroup contained in $Z_{l-1}(P)$.

\bigskip

\noindent \textit{Subclaim 1:} For $a\in B$ such that $a^{\bp}=1$, we have that $[a,g,g]=1$.

\begin{proof}[Subproof]
We have that  $g$ normalizes the subgroups $K=\langle a\in B\mid a^{\bp}=1\rangle$ and $[K,g]$. We also have that
\begin{equation}
\label{eq-main}
[K,g]=\langle [a,g]^b\mid a,b\in K\ \text{and}\ a^{\bp}=1\rangle .
\end{equation}
Take $a\in B$ such that $a^{\bp}=1$. Since $P$ controls fusion of $\Cp$-groups, there exists $x\in P$  such that $a^g=a^x$. In particular $[a,g]=[a,x]\in Z_{l-1}(P)$. Therefore,
by \eqref{eq-main}, $[K,g]=Z_{l-1}(P)$. Since $g$ normalices  $[K,g]$ and by induction  hypothesis we have that $[K,g,g]=1$.
\end{proof}

\noindent \textit{Subclaim 2:} $g\in C_G(B)$.

\bigskip

\noindent \textit{Subproof.}
Take $a\in B$ such that $a^{\bp}=1$ and put $p^e$ the exponent of $B$. Consider the subgroup $H=\langle g,[a,g]\rangle$.  By the Subclaim  1, $\gamma_2(H)=1$. Then, by
 \cite[Chap. III, Theorem 9.4]{hupp1:grp}, we have that $[a,g^{p^e}]= [a,g]^{p^e}=1$.
 Since $g$ is a $p^\prime$-element of $G$, $g$ centralices all elements of order $p$ ($2$ and $4$ in case $p=2$) in $B$. Thus, by
\cite[Chap. V Lemma 5.12]{hupp1:grp},  $g$ centralices $B$.

This ends the proof.
\end{proof}

As a consequence to this we have the following corollary.

\begin{cor}
\label{corolario1}
Let $G$ a finite group and $P$ a Sylow $p$-subgroup of $G$ such that
\begin{itemize}
\item[1.] $N_G(P)$ controls fusion of $\Cp$-groups and
\item[2.] $N_G(P)=C_G(P).P$.
\end{itemize}
Then $G$ is $p$-nilpotent.
\end{cor}

\begin{proof}
Let $A$ be a $\Cp$-group  and $g\in G$ such that $A,A^g\leq P$. Since $N_G(P)$ controls fusion
$\Cp$-groups, one has that $g\in C_G(A).N_G(P)=C_G(A).P$. Then $P$ controls fusion
of $\Cp$-groups  and, by Theorem \ref{main}, $G$ is $p$-nilpotent.
\end{proof}

\section{Some applications}

We now present the first application of Theorem \ref{main}. In \cite{hp:pcen} H-W. Henn
and S. Priddy proved that if a group $G$ has a Sylow $p$-subgroup $P$ such that
\begin{itemize}
\item[i)] if $p$ is odd, the elements of order $p$ of $P$ are in the center of $P$ and, if $p=2$,
the elements of order $2$ and $4$ are in the center of $P$,
\item[ii)] $\Aut (P)$ is a $p$-group,
\end{itemize}
then $G$ is $p$-nilpotent. This implies that "most" finite groups are
$p$-nilpotent (see \cite{hp:pcen}). The proof of Henn and Priddy is essentially topological.
In \cite{th} J. Thevenaz gave a group theoretical proof of this result using
Alperin's Fusion Theorem.  In fact Thevenaz proved that if $G$ satisfies condition i),
then $N_G(P)$ controls $p$-fusion in $G$. This, together with condition ii) above implies
that $P$ controls $p$-fusion in $G$ and therefore $G$ is $p$-nilpotent.
We now give a weaker version of Thevenaz result which
also implies that a group satisfying i) and ii) is $p$-nilpotent.

\begin{pro}
\label{priddy}
Let $G$ be a finite group and $P$ a Sylow $p$-subgroup of $G$.
Suppose that the elements of order dividing $p$ in $P$ (or $4$ in case $p=2$)
are in the center of $P$. Then
$N_G(P)$ controls fusion of $\Cp$-groups.
\end{pro}

\begin{proof}
Let $A$ be a  $\Cp$-group and $g\in G$ such that $A,A^g\leq N_G(P)$. In particular $A^g\leq  P$. Equivalently $A\leq P^{g^{-1}}$.
Hence, since the elements of $P$ of order $p$ (or $4$ in case $p=2$)  are in the center of $P$, we have that $P,P^{g^{-1}}\leq C_G(A)$. But, since $P$ and $P^{g^{-1}}$ are Syllow $p$-subgroups of $C_G(A)$, there exists $c\in C_G(A)$ such that $P=P^{g^{-1}c}$. Thus $g^{-1}c\in N_G(P)$ and $g\in N_G(P).C_G(A)$.
\end{proof}

\begin{cor}
Let $G$ a finite group and $P$ a Syllow $p$-subgroup of $G$ such that
\begin{itemize}
\item[1.] all elements of order dividing $p$ in $P$ (or $4$ in case $p=2$)
are in the center of $P$ and
\item[2.] $N_G(P)=P.C_G(P)$.
\end{itemize}
Then $G$ is $p$-nilpotent.
\end{cor}

\begin{proof}
It follows from Proposition \ref{priddy} and Corollary \ref{corolario1}.
\end{proof}

The second application of Theorem \ref{main} is a generalization to $p=2$ of the fact that if the elements of order $p$ of a finite group $G$ are in some upper center of $G$, then $G$ is $p$-nilpotent (see \cite{th:pcenp} and \cite{GSW}).

\begin{cor}
\label{pcentral}
Let $G$ a finite group such that $K=\langle x\in G\mid x^{\bp}=1\rangle\leq Z_n(G)$
for some $n\geq 1$ (here $\bp$ means $p$ in case $p$ is odd and $4$ in case $p=2$).
Then $G$ is $p$-nilpotent.
\end{cor}

