text:
\section{Introduction}
Let $\Sigma_{g}$ denote a closed, oriented surface of genus $g$, and let ${\cal M}_{g}$ denote its mapping class group, which is the group of homotopy classes of orientation-preserving homeomorphisms.

The study of generators of ${\cal M}_{g}$ was pioneered by Dehn.
Dehn [De] proved that ${\cal M}_{g}$ is generated by a finite set of Dehn twists.
Lickorish [Li] showed that $3g-1$ Dehn twists generate ${\cal M}_{g}$.
This number was improved to $2g+1$ by Humphries [Hu].
Humphries proved, moreover, that in fact the number $2g+1$ is minimal;
i.e. ${\cal M}_{g}$ cannot be generated by $2g$ (or less) Dehn twists.

It is classical problem to find small generating sets and torsion generating sets for ${\cal M}_{g}$.
Maclachlan [Ma] proved that the moduli space is simply connected as a topological space by showing that ${\cal M}_{g}$ is generated by torsion elements.
McCarthy and Papadopoulos [MP] proved that ${\cal M}_{g}$ is generated by infinitely many conjugates of a single involution for $g\geq 3$.
Luo [Luo] discovered a first finite set of involutions which generate ${\cal M}_{g}$ for $g\geq 3$.
Luo posed the question of whether there is a universal upper bound, independent of $g$, for the number of torsion elements needs to generate ${\cal M}_{g}$.
Brendle and Farb answered Luo's question.
Brendle and Farb [BF] proved that ${\cal M}_{g}$ is generated by 3 elements of order $2g+2$, $4g+2$ and $2$ (or $g$).
More, Korkmaz [Ko] showed that ${\cal M}_{g}$ is generated by 2 torsion elements, each of order $4g+2$.
Brendle and Farb [BF] also constructed a generating set of ${\cal M}_{g}$ for $g\geq 3$ consisting of 6 involutions.
Kassabov [Ka] improved their method to show that ${\cal M}_{g}$ is generated by 4 involutions if $g\geq 7$, 5 involutions if $g\geq 5$ and 6 involutions if $g\geq 3$.

For all $g\geq 1$, the elements of order 2, 3 and 4 are in ${\cal M}_{g}$.
In the present paper, we construct a generating set of ${\cal M}_{g}$ consisting of elements of small order($>2$).
\begin{maintheorem}
For $g\geq 3$, ${\cal M}_{g}$ can be generated by 3 elements of order 3 and by 4 elements of order 4.
\end{maintheorem}

\section{Preliminaries}
Let $c$ be a simple closed curve on $\Sigma_{g}$.
Then the (right hand) Dehn twist $T_{c}$ about $c$ is the homotopy class of the homeomorphism obtained
by cutting $\Sigma_{g}$ along $c$, twisting one of the side by $360^{\circ}$ to the right and gluing two sides of $c$ back to each other. Figure~\ref{fig1} shows the Dehn twist about the curve $c$.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=8cm]{t.eps}
 \end{center}
 \caption{The Dehn twist}
 \label{fig1}
\end{figure}
We will denote by $T_{c}$ the Dehn twist about the curve $c$.

We recall the following lemmas and theorems.
\begin{lemma}\label{lem1}
For any homeomorphism $h$ of the surface $\Sigma_{g}$, the Dehn twists around the curves $c$ and $h(c)$ are conjugate in the mapping class group ${\cal M}_{g}$,
\begin{eqnarray*}
T_{h(\textit{c})}=hT_{\textit{c}}h^{-1}.
\end{eqnarray*}
\end{lemma}
\begin{lemma}\label{lem2}
Let $c$ and $d$ be two simple closed curves on $\Sigma_{g}$. If $c$ is disjoint from $d$, then
\begin{eqnarray*}
T_{c}T_{d}=T_{d}T_{c}
\end{eqnarray*}
\end{lemma}
\begin{lemma}\label{lem3}
If the geometric intersection number of $c$ and $d$ is one, then
\begin{eqnarray*}
T_{c}T_{d}T_{c}=T_{d}T_{c}T_{d}
\end{eqnarray*}
\end{lemma}
\begin{theorem}[Lickorish]
We denote the curves $\alpha_{i}$, $\beta_{i}$, $\gamma_{i}$ as shown in Figure~\ref{fig2}. Then ${\cal M}_{g}$ is generated by $T_{\alpha_{1}},\cdots,T_{\alpha_{g}}, T_{\beta_{1}},\cdots,T_{\beta_{g}}, T_{\gamma_{1}},\cdots,T_{\gamma_{g-1}}$.
\end{theorem}
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=7cm]{s1.eps}
 \end{center}
 \caption{Lickorish's generators}
 \label{fig2}
\end{figure}
We call $T_{\alpha_{1}},\cdots,T_{\alpha_{g}}, T_{\beta_{1}},\cdots,T_{\beta_{g}}, T_{\gamma_{1}},\cdots,T_{\gamma_{g-1}}$ Lickorish's generators.
\begin{theorem}[Humphries]
${\cal M}_{g}$ is generated by $T_{\alpha_{1}}$, $T_{\alpha_{2}}$, $T_{\beta_{1}},\cdots,T_{\beta_{g}}$, $T_{\gamma_{1}},\cdots,T_{\gamma_{g-1}}$.
\end{theorem}
We call $T_{\alpha_{1}}, T_{\alpha_{2}}, T_{\beta_{1}},\cdots,T_{\beta_{g}}, T_{\gamma_{1}},\cdots,T_{\gamma_{g-1}}$ Humphries's generators.