\begin{proof}
The subgroup $K$ is nilpotent of class at most $n$, and therefore
a finite $p$-group. Let $p^e$ be the exponent of $K$.
Then, by Hall-Petrescu collection formula (see \cite[Theorem 2.1]{gust:omega}),
for any $y\in K$ and $x\in G$
\begin{equation}
\label{eq:bas-0}
[y,x^{p^{e+n}}]\in
\prod_{0\leq i\leq e+n}[K,\overbrace{G,\ldots, G}^{p^i}]^{p^{e+n-i}}=1.
\end{equation}
Therefore one has that $G^{p^{e+n}}\leq C_G(K)$.
Moreover, for any Sylow $p$-subgroup $P$ of $G$ one has
$G=P.G^{p^{e+n}}=P.C_G(K)$.  In particular
$P$ controls fusion of $\Cp$-groups. Hence, by Theorem \ref{main}, $G$ is $p$-nilpotent.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
\section{Introduction}
\indent Since the work of S. Abhyankar, F.-V. Kuhlmann, V. Cossart and O. Piltant, we know the importance to studying the decomposition, ramification and inertia groups of a valuation in the problem of local uniformization. In \cite{jccar0}, we saw the importance of the key polynomials to obtain a local uniformization for a valuation of rank one. If the characteristic of the residual field is $0$ then we obtain the result, otherwise it is sufisent to monomialize the first limit key polynomial wich is of the form:
$$Q_\omega=X^{p^e}+\sum\limits_{i=0}^{e-1}c_{p^i}X^{p^i}+c_0.$$
\\ \\\indent In \cite{jcdefaut}, we related the defect of a simple and finite extension of valued fields with the effective degree of the key polynomials. For example, if the valuation is of rank $1$ and there is only one limit key polynomial with the others of degree$1$, the defect is exactly the degree of the limit key polynomial.
\\\indent This paper complete \cite{jcdefaut} in our program to connect the the ramification theory, the defect of an finite extension and the works of F.-V. Kuhlmann in local uniformization with the key polynomials. For a simple, normal and finite extension of a valued field, we prove that we can related the order of the ramification group of the field extension and the set of key polynomials associated to the extension of the valuation.
\\ \\\indent In the first section we recall the definitions of the  decomposition, ramification and inertia groups. We also give some relations between the dimension of the different fields and the order of the groups of valuation.
\\\indent In the second chapter we recall the results of \cite{jcdefaut} and interpret them in terms of the order of ramification group. In particular we give a condition on the order of the ramification group to have no defect or no limit key polynomials.
\\\indent In the last section we give some conditions to obtain a local uniformization theorem with asumptions on the group of ramification.
\\ \\\noindent\textbf{Notation.} Let $\nu$ be a valuation of a field $K$. We write $R_\nu=\lbrace f\in K\:\vert\:\nu(f)\geqslant 0\rbrace$, this is a local ring whose maximal ideal is $m_\nu=\lbrace f\in K\:\vert\:\nu(f)> 0\rbrace$. We then denote by $k_\nu=R_\nu/m_\nu$ the residue field of $R_\nu$ and $\Gamma_\nu=\nu(K^*)$.
\\For a field $K$, we will denote by $\overline K$ an algebraic closure of $K$. For an algebraic extension $L\vert K$ we will denote by $(L\vert K)^{sep}$ a separable closure of $K$ in $L$ (or more simply by $K^{sep}$ if no confusion is possible) and by $Aut(L\vert K)$
 the group of automorphisms of $L\vert K$ (or $Gal(L\vert K)$ if $L\vert K$ is a Galois extension). \\For a finite extension $L\vert K$, we denote by $[L:K]_{sep}$ the \textit{separable degree} of $L\vert K$ and by $[L:K]_{ins}=[L:K]/[L:K]_{sep}$ the \textit{inseparable degree} of $L\vert K$; this a power of $car(K)$.
\\If $R$ is a ring and $I$ an ideal of $R$, we will denote by $\widehat{R}^I$ the \textit{ $I$-adic completion} of $R$. When $(R,\mathfrak m)$ is a local ring, we will say the \textit{completion} of $R$ instead of the $\mathfrak m$-adic completion  of $R$ and we will denote it by $\widehat{R}$.
\\For all $P\in Spec(R)$, we note $\kappa(P)=R_P/PR_P$ the residue field of $R_P$.
\\For $\alpha\in\mathbb Z^n$ and $u=(u_1,...,u_n)$ a $n$-uplet of elements of $R$, we write:
\[u^\alpha=u_1^{\alpha_1}...u_n^{\alpha_n}.\]
For $P,Q\in R\left[ X\right] $ with $P=\sum\limits_{i=0}^{n}a_{i}Q^{i}$ and $a_{i}\in R[X]$ such that the degree of $a_{i}$ is strictly less than $Q$, we write:
\[d_{Q}^{\:\circ}(P)=n.\]
If $Q=X$, we will simply write $d^{\:\circ}(P)$ instead of $d_{X}^{\:\circ}(P)$.
\\Finally, if $R$ is a domain, we denote by $Frac(R)$ its quotient field.
\section{Decomposition, inertia and ramification groups}
\indent We follow the definitions of \cite{kulhlivre}, Chapter 7.
\begin{defi}
Let $(K,\nu)\hookrightarrow (L,\mu)$ be a normal algebraic extension of valued fields.
\begin{enumerate}
\item The \textbf{decomposition group of} $\boldsymbol{L\vert K}$ is:
\[G^d(L\vert K,\mu)=\{\sigma\in Aut(L\vert K)\:\vert\:\forall\alpha\in L,\:\mu(\sigma (\alpha))=\mu(\alpha)\}.\]
\item The \textbf{inertia group of} $\boldsymbol{L\vert K}$ is:
\[G^i(L\vert K,\mu)=\{\sigma\in Aut(L\vert K)\:\vert\:\forall\alpha\in R_\mu,\:\mu(\sigma (\alpha)-\alpha)>0\}.\]
\item The \textbf{ramification group of} $\boldsymbol{L\vert K}$ is:
\[G^r(L\vert K,\mu)=\{\sigma\in Aut(L\vert K)\:\vert\:\forall\alpha\in R_\mu,\:\mu(\sigma (\alpha)-\alpha)>\mu(\alpha)\}.\]
\end{enumerate}
When no confusion is possible, we will denote respctiveley by $G^d$, $G^i$ and $G^r$ the groups $G^d(L\vert K,\mu)$, $G^i(L\vert K,\mu)$ and $G^r(L\vert K,\mu)$.
\end{defi}
\begin{rem}\label{normgroup}
\textup{$G^r(L\vert K,\mu)\lhd G^i(L\vert K,\mu)\lhd G^d(L\vert K,\mu)\lhd Aut(L\vert K)$.}
\end{rem}
\begin{defi}
Let $(K,\nu)\hookrightarrow (L,\mu)$ be a normal algebraic extension of valued fields.
\begin{enumerate}
\item The fixed field of $G^d(L\vert K,\mu)$ in the separable closure of $K$ in $L$ is called the \textbf{decomposition field of} $\boldsymbol{K}$ \textbf{in} $\boldsymbol L$ and will be denote by $K^d$. We also write $\mu^d=\mu_{\vert K^d}$.
\item The fixed field of $G^i(L\vert K,\mu)$ in the separable closure of $K$ in $L$ is called the \textbf{inertia field of} $\boldsymbol{K}$ \textbf{in} $\boldsymbol L$ and will be denote by $K^i$. We also write $\mu^i=\mu_{\vert K^i}$.
\item The fixed field of $G^r(L\vert K,\mu)$ in the separable closure of $K$ in $L$ is called the \textbf{ramification field of} $\boldsymbol{K}$ \textbf{in} $\boldsymbol L$ and will be denote by $K^r$. We also write $\mu^r=\mu_{\vert K^r}$.
\end{enumerate}
\end{defi}
\begin{rem}
\textup{The ramification, inertia and decomposition fields are separable over $K$. By the remark \ref{normgroup}, the extensions $K^i\vert K^d$ and $K^r\vert K^d$ are Galois extensions.
\\If $L=\overline K$ then $K^d$ is an henselization of $K$ denote by $K^h$.}
\end{rem}
\begin{thm}\label{thmgal}
Let $(K,\nu)\hookrightarrow (L,\mu)$ be a normal and finite extension of valued fields. Write $e=[\Gamma_\mu:\Gamma_\nu]$, $f=[k_\mu:k_\nu]$, $p=char(k_\nu)$ and $G=Aut(L\vert K)$. Let $g$ be the number of distinct extensions of $\nu$ from $K$ to $L$ and $d=d_{L\vert K}(\mu,\nu)$ be the defect of the extension $L\vert K$ in $\mu$ (see Definition 2.3 of \cite{jcdefaut}).
\begin{enumerate}
\item $G=Gal(K^{sep}\vert K)$, $G^d=Aut(L\vert K^d)=Gal(K^{sep}\vert K^d)$, $G^i=Aut(L\vert K^i)=Gal(K^{sep}\vert K^i)$, $G^r=Aut(L\vert K^r)=Gal(K^{sep}\vert K^r)$.
\item $[L:K]=g\times d\times e\times f$.
\item $K^d\vert K$ is an immediate extension and $g=[G:G^d]=[K^d:K]$.
\item $K^i\vert K^d$ is a Galois extension, $[\Gamma_{\mu^i}:\Gamma_{\mu^d}]=1$,
$[k_{\mu^i}:k_{\mu^d}]=[K^i:K^d]=f_0$.
\item $K^r\vert K^i$ is an abelian extension, $[\Gamma_{\mu^r}:\Gamma_{\mu^i}]=[K^r:K^i]=e_0$, $[k_{\mu^r}:k_{\mu^i}]=1$.
\item $K^{sep}\vert K^r$ is a $p$-extension, $[\Gamma_{\mu}:\Gamma_{\mu^r}]=p^t$, $[k_{\mu}:k_{\mu^r}]=p^s$ with $t,s\geqslant 0$ and $[K^{sep}:K^r]=\vert G^r\vert=p^u$.
\end{enumerate}
\end{thm}
\noindent\textit{Proof}: For a detailed proof, one can consult \cite{kulhlivre}, Chapter 7. We will give some ideas of proof.
\begin{enumerate}
\item It is clear because $L\vert K^{sep}$ is a purely inseparable extension and by using Galois theory.
\item By Lemma 7.46 of \cite{kulhlivre}, we know that:
\[\left[L:K\right]=\sum\limits_{i=1}^gd_ie_if_i,\]
where $d_i=d_{L\vert K}(\mu_i,\nu)$, $e_i=\left[\Gamma_{\mu_i}:\Gamma_\nu\right]$, $f_i=\left[k_{\mu_i}:k_\nu\right]$ and $\mu_i$ all the extensions of $\nu$ to $L$, $i\in\{1,...,g\}$. The extension $L\vert K$ is a normal extension so, for $i\neq j$, $e_i=e_j$, $f_i=f_j$ and $d_i=d_j$. Then we deduce the equality.
\item To prove that $g=[G:G^d]$, it is suffice to observe that, for $\sigma,\sigma'\in G$,  $\mu(\sigma(\alpha))=\mu(\sigma'(\alpha))$ if and only if $\sigma\sigma'^{-1}\in G^d$. The second equality comme from 1. and Galois theory.
\\To show that $K^d\vert K$ is an immediate extension, we need to prove that: $$[k_{\mu^d}:k_{\nu}]=1=[\Gamma_{\mu^d}:\Gamma_{\nu}].$$ The first equality come from the fact that $\overline\alpha=\overline{Tr_{K^d\vert K}(\alpha)}$ for $\overline\alpha\in k_{\mu^d}\setminus\{0\}$. To prove the second equality, take $\gamma\in\Gamma_{\mu^d}$ and construct an element of $K^d$ such that is minimal polynomial over $K$ have only one root of value $\gamma$: this element.
\item $K^i\vert K^d$ is a Galois extension by Galois theory because  $G^i(L\vert K,\mu)\lhd G^d(L\vert K,\mu)$ and we have $Gal(K^i\vert K^d)\simeq G^d/G^i$. Using the separable closure of $k_{\mu^d}$ in $k_\mu$ (which is $k_{\mu^i}$), we can prove that $G^d\rightarrow Aut(k_\mu\vert k_\nu)$ is surjective. So, by definition of $G^i$, $Gal(K^i\vert K^d)\simeq G^d/G^i\simeq Gal(k_\mu\vert k_{\mu^d})=Aut(k_\mu\vert k_\nu)$. In the same way, we can show that $Gal(k_\mu\vert k_{\mu^d})\simeq Gal(k_{\mu^i}\vert k_{\mu^d})$. Finally from the fundamental inequality, we obtain:
$$[K^i:K^d]=\left\vert Gal(K^i\vert K^d)\right\vert=\left\vert Gal(k_{\mu^i}\vert k_{\mu^d})\right\vert\leqslant [k_{\mu^i}: k_{\mu^d}]\leqslant [K^i:K^d].$$
\item$K^r\vert K^i$ is a Galois extension by Galois theory because  $G^r(L\vert K,\mu)\lhd G^i(L\vert K,\mu)$ and we have $Gal(K^r\vert K^i)\simeq G^i/G^r$. Assuming that 6. is true, we can show that there exists an embedding between $Gal(K^r\vert K^i)$ and $Hom((\Gamma_{\mu^r}/\Gamma_{\mu^i})_{p'},k_{\mu^r}^\times)$ where $\Gamma_{\mu^r}/\Gamma_{\mu^i}=(\Gamma_{\mu^r}/\Gamma_{\mu^i})_{p}\oplus (\Gamma_{\mu^r}/\Gamma_{\mu^i})_{p'}$ with $(\Gamma_{\mu^r}/\Gamma_{\mu^i})_{p}$ a $p$-group and $(\Gamma_{\mu^r}/\Gamma_{\mu^i})_{p'}$ a torsion group wich the orders of all elements are prime to $p$. We deduce from this that $K^r\vert K^i$ is an abelian extension. Using the fundamental inequality, we obtain:
\begin{align*}
[K^r:K^i]=\vert Gal(K^r\vert K^i)\vert&\leqslant\vert Hom((\Gamma_{\mu^r}/\Gamma_{\mu^i})_{p'},k_{\mu^r}^\times)\vert\\&\leqslant \left\vert Hom\left((\Gamma_{\mu^r}/\Gamma_{\mu^i})_{p'},\overline{k_{\mu^r}}^\times\right)\right\vert=\vert (\Gamma_{\mu^r}/\Gamma_{\mu^i})_{p'}\vert\\&\leqslant\vert\Gamma_{\mu^r}/\Gamma_{\mu^i}\vert\\&\leqslant [K^r:K^i].
\end{align*}
We have proved that $[\Gamma_{\mu^r}:\Gamma_{\mu^i}]=[K^r:K^i]$ and using the fundamental inequality, we obtain that $[k_{\mu^r}:k_{\mu^i}]=1$.
\item By Lemma 7.15 of \cite{kulhlivre}, it is sufficient to show that $G^r$ is a $p$-group, and we will have $[\Gamma_{\mu}:\Gamma_{\mu^r}]=p^t$. Since $k_{\mu^r}=k_{\mu^i}=k_{\nu}^{sep}$, then $[k_{\mu}:k_{\mu^r}]=[k_{\mu}:k_{\nu}]_{ins}$ is a power of $p$. To show that $G^r$ is a $p$-group, proceed by contradiction. Take an element of order a prime $q\neq p$. Consider the fixed field of this element, then $K^r$ is a cyclic extension of this field, of degree $q$. Let $x$ a primitive element, we can suppose that this trace is $0$ because $q\neq 0$. On the other hand, the sum of the residue of all the conjugates of $x$ divide by $x$ is $q$, so it is a contradiction with we consider his trace.
\end{enumerate}\qed
\\ \\As in \cite{kulhlivre}, Chapter 7, we can summarize the Theorem \ref{thmgal} in this table with $n=[L:K]$:
\newline
\begin{center}
\begin{tabular}{cccc}