\section{Generating the mapping class group by 3 elements of order 3}
In this section we prove that the mapping class group ${\cal M}_{g}$ is generated by 3 elements of order 3.
We assume that $g\geq 3$.

\subsection{Construction of elements of order 3}
We construct two elements of order 3 by cutting and gluing surfaces.
We take the curves $x_{1}$, $x_{2}$ and the separating curve $\delta$ like Figure~\ref{fig a}.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=7cm]{s2.eps}
 \end{center}
 \caption{the curves $x_{1}$, $x_{2}$, $\delta$}
 \label{fig a}
\end{figure}
\subsubsection{Odd genus}
We assume that $g$ is odd.
We construct an element $f$ of order 3.

We cut $\Sigma_{g}$ along the curves $\gamma_{1}, \gamma_{2}, \alpha_{3}$ and $\alpha_{2k}, \gamma_{2k}, \alpha_{2k+1}$ $(k=2,\ldots, \frac{g-1}{2})$ to obtain $\frac{g-1}{2}$ surfaces $S_{1}$, $S_{2}$,\ldots, $S_{\frac{g-1}{2}}$ as shown in Figure~\ref{fig3}.
$S_{1}$ is a sphere with $\frac{3(g+1)}{2}$ boundary components and $S_{k}$ $(k=2,\ldots, \frac{g-1}{2})$ is a pair of pants bounded by $\alpha_{2k}, \gamma_{2k}, \alpha_{2k+1}$.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=8cm]{s3.eps}
 \end{center}
 \caption{Cutting the surface}
 \label{fig3}
\end{figure}

We embed $S_{1}$, $S_{2}$,\ldots, $S_{\frac{g-1}{2}}$ in $\mathbb{R}^{3}$ so that they are invariant under $\frac{2\pi}{3}$-rotations $f_{1}$, $f_{2}$,\ldots, $f_{\frac{g-1}{2}}$, respectively (cf. Figure~\ref{fig4}).
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=6cm]{s4.eps}
 \end{center}
 \caption{$\mathbb{Z}_{3}$-symmetry of $\Sigma_{g}$}
 \label{fig4}
\end{figure}

Since the homeomorphisms $f_{1}$, $f_{2}$,\ldots, $f_{\frac{g-1}{2}}$ coincide on the boundaries, they naturally define a homeomorphism $f : \Sigma_{g}\rightarrow \Sigma_{g}$ of order 3.
$f$ acts on the curves on $\Sigma_{g}$ as follows:
\begin{center}
\begin{tabular}{lcr}
$f^{2}(\gamma_{1})=f(\gamma_{2})=\alpha_{3},$& $f(\beta_{2})=\beta_{3},$& $$\\
$f^{2}(\alpha_{2})=f(x_{2})=x_{1},$& $$& $$\\
$f^{2}(\alpha_{2k})=f(\gamma_{2k})=\alpha_{2k+1},$& $f(\beta_{2k})=\beta_{2k+1}$& $(k=2,\ldots, \frac{g-1}{2}).$
\end{tabular}
\end{center}

\

We construct an element $h$ of order 3.

We cut $\Sigma_{g}$ along the curves $\alpha_{2j-1}, \gamma_{2j-1}, \alpha_{2j}$ $(j=1,\ldots, \frac{g-1}{2})$ and $\delta$ to obtain $\frac{g+3}{2}$ surfaces $S_{1}^{\prime}$, $S_{2}^{\prime}$,\ldots, $S_{\frac{g+1}{2}}^{\prime}$, $S_{\frac{g+3}{2}}^{\prime}$ as shown in Figure~\ref{fig5}.
$S_{1}^{\prime}$ is a sphere with $\frac{3g+1}{2}$ boundary components, $S_{j}^{\prime}$ $(j=1,\ldots, \frac{g-1}{2})$ is a pair of pants bounded by $\alpha_{2j-1}, \gamma_{2j-1}, \alpha_{2j}$ and $S_{\frac{g+3}{2}}^{\prime}$ is a torus bounded by $\delta$.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=8cm]{s5.eps}
 \end{center}
 \caption{Cutting the surface, I\hspace{-.1em}I }
 \label{fig5}
\end{figure}

We embed $S_{1}^{\prime}$, $S_{2}^{\prime}$,\ldots, $S_{\frac{g+1}{2}}^{\prime}$ in $\mathbb{R}^{3}$ so that they are invariant under $\frac{2\pi}{3}$-rotations $h_{1}$, $h_{2}$,\ldots, $h_{\frac{g+1}{2}}$, respectively (cf. Figure~\ref{fig6}).
We define that $h_{\frac{g+3}{2}}=(T_{\beta_{g}}T_{\alpha_{g}})^{2}$.
Note that $h_{1}^{3}=T_{\delta}^{-1}$, $h_{\frac{g+3}{2}}^{3}=T_{\delta}$ and $h_{\frac{g+3}{2}}(\alpha_{g})=\beta_{g}$.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=6cm]{s6.eps}
 \end{center}
 \caption{$\mathbb{Z}_{3}$-symmetry of $\Sigma_{g}$, I\hspace{-.1em}I}
 \label{fig6}
\end{figure}

Since the homeomorphisms $h_{1}$, $h_{2}$,\ldots, $h_{\frac{g+1}{2}}$, $h_{\frac{g+3}{2}}$ coincide on the boundaries, they naturally define a homeomorphism $h:\Sigma_{g}\rightarrow \Sigma_{g}$ of order 3.
$h$ acts on the curves on $\Sigma_{g}$ as follows:
\begin{center}
\begin{tabular}{lcr}
$h(\alpha_{g})=\beta_{g},$& $$& $$\\
$h^{2}(\alpha_{2j-1})=h(\gamma_{2j-1})=\alpha_{2j},$& $h(\beta_{2j-1})=\beta_{2j}$& $(j=1,\ldots, \frac{g-1}{2}).$
\end{tabular}
\end{center}

\subsubsection{Even genus}
We assume that $g$ is even.
By the similar arguments of the case of odd genus we construct $f$ and $h$ which are order 3.

We cut $\Sigma_{g}$ along the curves $\gamma_{1}, \gamma_{2}, \alpha_{3}$, $\delta$ and $\alpha_{2k}, \gamma_{2k}, \alpha_{2k+1}$ $(k=2,\cdots, \frac{g-2}{2})$ to obtain $\frac{g}{2}$ surfaces $S_{1}$, $S_{2}$,\ldots, $S_{\frac{g-2}{2}}$, $S_{\frac{g}{2}}$ as shown in Figure~\ref{fig b}.
$S_{1}$ is a sphere with $\frac{3g+2}{2}$ boundary components, $S_{k}$ $(k=2,\cdots, \frac{g-2}{2})$ is a pair of pants bounded by $\alpha_{2k}, \gamma_{2k}, \alpha_{2k+1}$ and $S_{\frac{g}{2}}$ is a torus bounded by $\delta$.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=8cm]{s7.eps}
 \end{center}
 \caption{Cutting the surface, I\hspace{-.1em}I\hspace{-.1em}I}
 \label{fig b}
\end{figure}

We embed $S_{1}$, $S_{2}$,\ldots, $S_{\frac{g-2}{2}}$ in $\mathbb{R}^{3}$ so that they are invariant under $\frac{2\pi}{3}$-rotations $f_{1}$, $f_{2}$,\ldots, $f_{\frac{g-2}{2}}$, respectively (cf. Figure~\ref{fig7}).
We define that $f_{\frac{g}{2}}=(T_{\beta_{g}}T_{\alpha_{g}})^{2}$.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=6cm]{s8.eps}
 \end{center}
 \caption{$\mathbb{Z}_{3}$-symmetry of $\Sigma_{g}$, I\hspace{-.1em}I\hspace{-.1em}I}
 \label{fig7}
\end{figure}