Galois group&field extension& value group & residue field\\
\begin{tikzpicture}[node distance = 1.5cm, auto]
      \node (G) {$G$};
      \node (Gd) [above of=G] {$G^d$};
      \node (Gi) [above of=Gd] {$G^i$};
      \node (Gr) [above of=Gi] {$G^r$};
      \node (id) [above of=Gr] {$\{id\}$};
      \draw[-] (G) to node {$g$} (Gd);
      \draw[-] (Gd) to node {$f_0$} (Gi);
      \draw[-] (Gi) to node {$e_0$} (Gr);
      \draw[-] (Gr) to node {$p^u$} (id);

      \end{tikzpicture}
      &
      \begin{tikzpicture}[node distance = 1.5cm, auto]
      \node (K) {$K$};
      \node (Kd) [,above of=K] {$K^d$};
      \node (Ki) [above of=Kd] {$K^i$};
      \node (Kr) [above of=Ki] {$K^r$};
      \node (Ksep) [above of=Kr] {$K^{sep}$};
      \node (L) [above of=Ksep] {$L$};
      \draw[-] (K) to node {$g$} (Kd);
      \draw[-] (Kd) to node {$f_0$} (Ki);
      \draw[-] (Ki) to node {$e_0$} (Kr);
      \draw[-] (Kr) to node {$p^u$} (Ksep);
      \draw[-] (Ksep) to node {$p^l$} (L);
      \draw (K) -- (0.8,0);
       \draw[-] (0.8,0)  to node[swap] {$n$} (0.8,7.5);
      \draw (0.8,7.5) -- (L);
      \end{tikzpicture}
      &\begin{tikzpicture}[node distance = 1.5cm, auto]
      \node (g) {$\Gamma_{\nu}$};
      \node (gd) [above of=g] {$\Gamma_{\mu^d}$};
      \node (gi) [above of=gd] {$\Gamma_{\mu^i}$};
      \node (gr) [above of=gi] {$\Gamma_{\mu^r}$};
      \node (gsep) [above of=gr] {};
      \node (gmu) [above of=gsep] {$\Gamma_{\mu}$};
      \draw[-] (g) to node {$1$} (gd);
      \draw[-] (gd) to node {$1$} (gi);
      \draw[-] (gi) to node {$e_0$} (gr);
       \draw[-] (gr) to node {$p^t$} (gmu);
        \draw (g) -- (0.8,0);
       \draw[-] (0.8,0)  to node[swap] {$e$} (0.8,7.5);
      \draw (0.8,7.5) -- (gmu);
      \end{tikzpicture}
      &\begin{tikzpicture}[node distance = 1.5cm, auto]
      \node (g) {$k_{\nu}$};

      \node (gd) [above of=g] {$k_{\mu^d}$};
      \node (gi) [above of=gd] {$k_{\mu^i}$};
      \node (gr) [above of=gi] {$k_{\mu^r}$};
      \node (gsep) [above of=gr] {};
      \node (gmu) [above of=gsep] {$k_{\mu}$};
      \draw[-] (g) to node {$1$} (gd);
      \draw[-] (gd) to node {$f_0$} (gi);
      \draw[-] (gi) to node {$1$} (gr);
       \draw[-] (gr) to node {$p^s$} (gmu);
      \draw (g) -- (0.8,0);
       \draw[-] (0.8,0)  to node[swap] {$f$} (0.8,7.5);
      \draw (0.8,7.5) -- (gmu);
      \end{tikzpicture}

\end{tabular
\end{center}
~
\begin{coro}\label{corogaldef}
Under the same assumptions and notations of the Theorem \ref{thmgal}, we have:
\begin{enumerate}
\item $e=e_0\times p^t$ with $p\not\vert\; e_0$ and $f=f_0\times p^s$.
\item $d=p^{u+l-s-t}$ where $p^l=[L:K]_{ins}$.
\end{enumerate}
\end{coro}
\begin{rem}
\textup{If $L\vert K$ is a finite Galois extension, then $l=0$ an the Corollary \ref{corogaldef} is the same as the Corollary of Theorem 25, Ch. VI of \cite{zarsam2}. }
\end{rem}
\noindent\textit{Proof}: $e=e_0\times p^t$ and $f=f_0\times p^s$ come from the precedent table and $p\not\vert\; e_0$ because $G^r$ is the unique $p$-Sylow of $G^i$ and $\vert G^i\vert=e_0\times p^u$. Finally by 1. of Theorem \ref{thmgal} and the precedent table, since:
\[g\times d\times e\times f=[L:K]=p^{l+u}\times e_0\times f_0\times g,\]
then:
$$d=p^{u+l-s-t}.$$
\qed