Since the homeomorphisms $f_{1}$, $f_{2}$,\ldots, $f_{\frac{g-2}{2}}$, $f_{\frac{g}{2}}$ coincide on the boundaries, they naturally define a homeomorphism $f:\Sigma_{g}\rightarrow \Sigma_{g}$ of order 3.
$f$ acts on the curves on $\Sigma_{g}$ as follows:
\begin{center}
\begin{tabular}{lcr}
$f^{2}(\gamma_{1})=f(\gamma_{2})=\alpha_{3},$& $f(\beta_{2})=\beta_{3},$& $$\\
$f^{2}(\alpha_{2})=f(x_{2})=x_{1},$& $f(\alpha_{g})=\beta_{g},$& $$\\
$f^{2}(\alpha_{2k})=f(\gamma_{2k})=\alpha_{2k+1},$& $f(\beta_{2k})=\beta_{2k+1}$& $(k=2,\cdots, \frac{g-2}{2}).$
\end{tabular}
\end{center}

\

We cut $\Sigma_{g}$ along the curves $\alpha_{2j-1}, \gamma_{2j-1}, \alpha_{2j}$ $(j=1,\cdots, \frac{g}{2})$ to obtain $\frac{g+2}{2}$ surfaces $S_{1}^{\prime}$, $S_{2}^{\prime}$,\ldots, $S_{\frac{g+2}{2}}^{\prime}$ as shown in Figure~\ref{fig c}.
$S_{1}^{\prime}$ is a sphere with $\frac{3g}{2}$ boundary components and $S_{j+1}^{\prime}$ $(j=1,\cdots, \frac{g}{2})$ is a pair of pants bounded by $\alpha_{2j-1}, \gamma_{2j-1}, \alpha_{2j}$.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=7cm]{s9.eps}
 \end{center}
 \caption{Cutting the surface, I\hspace{-.1em}V}
 \label{fig c}
\end{figure}

We embed $S_{1}^{\prime}$, $S_{2}^{\prime}$,\ldots, $S_{\frac{g+2}{2}}^{\prime}$ in $\mathbb{R}^{3}$ so that they are invariant under $\frac{2\pi}{3}$-rotations $h_{1}$, $h_{2}$,\ldots, $h_{\frac{g+2}{2}}$, respectively (cf. Figure~\ref{fig8}).
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=6cm]{s10.eps}
 \end{center}
 \caption{$\mathbb{Z}_{3}$-symmetry of $\Sigma_{g}$, I\hspace{-.1em}V}
 \label{fig8}
\end{figure}

Since the homeomorphisms $h_{1}$,$h_{2}$,\ldots, $h_{\frac{g+2}{2}}$ coincide on the boundaries, they naturally define a homeomorphism $h:\Sigma_{g}\rightarrow \Sigma_{g}$ of order 3.
$h$ acts on the curves on $\Sigma_{g}$ as follows:
\begin{center}
\begin{tabular}{lcr}
$h^{2}(\alpha_{2j-1})=h(\gamma_{2j-1})=\alpha_{2j},$& $h(\beta_{2j-1})=\beta_{2j}$& $(j=1,\cdots, \frac{g}{2}).$
\end{tabular}
\end{center}

\subsection{Generating the Dehn twist by 3 elemensts of order 3}
We generate the Dehn twist by 3 elements of order 3.
The basic idea is to use the \textit{lantern relation} which was discovered by Dehn and rediscovered by Johnson [Jo].
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=2cm]{sl.eps}
 \end{center}
 \caption{The Lantern Relation}
 \label{fig9}
\end{figure}

The \textit{lantern relation} is read as follows :
\begin{eqnarray*}
T_{\alpha_{1}}T_{\gamma_{1}}T_{\gamma_{2}}T_{\alpha_{3}}=T_{\alpha_{2}}T_{x_{1}}T_{x_{2}}.
\end{eqnarray*}
where the curves $\alpha_{1}$, $\alpha_{2}$, $\alpha_{3}$, $\gamma_{1}$, $\gamma_{2}$, $x_{1}$ and $x_{2}$ are shown in Figure~\ref{fig a} and Figure~\ref{fig9}.
Since $\alpha_{1},\gamma_{1},\gamma_{2}$ and $\alpha_{3}$ are disjoint each other and $\alpha_{2}, x_{1}$ and $x_{2}$, by Lemma~\ref{lem2} we can rewrite the relation as
\begin{eqnarray}\label{eq1}
T_{\alpha_{1}}=(T_{\alpha_{2}}T_{\gamma_{1}}^{-1})(T_{x_{1}}T_{\alpha_{3}}^{-1})(T_{x_{2}}T_{\gamma_{2}}^{-1}).
\end{eqnarray}
We can find that $f^{2}(\alpha_{2})=x_{1}$, $f^{2}(\gamma_{1})=\alpha_{3}$, $f(\alpha_{2})=x_{2}$ and $f(\gamma_{1})=\gamma_{2}$ from the argument of Section 3.1.
By using Lemma~\ref{lem1} we see that
\begin{eqnarray*}
(T_{x_{1}}T_{\alpha_{3}}^{-1})=f^{2}(T_{\alpha_{2}}T_{\gamma_{1}}^{-1})f^{-2}\\
(T_{x_{2}}T_{\gamma_{2}}^{-1})=f(T_{\alpha_{2}}T_{\gamma_{1}}^{-1})f^{-1}.
\end{eqnarray*}
Since $h$ maps $\gamma_{1}$ to $\alpha_{2}$, we see that
\begin{eqnarray*}
T_{\alpha_{2}}=hT_{\gamma_{1}}h^{-1}
\end{eqnarray*}
and
\begin{eqnarray*}
T_{\alpha_{2}}T_{\gamma_{1}}^{-1}=hT_{\gamma_{1}}h^{-1}T_{\gamma_{1}}^{-1}=h(T_{\gamma_{1}}h^{-1}T_{\gamma_{1}}^{-1}).
\end{eqnarray*}

Let $\bar{h}$ denote $T_{\gamma_{1}}h^{-1}T_{\gamma_{1}}^{-1}$.
We can now rewrite (\ref{eq1}) as
\begin{eqnarray}\label{eq4}
T_{\alpha_{1}}=(h\bar{h})(f^{2}h\bar{h}f^{-2})(fh\bar{h}f^{-1}).
\end{eqnarray}
and hence $T_{\alpha_{1}}$ is a product of 3 elements of order 3.

\subsection{Proof that 3 elements of order 3 generate}
We prove Theorem.
\begin{theorem}
If $g\geq 3$, ${\cal M}_{g}$ is generated by $f$, $h$ and $\bar{h}$.
\end{theorem}
\begin{proof}
Let $G_{1}$ denote the group generated by $f$, $h$ and $\bar{h}$.
By the relation~(\ref{eq4}), $T_{\alpha_{1}}$ is in $G_{1}$.

Let $\alpha$ and $\beta$ be simple closed curves on $\Sigma_{g}$.
The symbol $\alpha \overset{f}{\longrightarrow} \beta$ (resp. $\alpha \overset{h}{\longrightarrow} \beta$) means that $f(\alpha)=\beta$ (resp. $h(\alpha)=\beta$).

The Figure~\ref{fig10} shows that we can send $\alpha_{1}$ to all $\alpha_{i}$ and $\gamma_{i}$ by $f$ and $h$.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=10cm]{s11.eps}
 \end{center}
 \caption{}
 \label{fig10}
\end{figure}
Therefore, $T_{\alpha_{i}}, T_{\gamma_{i}}\in G_{1}$ for all $i$.

Since $h$ (resp. $f$) maps $\alpha_{g}$ to $\beta_{g}$ in the case of odd (resp. even) genus, $T_{\beta_{g}}\in G_{1}$.
As shown on Figure~\ref{fig11}, we can find that $\beta_{g}$ can be send to $\beta_{i}$ for all $i$ by $f$ and $h$.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=10cm]{s12.eps}
 \end{center}
 \caption{}
 \label{fig11}
\end{figure}
These shows that $T_{\beta_{i}}\in G_{1}$ for all $i$.

Since we show that all Lickorish's generators are in $G_{1}$, $G_{1}$ is equal to ${\cal M}_{g}$.
\end{proof}

\section{Generating the mapping class group by 4 elements by order 4}
In this section we prove that ${\cal M}_{g}$ can be generated by 4 elements of order 4.
The key point is to use \textit{chain relation}.
\subsection{Construction of elements of order 4}
We prepare to construct 3 elements of order 4.

We recall \textit{chain relation}.
We say that an ordered set of $c_{1}, \ldots, c_{n}$ of simple closed curves on $\Sigma_{g}$ forms an \textit{n-chain} if the geometric intersection $(c_{k}, c_{k+1})=1$ for $k=1, \ldots, n-1$ and $(c_{k}, c_{l})=0$ if $\vert k-l \vert \geq 2$.
If $n$ is odd, the boundary of a regular neighborhood of any $n$-chain has two componets $d_{1}$ and $d_{2}$.