\section{Key polynomials and ramification group}
\indent For a basic definition of key polynomials see Definition 3.1 of \cite{jcdefaut}, for more details see \cite{spivamahboub}. Here, we suppose known the theory of key polynomials, we only recall the link with the defect as in \cite{jcdefaut}.
\\\indent Let $\lbrace Q_{l}\rbrace_{l\in\Lambda}$ be a complete set of key polynomials, for $l\in\Lambda$ having a predecessor, write:
$\alpha_{l}=d_{Q_{l-1}}^{\:\circ}(Q_{l}).$
If $l=\omega n$ is a limit ordinal, $n\in\mathbb N^*$, denote by $\alpha_{l}=d_{Q_{l_0}}^{\:\circ}(Q_{l})$ where $l_0=\min\lbrace m\geqslant 1\:\vert\:\alpha_{\omega(n-1)+m}=1\rbrace$.
\begin{defi}
Let $K$ be a field and $\mu$ a valuation of $K[x]$. For $h\in K[x]$, consider its $i$-standard expansion $h=\sum\limits_{j=0}^{s_{i}}c_{j,i}Q_{i}^j$. We call by the $\boldsymbol{i}$\textbf{-th effective degree of h} the natural number:
\[\delta_i(h)=\max \lbrace j\in\lbrace 0,...,s_{i}\rbrace\:\vert\:j\beta_i+\mu\left(c_{j,i}\right)=\mu_i(h)\rbrace,\]
where:
\[\mu_{i}(h)=\min_{0\leqslant j\leqslant s_{i}}\lbrace j\mu(Q_{i})+\mu(c_{j,i})\rbrace.\]
By convention, $\delta_i(0)=-\infty$.
\end{defi}
\begin{rem}
\textup{Remind that, by Proposition 5.2 of \cite{spivamahboub}, for $l\in\Lambda$ an ordinal number, the sequence $(\delta_{l+i}(h))_{i\in\mathbb N^*}$ decreases. Thus there exists $i_0\in\mathbb N^*$ such that $\delta_{l+i_0}(h)=\delta_{l+i_0+i}(h)$, for all $i\geqslant 1$ and we denote this common value by $\delta_{l+\omega}(h)$. From here until the end, we write $\delta_{l+\omega}=\delta_{l+\omega}(Q_{l+\omega})$.}
\end{rem}
\begin{thm}(\cite{jcdefaut}, Corollary 4.5)\label{defautdelta}
Let $(K,\nu)$ be a valued field and $L$ be a finite and simple extension of $K$. Write $\mu^{(1)},...,\mu^{(g)}$ the different extensions of $\nu$ on $L$, its corresponds to a (pseudo-)valuation of $K\left[ x\right]$ denoted by the same way. Consider $\lbrace Q_l^{(i)} \rbrace_{l\in\Lambda^{(i)}}$ the set of key polynomials associated to $\mu^{(i)}$ and $n_0^{(i)}\in\mathbb N^*$ the smallest possible such that $\Lambda^{(i)}\leqslant \omega n_0^{(i)}$, $1\leqslant i\leqslant g$. Then:
\[d_{L\vert K}(\mu^{(i)},\nu)=\prod\limits_{j=1}^{n_0^{(i)}}d_{\omega j}^{(i)}.\]
We deduced that:
\[ \left[ L:K\right]=\sum\limits_{i=1}^ge_if_id_{\omega}^{(i)}d_{\omega 2}^{(i)}...d_{\omega n_0^{(i)}}^{(i)},\]
where $e_i=\left[\Gamma_{\mu^{(i)}}:\Gamma_\nu\right]$, $f_i=\left[k_{\mu^{(i)}}:k_\nu\right]$, $d_{\omega j}^{(i)}=\delta_{\omega j}^{(i)}$ for $j<n_0^{(i)}$ and:
\[d_{\omega n_0^{(i)}}^{(i)}=\left \{ \begin{array}{ccl}  \delta_{\omega n_0^{(i)}}^{(i)} & \textup{si} & \Lambda=\omega n_0^{(i)}\textup{ et } \sharp\lbrace m\geqslant 1\:\vert\:\alpha_{\omega(n_0^{(i)}-1)+m}^{(i)}=1\rbrace=+\infty \\  1  & \textup{si} & \Lambda<\omega n_0^{(i)} \textup{ ou } \Lambda=\omega n_0^{(i)}\textup{ et } \sharp\lbrace m\geqslant 1\:\vert\:\alpha_{\omega(n_0^{(i)}-1)+m}^{(i)}=1\rbrace<+\infty \end{array} \right.\]
\end{thm}
\begin{rem}
\textup{Denote by $p=char(k_\nu)$. By a suggestion of G. Leloup, since $d_{\omega j}^{(i)}\geqslant p$ for $j\in\{1,...n_0^{(i)}-1\}$ and $d_{\omega n_0^{(i)}}^{(i)}\geqslant 1$, we have:
\[d_{L\vert K}(\mu^{(i)},\nu)\geqslant p^{n_0^{(i)}-1}.\]
We deduce that:
\[n_0^{(i)}\leqslant\log_p\left(d_{L\vert K}(\mu^{(i)},\nu)\right)+1.\]
This result brings more precision than the inequality given in \cite{spivaherrera}.}
\end{rem}
From now until the end of this section we will use the notations of the previous section.
\begin{thm}\label{thmprinc}
Consider the same assumptions as Theorem \ref{defautdelta} and assume more that $L\vert K$ is a normal extension. Then, for all $i\in\{1,...,g\}$, we have:
\[\vert G^r\vert = p^{s+t-l}\times\prod\limits_{j=1}^{n_0^{(i)}}d_{\omega j}^{(i)}.\]
\end{thm}
\noindent\textit{Proof}: By 2. of Corollary \ref{corogaldef}:
$$\vert G^r\vert=p^{s+t-l}\times d_{L\vert K}(\mu^{(i)},\nu).$$
To conclude it is sufficient to apply Theorem \ref{defautdelta}.\\\qed
\begin{coro}
With the same assumptions of Theorem \ref{thmprinc}, $d_{L\vert K}(\mu^{(i)},\nu)=1$, for all $i\in\{1...,g\}$, if and only if $\vert G^r\vert = p^{s+t-l}$.
\end{coro}
\indent In some cases, we can express the order of the ramification group only in terms of key polynomials, essentially with the degree of the key polynomials and the effective degree.
\begin{coro}
Consider the same assumptions as Theorem \ref{defautdelta} and assume more that $L\vert K$ is a Galois extension. Assume that $K^r=K^d$ or equivalently $G^r=G^i=G^d$. Then, for all $i\in\{1,...,g\}$, there exists an index $i_0\in\Lambda^{(i)}$ having a predecessor, such that:
\[\vert G^r\vert = d^{\:\circ}\left(Q_{\omega(n_0^{(i)}-1)+i_0}^{(i)}\right)\times d_{\omega n_0^{(i)}}^{(i)}.\]
\end{coro}
\noindent\textit{Proof}: The extension $L\vert K$ is Galois then $l=0$. The assumption $K^r=K^d$ is equivalent to $e_0=f_0=1$. Thus $e=p^t$, $f=p^s$ and, by Theorem \ref{thmprinc}: $$\vert G^r\vert=\left(e\times f\times \prod\limits_{j=1}^{n_0^{(i)}-1}d_{\omega j}^{(i)}\right)d_{\omega n_0^{(i)}}^{(i)}.$$ But in the proof of Corollary 4.3 of \cite{jcdefaut}, we have seen that, as a consequence of Proposition 2.9 of \cite{vaquie3}, there exists an index $i_0\in\Lambda^{(i)}$ having a predecessor, such that:
$$d^{\:\circ}\left(Q_{\omega(n_0^{(i)}-1)+i_0}^{(i)}\right)=e\times f\times\prod\limits_{j=1}^{n_0^{(i)}-1}d_{\omega j}^{(i)}.$$\qed
\begin{thm}\label{pasdepolycle}
Consider the same assumptions as Theorem \ref{thmprinc} and assume more that $rk(\nu)=1$. If $\vert G^r\vert = p^{s+t-l}$ then the set of key polynomials $\lbrace Q_l^{(i)} \rbrace_{l\in\Lambda^{(i)}}$ associated to $\mu^{(i)}$ have no limit key polynomials, ie: $\Lambda^{(i)}\subseteq\mathbb N^*$.
\end{thm}
\noindent\textit{Proof}: If $\vert G^r\vert = p^{s+t-l}$ then, by definition, $n_0^{(i)}=1$ and $d_\omega^{(i)}=1$. We have three possibilities:
\begin{enumerate}
\item $\Lambda^{(i)}<\omega$, we have nothing to proove;
\item $\Lambda^{(i)}=\omega$ and $\sharp\lbrace m\geqslant 1\:\vert\:\alpha_{m}^{(i)}=1\rbrace<+\infty$, we conclude with the Proposition 3.19 of \cite{jcdefaut};
\item $\Lambda^{(i)}=\omega$ and $\sharp\lbrace m\geqslant 1\:\vert\:\alpha_{m}^{(i)}=1\rbrace=+\infty$, we conclude with the Proposition 3.20 and the Proposition 3.18 of \cite{jcdefaut} because $\delta_\omega^{(i)}=d_\omega^{(i)}=1$.
\end{enumerate}\qed
\begin{rem}
\textup{In the situation of the Theorem \ref{pasdepolycle}, the field is defectless and we can also apply directly the Proposition 5.1 of \cite{jcdefaut}.}
\end{rem}
\section{Ramification and local uniformization}
\indent Let $(R,\mathfrak{m},k)$ be a local complete regular equicharacteristic ring of dimension $n$ with $\mathfrak{m}=\left(u_1,...,u_{n}\right)$. Let $\nu$ be a valuation of $K=Frac(R)$, centered on $R$, of value group $\Gamma$ and $\Gamma_1$ the smallest non-zero isolated subgroup of $\Gamma$. Write:
\[H=\lbrace f\in R\:\vert\: \nu(f)\notin\Gamma_{1}\rbrace.\]
$H$ is a prime ideal of $R$ (see the proof of Theorem 6.2 of \cite{jcdefaut}). Moreover suppose that:
\[n=e(R,\nu)=emb.dim\left(R/H\right),\]
 that is to say:
 \[H\subset\mathfrak{m}^2.\]
Write $r=r(R,u,\nu)=\dim_\mathbb Q\left(\sum\limits_{i=1}^n\mathbb Q\nu(u_i)\right)$.
\\ The valuation $\nu$ is unique if $ht(H)=1$; it is the composition of the valuation $\mu:L^{*}\rightarrow\Gamma_{1}$ of rank $1$ centered on $R/H$, where $L=Frac(R/H)$, with the valuation $\theta :K^{*}\rightarrow \Gamma / \Gamma_{1}$, centered on $R_{H}$, such that $k_{\theta}\simeq \kappa(H)$.
\\By abuse of notation, for $f\in R$, we will denote by $\mu(f)$ instead of $\mu(f\mod H)$.
By the Cohen's theorem, we can suppose that $R$ is of the form:
\[R=k\left[ \left[ u_{1},...,u_{n}\right] \right].\]
For $j\in \lbrace r+1,...,n\rbrace$, write $\lbrace Q_{j,i}\rbrace_{i\in\Lambda_{j}}$ the set of key polynomials of the extension $k\left( \left( u_{1},...,u_{j-1}\right) \right)\hookrightarrow k\left( \left( u_{1},...,u_{j-1}\right) \right)(u_{j})$, $\textbf{Q}_{j,i}=\left\lbrace Q_{j,i'}\vert i'\in\Lambda_{j},i'<i\right\rbrace $, $\Gamma^{(j)}$ the value group of $\nu_{\vert k\left( \left( u_{1},...,u_{j}\right) \right)}$ and $\nu_{j,i}$ the $i$-troncation of $\nu$ for this extension.
\\ \\\indent For the definition of local framed sequences, one may consult D\'efinition 7.1 and the sections 4.1 and 4.2 of \cite{jccar0}.
\begin{thm}\label{thmeclatformcar0}
Suppose that, for $R_{n-1}=k\left[ \left[ u_{1},...,u_{n-1}\right] \right]$ we have:
\begin{enumerate}
\item
\begin{enumerate}
\item Or $H\cap R_{n-1}\neq (0)$ and there exists a local framed sequence $(R_{n-1},u)\rightarrow (R',u')$ such that: $$e(R',\nu)<e(R_{n-1},\nu);$$
\item Or $H\cap R_{n-1}=(0)$ and for all $f\in R_{n-1}$, there exists a local framed sequence $(R_{n-1},u)\rightarrow (R',u')$ such that $f$ is a monomial in $u'$ times a unit of $R'$.
\end{enumerate}
\item The local framed sequence $(R_{n-1},u)\rightarrow (R',u')$ of (1) can be chosen defined over $T$.
\end{enumerate}
Moreover suppose that the ramification group of the extension $k\left( \left( u_{1},...,u_{n-1}\right) \right)\hookrightarrow k\left( \left( u_{1},...,u_{n-1}\right) \right)\left[u_n\right]/H$ have order $p^{s+t-l}$ with $p^s=[k_{\mu}:k_{\mu^r}]$, $p^t=[\Gamma_{\mu}:\Gamma_{\mu^r}]$ and $p^l$ the inseparable degree of the extension. Then the assumptions 1. and 2. are true with $R$ instead of $R_{n-1}$.
\end{thm}
\noindent\textit{Proof}: The proof is the same as the proofs on Theorem 5.1 and 7.2 of \cite{jccar0}. With the assumptions of the Theorem \ref{thmeclatformcar0}, we can use the Proposition 5.2 of \cite{jccar0}: $H$ is generated by a irreducible monic polynomial in $u_n$. Since the order of the ramification group of the extension $k\left( \left( u_{1},...,u_{n-1}\right) \right)\hookrightarrow k\left( \left( u_{1},...,u_{n-1}\right) \right)\left[u_n\right]/H$ is $p^{s+t-l}$, by Theorem \ref{pasdepolycle}, the set of key polynomials $\lbrace Q_{j,i}\rbrace_{i\in\Lambda_{j}}$ has not limit key polynomial. To conclude it is sufficient to apply Theorem 7.2 of \cite{jccar0}.\\\qed
\begin{rem}
\textup{In \cite{jcthese} and \cite{jccar0} we saw that the problem of local uniformization is reduced to monomialize the first limit key polynomial $Q_\omega\in K[X]$. where $(K,\nu)$ is valuated field with $rk(\nu)=1$ and we have a local uniformization property on $K$. We know that we can suppose that:
$$Q_\omega=X^{p^e}+\sum\limits_{i=0}^{e-1}c_{p^i}X^{p^i}+c_0,$$
where $p^e$ is equal to the defect. In this situation, if we write $L=K[X]/(Q_\omega)$, we have for the extension $L\vert K$, $e=f=1$. This extension have defect and if we take a Galois closure of $L$, we obtain that the order of $G^r$ is exactely the defect. So, if we want to investigate a way to obtain a local uniformization theorem studying the ramification group of an extension, this the only situation which is problematic.}
\end{rem}