The \textit{chain relation} is read as follow :

For a given $n$-chain $c_{1}, \ldots, c_{n}$, if $n$ is odd we have
\begin{eqnarray*}
(T_{c_{1}}T_{c_{2}}\cdots T_{c_{n}})^{n+1}=T_{d_{1}}T_{d_{2}}
\end{eqnarray*}

We denote the simple closed curves $\alpha^{\prime}_{1},\ldots ,\alpha^{\prime}_{g}$, $s$ and the separating curves $\delta_{1},\ldots ,\delta_{g-1}$ as shown on Figure~\ref{fig13}.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=6cm]{s13.eps}
 \end{center}
 \caption{The simple closed curves $\alpha^{\prime}_{1},\ldots ,\alpha^{\prime}_{g}$, $s$ and the separating curves $\delta_{1},\ldots ,\delta_{g-1}$}
 \label{fig13}
\end{figure}

We note that $\alpha_{1}=\alpha^{\prime}_{1}$ and $\alpha_{g}=\alpha^{\prime}_{g}$.

\

For $i=1, 2,\ldots ,g-2$ we define
\begin{eqnarray*}
\rho_{i}=(T_{\alpha_{i+2}}T_{\beta_{i+2}}T_{\gamma_{i+1}}T_{\beta_{i+1}}T_{\gamma_{i}}T_{\beta_{i}}T_{\alpha^{\prime}_{i}})^{2}.
\end{eqnarray*}
We can find that the boundary components of a regular neighborhood of 7-chain
$\alpha_{i+2}$, $\beta_{i+2}$, $\gamma_{i+1}$, $\beta_{i+1}$, $\gamma_{i}$, $\beta_{i}$, $\alpha^{\prime}_{i}$ are $\delta_{i+2}$ and $\delta_{i-1}$ like Figure~\ref{fig14}.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=4cm]{s14.eps}
 \end{center}
 \caption{}
 \label{fig14}
\end{figure}
We note that $\delta_{0}$ and $\delta_{g}$ is trivial.
By \textit{chain relation}, we have
\begin{eqnarray*}
\rho_{i}^{4}=(T_{\alpha_{i+2}}T_{\beta_{i+2}}T_{\gamma_{i+1}}T_{\beta_{i+1}}T_{\gamma_{i}}T_{\beta_{i}}T_{\alpha^{\prime}_{i}})^{8}=T_{\delta_{i+2}}T_{\delta_{i-1}}.
\end{eqnarray*}

For each $i$ we see that $\rho_{i}$ acts on the curves on $\Sigma_{g}$ as follow:
\begin{eqnarray*}
\rho_{i}^{3}(\alpha_{i+2})=\rho_{i}^{2}(\gamma_{i+1})=\rho_{i}(\gamma_{i})=\alpha^{\prime}_{i}, && \rho_{i}^{2}(\beta_{i+2})=\rho_{i}(\beta_{i+1})=\beta_{i}.
\end{eqnarray*}
In paticular we note that $\rho_{1}(\alpha_{2})=x_{1}$.

\

We define
\begin{eqnarray*}
\rho^{\prime}=(T_{\gamma_{2}}T_{\beta_{3}}T_{\alpha_{3}}T_{s}T_{\gamma_{1}}T_{\beta_{1}}T_{\alpha^{\prime}_{1}})^{2}.
\end{eqnarray*}
We see that the boundary component of a regular neighborhood of 7-chain $\gamma_{2}$, $\beta_{3}$, $\alpha_{3}$, $s$, $\gamma_{2}$, $\beta_{2}$, $\alpha^{\prime}_{1}$ in Figure~\ref{fig15} is $\delta_{3}$.
By \textit{chain relation} we have $(\rho^{\prime})^{4}=T_{\delta_{3}}$.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=4cm]{s15.eps}
 \end{center}
 \caption{}
 \label{fig15}
\end{figure}

$\rho^{\prime}$ acts on the curves on $\Sigma_{g}$ as follow:
\begin{eqnarray*}
(\rho^{\prime})^{3}(\gamma_{2})=(\rho^{\prime})^{2}(\alpha_{3})=\rho^{\prime}(\gamma_{1})=\alpha^{\prime}_{1}, && (\rho^{\prime})^{3}(x_{2})=(\rho^{\prime})^{2}(\alpha_{2})=\rho^{\prime}(x_{2})=\alpha_{2}.
\end{eqnarray*}
In paticular we note that $\rho^{\prime}(\alpha_{2})=x_{2}$.

\

We define
\begin{eqnarray*}
\tau_{i}&=&(T_{\alpha^{\prime}_{i+2}}T_{\beta_{i+2}}T_{\gamma_{i+1}}T_{\beta_{i+1}}T_{\gamma_{i}}T_{\beta_{i}}T_{\alpha_{i}})^{2}\\
\end{eqnarray*}
We can find that the boundary components of a regular neighborhood of 7-chain
$\alpha^{\prime}_{i+2}$, $\beta_{i+2}$, $\gamma_{i+1}$, $\beta_{i+1}$, $\gamma_{i}$, $\beta_{i}$, $\alpha_{i}$ are $\delta_{i+2}$ and $\delta_{i-1}$ like Figure~\ref{fig16}.
By \textit{chain relation} we have $\tau_{i}^{4}$$=$$(T_{\alpha^{\prime}_{i+2}}T_{\beta_{i+2}}T_{\gamma_{i+1}}T_{\beta_{i+1}}T_{\gamma_{i}}T_{\beta_{i}}T_{\alpha_{i}})^{8}$$=$$T_{\delta_{i+2}}T_{\delta_{i-1}}$.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=4cm]{s16.eps}
 \end{center}
 \caption{}
 \label{fig16}
\end{figure}

For each $i$ we see that $\tau_{i}$ acts on the curves on $\Sigma_{g}$ as follow:
\begin{eqnarray*}
\tau^{3}(\alpha^{\prime}_{i+2})=\tau^{2}(\gamma_{i+1})=\tau(\gamma_{i})=\alpha_{i} && \tau^{2}(\beta_{i+2})=\tau(\beta_{i+1})=\beta_{i}.
\end{eqnarray*}

\

For $j=1, 2,\ldots, g$ we define
\begin{eqnarray*}
\sigma_{j}&=&T_{\alpha_{j}}T_{\beta_{j}}T_{\alpha^{\prime}_{j}}.
\end{eqnarray*}
We can find that the boundary components of a regular neighborhood of 3-chain $\alpha_{j}$, $\beta_{j}$, $\alpha^{\prime}_{j}$ like Figure~\ref{fig17} is $\delta_{j}$ and $\delta_{j-1}$.
By Lemma~\ref{lem3} and \textit{chain relation} we have $\tau^{4}_{j}=T_{\delta_{j}}T_{\delta_{j-1}}$.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=2cm]{s17.eps}
 \end{center}
 \caption{}
 \label{fig17}
\end{figure}

For each $j$ we see that $\sigma_{j}(\alpha_{j})$$=$$T_{\alpha_{j}}T_{\beta_{j}}T_{\alpha^{\prime}_{j}}(\alpha_{j})$$=$$\beta_{j}$.

\subsubsection{The genus is $3m$}
We assume that $g=3m$.
We construct an element $\phi$ of order 4.
We define
\begin{equation*}
\phi=
\begin{cases}
\rho_{g-2}^{-1}\rho_{g-5}\cdots \rho_{7}^{-1}\rho_{4}\rho_{1}^{-1} & \text{($m (=\frac{g}{3})$ is odd)} \\
\rho_{g-2}\rho_{g-5}^{-1}\cdots \rho_{7}^{-1}\rho_{4}\rho_{1}^{-1} & \text{($m$ is even)}.
\end{cases}
\end{equation*}

We can find that $\phi^{4}=T_{\delta_{g-3}}^{\mp 1}(T_{\delta_{g-3}}T_{\delta_{g-6}})^{\pm 1}\cdots (T_{\delta_{9}}T_{\delta_{6}})^{-1}(T_{\delta_{6}}T_{\delta_{3}})T_{\delta_{3}}^{-1}=1$.

$\phi$ acts on the curves on $\Sigma_{g}$ as follow:
\begin{center}
\begin{tabular}{lcr}
$\alpha_{2}=\phi(x_{1}),$& $$& $$\\
$\phi^{3}(\alpha^{\prime}_{1})=\phi^{2}(\gamma_{1})=\phi(\gamma_{2})=\alpha_{3},$& $\phi^{2}(\beta_{1})=\phi(\beta_{2})=\beta_{3}$& $$\\
$\alpha^{\prime}_{4}=\phi(\gamma_{4})=\phi^{2}(\gamma_{5})=\phi^{3}(\alpha_{6}),$& $\beta_{4}=\phi(\beta_{5})=\phi^{2}(\beta_{6})$& $$\\
$\vdots$&$\vdots$\\
$\phi^{3}(\alpha^{\prime}_{g-2})=\phi^{2}(\gamma_{g-2})=\phi(\gamma_{g-1})=\alpha_{g},$& $\phi^{2}(\beta_{g-2})=\phi(\beta_{g-1})=\beta_{g}$& $(m \ is \ odd),$\\
\\
$(\alpha^{\prime}_{g-2}=\phi(\gamma_{g-2})=\phi^{2}(\gamma_{g-1})=\phi^{3}(\alpha_{g}),$& $\beta_{g-2}=\phi(\beta_{g-1})=\phi^{2}(\beta_{g})$& $(m \ is \ even))$.
\end{tabular}
\end{center}