\bibliographystyle{plain}


Shannon Robson received a Scientist Development Grant from the American Heart Association to investigate the impact of a low energy dense diet in children. This type of diet includes foods with more water, fiber and air, and less fat. Such foods include fruits and vegetables. With the help of UD nutrition students, the Behavioral Health and Nutrition assistant professor is comparing a low energy dense diet to the standard MyPlate recommendations put out by the United States Department of Agriculture. Read the full article on UDaily.


The newest publication from local design team Tin Can Knits, Road Trip features 14 knitting patterns that are inspired by the open road. The collection contains patterns for toques, sweaters, cowls, mittens, and even a blanket - all the knitted things you might need when venturing out on a road trip of your own!
As is typical Tin Can Knits, the patterns are sized from baby to adult, and shown off to great advantage with lots of beautiful photographs. A digital download of the patterns is also included with the purchase of each book - just in case you want to take your patterns on the road with you, and not have to worry about damaging the printed book!


List of Dursley based agencies offering home, carpet & upholstery, office & window cleaning services.
At Dedman Contract Services, you'll find that we always try harder than most to do the simple things as well as the most exacting jobs for our customers. This means that we never fail to meet all of our cleaning service commitments - at competitive prices, too! Dedman Contract Services can handle any industrial or commercial cleaning contract large or small, regional or nationwide.


Next 5 Dual Champions Preview!
Dragon Front now available on PC for Facebook Gameroom!
Conquest Single Player Mode NOW AVAILABLE in Dragon Front!
Update as of July 25th-- Conquest battle nerfs and Daily Logins!
Upcoming Dragon Front 2.0 update--Postponed--Single Player Conquest Mode!
Update 4/18- New Masks, Portals appear, and the possibility of deep discounts!
Cannot forge spell card "Flourish"
Double XP Weekend! Feb 3- Feb 5!
Double XP this weekend, Double Gold next weekend, and New Dragon Front Content coming in January!


text:
Officer crashes, dies during California chase
Authorities say a police officer was killed when he lost control of his squad car and hit a palm tree as he chased a parolee driving a stolen car in California.
PALM SPRINGS, Calif. – Authorities say a police officer was killed when he lost control of his squad car and hit a palm tree as he chased a parolee driving a stolen car in California.
Cathedral City police said in a statement Saturday that 29-year-old Officer Jermaine Gibson's vehicle caught fire and he was declared dead at the scene Friday night.
Gibson was a Marine Corps veteran who earned a Purple Heart serving in the Iraq War. He had been an officer in Cathedral City for 18 months and had a 4-week-old baby.
Governor Jerry Brown issued a statement praising Gibson's sacrifice.
Police said that in Palm Springs, two suspects, one of them a parolee, hit another palm tree in a Ford Mustang that had been reported stolen. They were hospitalized with what police called major injuries.
Senator: Deactivating cellphones can deter theft
Tour bus hits truck on Pa. highway, driver killed
Sister of slain Las Vegas dancer curses defendant
Florida Doctor's Terrorism Conviction is Upheld
Wife defends guardsman charged in Afghan murder
Lawyer: Manic episode in NY hotel castration death

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: transform xml with xslt in swift I need to transform a xml string with xslt in Swift.
I am an android developer, so I am trying to find some code to do that in iOS with Swift.
I found this post Version of XSLT in iPhone
and this post How to process an XMLDocument using EXSLT or XSL 2 in Swift?
The second one seems most recent.
In particular what I need is a function (with 2 parameters, a xml string and a path for saving new file) that transforms xml, saves file on disk and returns the string transformed.
I know it's a bad help request, but I am not able to learn to program in Swift in short times.

redpajama_set_name: RedPajamaStackExchange


With more than 200,000 users in 180-plus countries, Kashoo is simple cloud accounting software for small businesses on the web and on the App Store (the most downloaded accounting iPad app), but it's more than just that. Kashoo is peace of mind. It's control. It's your business's financial health at your fingertips, whether you're working from your home office or meeting with a customer. Kashoo lets you run your business on your terms. Get started at kashoo.com or download the app for iOS and Android.
What to expect from white-label partnerships—true stories!
Chuck Clark is the CTO for Kashoo Cloud Accounting, the SaaS company behind the most downloaded iPad accounting application in the iTunes App Store. While delivering outstanding customer experiences over the last 20 years for sites like Orbitz, American Airlines, and Cheaptickets, Chuck routinely found himself as a developer hanging out with the operations team. As a technologist with a belief in the principles behind agile software development, it was a logical leap to embrace the devOps philosophy.
Prior to Kashoo, Chuck led product development at Vancouver-based Pulse Energy and before that was the chief architect for Orbitz Worldwide, based in Chicago, Illinois. When he's not in front of a keyboard, he can be found with his family, either on a bike or supporting his adopted home team, the Vancouver Whitecaps.
ayondo hires Peg Reed as Senior Vice President of Business Development.
School of Bookkeeping reviews Kashoo Simple Cloud Accounting.
Fintech Monthly focuses on recent fintech fundraising including recent investment in TransferWise.
ProfitStars adds billpay, invoice management, and mobile banking functionality to its ICBA Preferred Service Provider program.
Temenos teams up with Polytech Ventures to launch the first Swiss fintech incubator.
Expensify adds 4,000 new customers in 2014, including the BBB, ZenDesk, and Yahoo!
Coinbase introduces Google Now cards.
Forbes features Currency Cloud's cross border payment API.
Kony's partners now offering apps on its business mobile app marketplace.


"I'm proud to serve on a board with members who are dedicated to sharing their time and expertise to continuously improve fairs and festivals in the great State of Michigan," said Micheau.
Micheau fills a three-year term for the MAFE director-at-large position on the 11-member board with representatives from across the State. Ric Crawford, Shiawassee County Fair, did not seek reelection after serving in the position for nine years.
The Michigan Association of Fairs and Exhibitions promotes the development of agriculture and related industries, coordinates the objectives of fairs and exhibitions across the state, informs and educates its members through meetings and an annual convention and encourages participation of youth in these endeavors.


St. Petersburg Locksmith company is offering our customers 24 hour locksmith services. Here at St. Petersburg Locksmith Guys, we are equip to handle many different types of jobs. Call us today and join our team of happy customers at 727-674-1809.
St. Petersburg Locksmith - We are a 24 hour full mobile locksmith service, Call us anytime of the day for your locksmith services, at 727-674-1809.
They work hard and quickly to install new locks for me. The locks are quality locks and they installed them perfectly. Thanks so much for all the help!
This is a good company, I use them often and suggest them to friends. I like this company alot.There very professional. I used them twice and will use them again.
The locksmith that came out to help me was so friendly and polite. I really appreciate that he handled my property like it was his own. He was really respectful towards my door. He was able to install a new set of locks without damaging the door.
Posted by jmichaels_1 on January 16, 2015. Brought to you by localcom.
I was able to get a rekey & a lock change done all in meeting. They didn't waste time by having to go back & forth from the office, they were ready to work as soon as they knocked on my door.
Posted by by dollarsaywhat1 at Citysearch on September 26, 2011. Brought to you by Citysearch.


For the first time since 2009, H&R Block has bought a 60-second Super Bowl spot – just in time for tax prep.
The big buy follows a soft 2016 tax season when tax returns filed using H&R Block's service declined 4.5% due to competitors like Intuit's TurboTax, which "took share from all major branded competitors," CEO Bill Cobb noted during an earlier earnings call.
H&R Block is also partnering with IBM Watson to improve its in-store tax prep process.
Collins spoke with AdExchanger about H&R Block's investment in TV, Watson and how it's getting more targeted with consumers.
AdExchanger: What are you doing with IBM Watson?
KATHY COLLINS: We've worked hard over the last seven or eight months to do more research around what consumers were looking for in terms of tax prep. A lot of consumers were saying, "We just want to get our taxes done," which is what one of our competitors says. We thought there was more to it.
We had this moment where my boss [Bill Cobb] and our VP of advertising were sitting in a room with our ad agency and this whole idea of artificial intelligence came up. We called the folks at IBM and they were fascinated by this idea of man and machine, which was so similar to what we're all about – having the tax professional at the center of everything we do.
Instead of sitting across from the tax pro and watching them input your data, the client has their own monitor and Watson is built into that. It's not so much that the content is different from what the tax pro can offer, but Watson can scan hundreds of millions of data points and find every credit and deduction for that client in real time.
If, for instance, there's a farmer from Iowa as a client, Watson can look for other farmers from Iowa and prompt the tax pro to ask additional questions to ensure there aren't other deductions they can claim. And the client can watch in real time.
Did Watson influence your decision to invest in a Super Bowl ad?
We knew we had a huge opportunity. We were transforming the category and the timing was perfect. The beginning of February works very well for getting this kind of message out and in what better venue than the Super Bowl? We, of course, hope to get people's attention, but it's really more strategic to us. We're thinking two to three years out.
So the Super Bowl is an audience reach play for you, but what about digital?
Traditional media still works very well for us, to be honest. TV and radio are very important to our mix. We, however, invest a great deal in digital media and have gotten a lot savvier this year becoming more targeted in our digital video and banner ads.
We have slightly different businesses. We have an online, do-it-yourself product and we have a promotion there targeting the DIY'er. We also have retail offers we've had out there like the refund advance. And we have a free, federal 1040EZ product, which targets someone with a fairly simple federal return. We have many audiences, and some trend younger and a little more digitally savvy.
We know where they're browsing and we're able to be a little more targeted because of it. We do see a big spike in some of our social activity around our [TV ads], so our digital investment is really important to us.
What are you getting better at in digital?
Targeting acquisition or retention-based clients based on where they are online. If we know they're already an H&R Block client and what they're browsing for online, we can target them differently than someone who's never come to our site before.
That data is power and then you think about, on the retail side of our business, we completed over 700 million tax returns in our history and we have a lot of data on people.
And of course, we're very protective of that data, but we do know who they are, when they like to file, if they're a retail client or DIY client, so that even with things like email, direct mail and social, we can be very targeted based on who they are.
"Get your taxes won" (H&R Block's campaign featuring Jon Hamm) and the partnership with Watson are a long-term play. It's very strategic. It's not a promotion for us where we'll just measure to see if we got eyeballs. In terms of our advertising, we track it on a continuum across likability, memorability, brand and message recall, but to your point about online and digital – we can follow those people to see if it's working.
We can see if they're coming to our website and then converting. We can see if they click on an ad and then go to a store versus registering online as a client. All of those metrics we roll into one – traditional, brand metrics, but also conversion-based metrics – to make sure we're doing more than just entertaining people with our ads.


TUESDAY, JUNE 26, 2012, 6:00 P.M.
The Urban Forestry Council's Regular Meeting of Tuesday, June 26, 2012, at 6:00 p.m. has been CANCELLED. The next regular meeting is scheduled for Tuesday, August 28, 2012 at 6:00 p.m. and will be held at City Hall, Room 416, One Dr. Carlton B. Goodlett Jr. Place, San Francisco, CA 94102. An agenda will be prepared and distributed at least 72 hours prior to the next scheduled meeting. For additional information or to be placed on the agenda mailing list, please contact Monica Fish, Commission Secretary, at (415) 355-3709 or by email [email protected] or Mei Ling Hui, Urban Forestry Council Coordinator at (415) 355-3731 or by email at [email protected]. Agendas may also be accessed on the Urban Forestry Council's website http://www.sfenvironment.org/about/taskforce/urban-forestry-council/agendas.


Ever wonder what it would be like to walk across a high wire in a circus? Or how a performer is able to swallow a sword without hurting their mouth and throat?
Photo courtesy of Telus Spark.
Telus Spark's Circus! Science Under the Big Top is on from February 2 to June 9 and allows you to look inside the science behind the human body's ability to withstand such perceived torture. You'll also be able to smell the circus smells and talk the circus talk. With 20 displays to choose from, you might want to come back and experience this exhibit more than once.
Dornian says it will be fun to try some flips while strapped into a harness in a safe way. The whole exhibit is hands on, and while you might not be able to swallow a sword yourself—which is probably for the best—you will be able to learn about how the human body works to allow itself to carry out a most unlikely task. Here are five experiences to check out at Circus! Science Under the Big Top.
Just what is involved in a flea circus, anyway? This exhibit allows you to look at the inner workings of a flea and shows you how a flea circus actually works. Flea circuses are tricks of the imagination and in this flea circus exhibit, you'll be able to take a peek behind the curtain at the scientific workings of a flea circus. No need to worry, there are no live fleas at Telus Spark.
If we want to make our thoughts known, we will open our mouths and speak to each other. But how do animals communicate, if they don't have a voice? In this experience, you will get to listen to animal sounds and learn how animals communicate with their mobilization and actions. And while there are no live animals in the exhibit, kids will be able to dress up like circus themed animals and put on a performance for their parents.
Enliven those memories of circus experiences by nosing circus smells. You'll be able to get a whiff of your favourite circus foods like cotton candy and popcorn. But don't forget those animals — you'll also have the chance to take in those famous circus animal smells at Circus! Science Under the Big Top.
Ouch! Don't worry about pricking yourself at the end of a sword tip — this exhibit is totally safe.
Even though you won't experience this one on a hands-on level, you'll get to see an x-ray version of how a professional sword swallower is able to swallow a sword. Circus professionals spend a lot of time training and practicing, and some things are best left to the people with the practice.
What is the connection between clowns and science? Clowns are good at making people laugh and in this exhibit, you can learn about the science behind laughter and the role clowns play in creating that laughter. Either love or hate them, there will be no live clowns at Circus! Science Under the Big Top.


It was all for Steve.
Everything Bucky did was so that Steve would be happy.
Jack and Mark have been living together for a couple of months, and for several of those months, Jack's loved Mark, who seemingly doesn't reciprocate his feelings. His unreciprocated feelings just serve to fuel his depression.
Mark is worried. Jack isn't smiling very much lately, and he's scared Jack will do something drastic.
You and Mark are dating and you live together in your suburban house. At the moment, Jack is staying with you two.
Life feels so perfect, and you want to stay like this forever, but nothing ever lasts very long. Especially not the good things.
This is me writing out the animation series in a story form. Longer episodes will be in multiple parts because they'll take longer to write them out.
If there's typos, please let me know.
Roman's first mistake was taking Mr. Sanders History class, expecting an easy ride. His second mistake?
Deceit never can resist flirting with the other sides. The only problem is when they flirt back.


text:
DHC-3 Turbine Otter (N725TH) c...
InvestigationsCurrently selected
Accident Dockets
Aviation Database
Investigative Detail Image
An image of the wreckage on the seafloor, computer processed from data taken on September 10 by the University of Washington's vessel using 3D synthetic aperture sonar system.
DHC-3 Turbine Otter (N725TH) crash
Investigation Details
We are investigating the DHC-3 Turbine Otter (N725TH) crash in Mutiny Bay, WA that happened on 9/4/2022.
Update 10/24/2022: Investigative Update released
Update 9/27/2022: Recovery Operations Begin for Mutiny Bay Plane Crash
Update 9/16/2022: We have released the preliminary report on this investigation.
Update 9/12/2022: NTSB has located the wreckage of the DHC-3 Turbine Otter that crashed off Whidbey Island, Washington on September 4. NTSB coordinated with the National Oceanic and Atmospheric Administration (NOAA) and the University of Washington's Applied Physics Laboratory to use side scan sonar, multibeam sonar, and 3D instruments to locate the wreckage. The University of Washington's vessel scanned the area identified from the NOAA multibeam data. Using all available data, investigators concluded the targets they identified were from the aircraft.
Due to the depth of the water (approximately 190 feet) and the current (3-5 knots), the most suitable tool for recovery is a work class remotely operated vehicle (ROV). NTSB continues to be in communication with federal agencies and local companies to obtain a work class ROV.
An image of the wreckage on the seafloor taken on September 10 by the University of Washington's vessel using side scan sonar.
Member Thomas B. Chapman was the spokesperson on-scene.
Update 9/9/2022: The search is ongoing. Yesterday, NTSB and NOAA used multibeam and side scan sonar to map and survey the seafloor. The NOAA and NTSB team used the aircraft's last known location, time of entry, and tides and currents to narrow the search area. The team collected data for a 1.75 x 0.75-mile area. The mapping effort identified targets on the seafloor in the area we expected to find airplane debris. It has not been confirmed at this time if they are part of the aircraft. The next phase of the search is visual confirmation that the targets identified yesterday are or are not part of the aircraft. Due to the depth of the water (100-200 feet) and the current (3-5 knots), the most suitable tool for visual confirmation is a remotely operated vehicle (ROV). NTSB is coordinating with our federal partners and private companies to use an ROV in the search.
What We Recommended
Ongoing Investigation
Accident No
DCA22MA193
Accident Date
Mutiny Bay, WA
Family Assistance Contact
Sarah Quinn
assistance@ntsb.gov
Public Affairs Contact
NTSB Media Relations
Jennifer Gabris
A-22-023
AIR-22-08
NTSB Issues Urgent Aviation Safety Recommendation
WASHINGTON (Oct. 27, 2022) — The National Transportation Safety Board today issued an urgent safety recommendation to the Federal Aviation…
NTSB Identifies Possible Safety Issue in Mutiny Bay Plane Crash
Investigative Update Issued
WASHINGTON (Oct. 24, 2022) — National Transportation Safety Board investigators identified a possible safety issue with…


A collection of 19th century English framed plaster reliefs depicting images from the grand Tour as well as a circular portrait of the Holy Family and one other depicting William Pitt, the Elder. These miniatures are by the sculpture/artist John Henning who was responsible for the fine plaster frieze at the Athenaeum Club together with a frieze on a screen erected at Hyde Park Gate in the 19th century. He also created plaster reliefs of the Parthenon. There are a selection of miniatures here at Mary Wise Antiques and not all have been photographed. All are framed and for sale and details will be sent on request.
The pieces illustrated are all framed.


 GADIWA-3161,GADIWA Series,Power Supply,Industrial Computer- Portwell, Inc.
GADIWA-3161 is a wide input board type converter. It normally support 128 Watts and peak 160 Watts. GADIWA-3161 can save more space and cost, it's not only capability for fan-less system but also suitability for different application. Besides, the converter is made and tested by automatic production line; therefore, it can provide high quality and performance.


text:
Napomyza minuta este o specie de muște din genul Napomyza, familia Agromyzidae, descrisă de Spencer în anul 1981.

Este endemică în California. Conform Catalogue of Life specia Napomyza minuta nu are subspecii cunoscute.

Referințe

Napomyza

redpajama_set_name: RedPajamaWikipedia


text:
În tehnologia informației, accesul secvențial, spre deosebire de accesul aleatoriu, este o metodă de înregistrare și regăsire a informației, în care timpul necesar memorării sau citirii datelor depinde de poziția datei solicitate și a datei adresate anterior. E întâlnit frecvent la memoriile cu propagare și la cele cinematice. De exemplu, în cazul utilizării benzii magnetice datele sunt disponibile în ordinea secvențială, pe măsură ce banda se deplasează prin dreptul dispozitivului magnetic de scriere sau citire a informației.

Vezi și
 Memorie cu acces secvențial

Memorii
Date de calculatoare


text:
CONVICTED: A woman has been convicted and fined for an 'unusual' domestic violence order breach.
Woman fined for 'unusual' DVO breach
liana walker, liana.walker@gladstoneobserver.com.au
9th Aug 2020 12:00 PM
A GLADSTONE woman has been hit with a fine after she breached a domestic violence order under "unusual" circumstances.
The woman, 42, pleaded guilty in Gladstone Magistrates Court on Tuesday to contravention of a domestic violence order.
READ MORE: Sex toy thief caught with $4k worth of meth
READ MORE: Man claims he 'sarcastically' threatened to kill DV victim
The court was told the order had the condition the woman could not enter or attempt to enter the victim's house without prior consent.
The prosecution told the court police were called to an address on June 19 in relation to a disturbance where they located the defendant on the driveway of the victim's house.
Police confirmed with the victim the woman had not been provided consent to be there.
Defence lawyer Jun Pepito said the reason his client attended the property was because there had been a death in the victim's family.
He said his client had intended to attend the address to console him, but it didn't go that way.
Mr Pepito said both the defendant and the victim were emotional at the time and that was when the police became involved.
Magistrate John Milburn acknowledged the couple had a regime that was working.
"I do accept this was an unusual circumstance but it is of course a breach of the order," Mr Milburn said.
The woman was convicted and fined $350, a conviction was recorded.
Premium Content Dad sentenced to jail time as partner goes into labour
Actor charged with domestic violence
gladstonecourt gladstone domestic violence gladstone magistrates court


Oceana Gold's mine at Waihi in the central North Island has the green light for its underground Martha project, after no appeals were filed against its recent resource consent application.
Oceana chief executive Mick Wilkes said the permitting process had concluded with no appeals filed and the company had received its resource consents to start operations.
"The Waihi operation has had a long, rich history of operating to the highest of environmental and social standards globally while contributing significant socio-economic benefits to Waihi and the country.
"We're excited to be granted the consents and for the opportunity to continue over 140 years of gold mining history at Waihi," Mr Wilkes said in an ASX update yesterday.
Waihi employs up to 350 staff, including contractors. Earlier this week, Oceana outlined its capital spending programme for its four mines for calendar 2019.


Fluoride helps to prevent tooth decay by making your teeth more resistant to plaque. When combined with good dental habits, fluoride can dramatically reduce cavities and keep your mouth healthy. For most patients, we recommend fluoride treatments once or twice a year.


Sunbelt is scheduling interviews for a part time Physical Therapist in the Mission Viejo, California area to work the rest of the school year.
Sunbelt is currently looking to hire on Full time Physical Therapist at a school district near Princeton, NJ.
Sunbelt is currently looking to hire on Full time Physical Therapist at a school district near Pomona, CA.
I am a hiring manager looking to find a candidate for a Full Time Job as a Physical Therapist at a school district near Chino, CA.
I am a Hiring Manager working directly with a school district in Covina, CA to bring on a Physical Therapist in a School Based Job.
I am a Hiring Manager working directly with a school district in Covina, CA to bring on a PT to provide Physical Therapy in a School Based Job.
I am a Hiring Manager working directly with a school district near Banning, CA to bring on a PT to provide Physical Therapy in a School Based Job.
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text:
Midway industrial park plans push forward despite lawsuit
Gerald Witt
USA TODAY NETWORK – Tennessee
An Aug. 1 court date has been set for the lawsuit against the proposed Midway Road business park, but Knox County's plans to open the facility are slowly moving forward.
Todd Napier, president of the Knox County Development Corp., said all the legislative approvals needed to zone and plan the facility are in hand.
"We're continuing to be in the engineering phase, trying to finalize our work on engineering and design plans so we can ultimately get those in the position to be out for bid," he said.
That could happen as soon as September.
Meanwhile, the lawsuit against the proposed 345-acre site located by Thorngrove Pike and Midway Road is still in play.
"There's an Aug. 1 trial date, and a March 21 date, tentatively, for a hearing on the Development Corporation and Knox County's partial motion to dismiss," said Jason Long, an attorney representing residents who live near the business park.
Members of the 8th District Preservation Association, the group Long represents, have appeared numerous times at Knox County Commission and Metropolitan Planning Commission meetings to oppose rezoning and other action for the business park.
"Amongst my clients, I would say there's been no change in that they believe this development is out of character with the area it's located in," Long said. "It's not appropriate in any area."
The business park location is near the Interstate 40 interchange with Midway Road, one that's largely undeveloped, in East Knox County. Although it's 15 minutes from Knoxville's urban center, the area offers a bucolic scene of gently rolling hills about 3 miles north of the French Broad River.
Nearby residents have resisted development at the site for many years, successfully thwarting a past effort.
Commission in 2006 rezoned land for development of the business park. Residents sued and won. The Development Corp. made another shot at rezoning in 2010 but Commission rejected it.
In 2015 Knox County Mayor Tim Burchett pushed the park with a public relations campaign including a picnic barbecue dinner, a bus tour and two community meetings.
The Development Corp. also beefed up its rezoning request, which Commission approved in December 2015.
The latest lawsuit challenges how "substantial" the changes are, and called the rezoning a result of political bargaining.
"And there is a concern, given the number of sink holes, that the groundwater supply won't be protected, and it will be impossible, if not difficult, to protect the groundwater supply," Long said. "They don't believe that anything has changed in the neighborhood in the past 20 years that would justify this business park."
Bob Thomas, a Knox County Commissioner and mayoral candidate for 2018, has supported the project along with others, including Burchett and Commission Chairman Dave Wright.
Thomas also has been on the Development Corp. board for critical decisions on the business park. He said that they're beginning to spend "real money" on the park.
"I think that in the summer of '18 … utilities will be at the entrance of Midway, and in December of '18 Midway will basically be ready to go," Thomas said.
He said that the area is ready for such a development like Midway, and that nearby construction of subdivisions - and a new middle school in-progress for Gibbs - are proof.
The kinds of jobs Midway would attract would pay well, he added. And the park could one day have a medical facility to treat people in the area.
"Once the community sees what this is," he said, "it's going to be the nicest industrial park in the entire county. This is going to be very impressive."


text:
Mooney on Theatre
Contact MoT
Theatre that Caught Our Eye
Cheap Theatre in Toronto
Review: The Humans (Canadian Stage and Citadel Theatre)
February 10, 2018 Sam Mooney
Tony-winning play The Humans takes to the Toronto stage!
Stephen Karam's Tony award winning play, The Humans, opened at the Bluma Appel Theatre on Thursday evening. It's a comedy/drama, funnier earlier on and more dramatic later, about a family Thanksgiving dinner that unfolds in real time, at just under two hours.
We've probably all experienced special occasion family dinners like this. Mom, Dad, Grandma (in this case Momo), and the adult kids get together. There's joking and teasing and bickering and unsolicited advice and sometimes real tension. People talk over each other, little groups form, break apart, and form new groups; people laugh, sometimes they yell, sometimes they cry. They did in my family. And that's what my friend Patricia and I loved about The Humans; they seemed like a real family.
Diedre (Laurie Paton) and Eric (Ric Reid) Blake and Momo (Maralyn Ryan), Eric's mother, have driven from Scranton to New York to have Thanksgiving dinner with their daughter Brigid (Sara Farb) and her partner Richard (Richard Lee) in their new two level basement apartment in Chinatown. Aimee (Alana Hawley Purvis), their older daughter, has come from Philadelphia to join them.
But life isn't quite working out the way the Blakes expected. Momo is lost in dementia and Deidre and Eric care for her at home because they can't afford a care facility. They still have a mortgage and car payments.
Brigid is a composer, burdened with student debt, who works two bartending jobs under the table and collects unemployment. She's just found out why she hasn't been offered any of the fellowships she's applied for.
Aimee is a lawyer who has been dumped by her firm and her partner and has ulcerative colitis.
Richard, meanwhile, seems to be doing ok. He's older than Brigid, 38, a student who will inherit a trust fund when he's 40. His biggest issue is hoping that the Blakes will like him.
If it all sounds grim, it isn't. It's life in the post 9/11 American middle class.
The play shifts from funny to sad to funny to poignant to angry to funny very quickly. One minute Brigid and Deidre are sniping at each other and the next they're all holding hands and saying grace when Momo suddenly joins in. She joins in when they clap as well, and then picks up a serving spoon and prepares to eat from the serving dish.
This is very much an ensemble piece and the cast is terrific together. They pat a hand or arm in passing, there are rolled eyes and raised eyebrows off to the side, they interrupt each other, they anticipate each other's responses. Through it all, Deidre and Erik hint at a deeper issue, and we kept waiting for it to be revealed.
There were some moments that stood out for me. Purvis, as Aimee, phoning her ex-girlfriend; the one-sided phone conversation was heart breaking as she struggled to sound upbeat and not cry.
Ryan, as Momo, who spent most of the play in a wheelchair occasionally repeating the same sentence (which I could not really understand), becoming suddenly enraged, screaming and throwing things. I think that must be the rage that people with dementia can sometimes feel when they realized they're trapped in their body by their brain.
Paton, as Deidre, reading Momo's email, a tradition at family gatherings. The one she sent to Aimee and Brigid four years earlier when the dementia was getting serious, saying goodbye; telling them not to mourn her, to dance more than she did, to drink less than she did. It made me think of my grandmother telling me goodbye when she was at about the same stage of dementia.
There was one thing about the play that I'm not sure I liked. There were elements that I think were supposed to be supernatural. Banging on the ceiling that was initially explained away as the lady upstairs, an all enveloping noise that was attributed to the trash compactor, lights that flickered off and on and then finally just stopped working.
I didn't hate that part of it, but it didn't add anything for me. They all just seemed like things that could happen in a grungy apartment building. I'm not so much for supernatural.
I also thought the ending was really pushing a metaphor.
Having said that, I loved The Humans. In the same way that Judith Bowden's set took the front off the apartment so we could see inside, the play took the front off the Blakes so we could see inside them. Definitely worth seeing.
The Humans is playing until February 25 at the Bluma Appel Theatre (27 Front St East)
Showtimes are Tuesday through Thursday and Saturday at 8:00pm, Friday at 7:00pm, Wednesday, Saturday, and Sunday at 1:00pm
Tickets range from $39.00 to $99.00 and are available online, by phone at 416.368.3110, and in person at the box office.
Photo of Richard Lee, Maralyn Ryan, Sara Farb, Laurie Paton, Alana Hawley Purvis, and Ric Reid by Epic Photography
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text:
permalink: '/11ty-eleventy-introduction/'
title:  'Introduction to Eleventy (11ty)'
meta:
  description: This is a simple introduction to Eleventy (11ty), a popular static site generator.
type: Tutorial
date: 2021-10-03
tags:
  - Eleventy
  - Static Site Generators
---

{% include atoms/figure.html, src: "/images/posts/eleventy-introduction/homepage.png", alt: "Screenshot of Eleventy homepage", caption: "", source_title: "https://11ty.dev", source_link: "https://11ty.dev" %}

[Eleventy](https://www.11ty.dev/) (11ty) is a static site generator (SSG) built on top of [Node.js](https://nodejs.org/en/) that compiles static website assets (i.e., HTML files) using content inside of various "source" files (e.g., markdown, templates, JSON, etc.). Eleventy provides a platform that helps web developers create organized and [DRY](https://en.wikipedia.org/wiki/Don't_repeat_yourself) codebases through its support for [templating languages](https://www.11ty.dev/docs/languages/), a robust [templating engine](https://www.11ty.dev/docs/templates/), flexible [data models](https://www.11ty.dev/docs/data/), and [plugins](https://www.11ty.dev/docs/plugins/).

Eleventy is designed as a [zero-configuration](https://www.11ty.dev/docs/glossary/#zero-config) platform that provides as much or as little overhead as you would like. This preference for simplicity, combined with a philosophy of tooling agnosticism, offers a flexible and empowering developer experience.

Large websites won't only benefit from what Eleventy offers. Unless you are building the most straightforward low-maintenance websites, odds are that some aspect of Eleventy or another SSG will improve your workflow and codebase.

## SSG Ecosystem

Eleventy exists in an expansive ecosystem of SSGs, alongside [Next.js](https://nextjs.org/), [Jekyll](https://jekyllrb.com/), [Hugo](https://gohugo.io/), [Gatsby](https://www.gatsbyjs.org/), and [so many others](https://jamstack.org/generators/). In fact, Eleventy might seem like an underdog among these. However, its frequency of updates and thriving community provide me with confidence in using it for my projects.

I don't plan on covering the similarities and differences between different SSGs in this article. Instead, here's a list of resources you can consult if you're evaluating this type of software.

- [Defining metrics that help with static site generator evaluation](/evaluating-static-site-generators)
- [Questions to ask before choosing a static site generator](https://www.ample.co/blog/questions-to-ask-before-choosing-a-static-site-generator)
- [A List of Static Site Generators for Jamstack Sites](https://jamstack.org/generators/)
- [Jekyll alternatives: The benefits to JavaScript static site generators](https://www.takeshape.io/articles/jekyll-alternatives-the-benefits-to-javascript-static-site-generators/)
- [Comparing Static Site Generator Build Times](https://css-tricks.com/comparing-static-site-generator-build-times/)
- [11ty: The Good, the Bad, and the... Possum?](https://www.aleksandrhovhannisyan.com/blog/eleventy-the-good-the-bad-and-the-possum/)

## My Take

**I use Eleventy for _a lot_ of projects**

I've used different SSGs in the past, predominantly Jekyll and Hugo. While I enjoyed the ubiquity of Jekyll and the speed of Hugo, what keeps me coming back to Eleventy is its developer experience when used across a spectrum of web projects. About 90% of the projects I work on right now can be built with Eleventy, making it my go-to software for building websites of all shapes and sizes.

{% include type/p_large.html, content: "The most important advice I would give those choosing software for any project is to be mindful about what you're choosing and why." %}

Understand what problems you're solving, determine which of these are most important, then choose a piece of software that solves your most important problems in an ideal way.

**If you're choosing the right tool for the job or at least _trying_ to, that's what matters.**

{% include type/tip.html, content: "Don't buy into the hype of software X, paradigm Y, or language Z, and instead, focus on understanding your problem domain and using that as your north star." %}

## Getting Started

I plan on writing more about how to use Eleventy for web development projects, but you needn't wait for me to shill my articles when there are so many excellent resources available.

#### [Eleventy documentation](https://www.11ty.dev/docs/getting-started/)

Documentation is _always_ my first stop when learning new software. I start by seeing how the documentation advises I get started and then reach for external resources as needed.

Overall, I think the Eleventy documentation is great. The introductory "Getting Started" resource that I've included hits core concepts at a high level, making it suitable for folks who already have decent technical know-how with similar software.

#### [Learn Eleventy From Scratch](https://piccalil.li/course/learn-eleventy-from-scratch/)

This course—formerly paid but now free—is currently the best resource for learning how to use Eleventy from the ground up.

Taught by [Andy Bell](https://twitter.com/piccalilli_), this course teaches you the ins and outs of Eleventy without glossing over the more minor details.

#### [Beginner's Guide to Eleventy](https://www.tatianamac.com/posts/beginner-eleventy-tutorial-parti/)

This tutorial, written by [Tatiana Mac](https://twitter.com/TatianaTMac), is another excellent resource for anybody getting started with Eleventy. The first part of the series covers how Eleventy fits into the world of SSGs, while the second part goes incredibly deep on installing the software.


text:
Johnny Mathis ist der Name folgender Personen:

* Johnny Mathis (Countrysänger) ("Country" Johnny Mathis; 1933–2011), US-amerikanischer Countrysänger
 Johnny Mathis (Popsänger) (* 1935), US-amerikanischer Popsänger