\

We construct an element $\psi$ of order 4.
We define
\begin{equation*}
\psi=
\begin{cases}
\rho_{g-2}^{-1}\rho_{g-5}\cdots \rho_{7}^{-1}\rho_{4}\rho^{\prime -1} & \text{($m (=\frac{g}{3})$ is odd)} \\
\rho_{g-2}\rho_{g-5}^{-1}\cdots \rho_{7}^{-1}\rho_{4}\rho^{\prime -1} & \text{($m$ is even)}.
\end{cases}
\end{equation*}

We can find that $\psi^{4}=1$.
Since $\rho^{\prime}(\alpha_{2})=x_{2}$ and $\rho^{\prime}(\gamma_{2})=\alpha_{3}$, we can easily find that $\psi^{-1}(\alpha_{2})=x_{2}$ and $\psi^{-1}(\gamma_{2})=\alpha_{3}$.

\

We construct an element $\omega$ of order 4.
We define
\begin{equation*}
\omega=
\begin{cases}
\sigma_{g}^{-1}\sigma_{g-1}\tau_{g-5}^{-1}\tau_{g-8}\cdots \tau_{8}\tau_{5}^{-1}\tau_{2}\sigma_{1}^{-1} & \text{($m (=\frac{g}{3})$ is odd)} \\
\sigma_{g}\sigma_{g-1}^{-1}\tau_{g-5}\tau_{g-8}^{-1}\cdots \tau_{8}\tau_{5}^{-1}\tau_{2}\sigma_{1}^{-1} & \text{($m$ is even)}.
\end{cases}
\end{equation*}
We can find that $\omega^{4}=1$.

$\omega$ acts on the curves on $\Sigma_{g}$ as follow:
\begin{center}
\begin{tabular}{lcr}
$\alpha_{1}=\omega(\beta_{1}),$\\
$\omega^{3}(\alpha_{2})=\omega^{2}(\gamma_{2})=\omega(\gamma_{3})=\alpha^{\prime}_{4},$& $\omega^{2}(\beta_{2})=\omega(\beta_{3})=\beta_{4},$\\
$\alpha_{5}=\omega(\gamma_{5})=\omega^{2}(\gamma_{6})=\omega^{3}(\alpha^{\prime}_{7}),$& $\beta_{5}=\omega(\beta_{6})=\omega^{2}(\beta_{7}),$\\
$\ \ \ \ \ \ \vdots$&$\vdots$\\
$\omega^{3}(\alpha_{g-4})=\omega^{2}(\gamma_{g-4})=\omega(\gamma_{g-3})=\alpha^{\prime}_{g-2},$& $\omega^{2}(\beta_{g-4})=\omega(\beta_{g-3})=\beta_{g-2}$& $(m \ is \ odd),$\\
\\
$(\alpha_{g-4}=\omega(\gamma_{g-4})=\omega^{2}(\gamma_{g-3})=\omega^{3}(\alpha^{\prime}_{g-2}),$& $\beta_{g-4}=\omega(\beta_{g-3})=\omega^{2}(\beta_{g-2})$& $(m \ is \ even)).$
\end{tabular}
\end{center}

\subsubsection{The genus is $3m+1$}
We assume that $g=3m+1$.

We define
\begin{equation*}
\phi=
\begin{cases}
\sigma_{g}\rho_{g-3}^{-1}\rho_{g-6}\cdots \rho_{7}^{-1}\rho_{4}\rho_{1}^{-1} & \text{($m (=\frac{g-1}{3})$ is odd)} \\
\sigma_{g}^{-1}\rho_{g-3}\rho_{g-6}^{-1}\cdots \rho_{7}^{-1}\rho_{4}\rho_{1}^{-1} & \text{($m$ is even)}.
\end{cases}
\end{equation*}

$\phi$ acts on the curves on $\Sigma_{g}$ as follow:
\begin{center}
\begin{tabular}{lcr}
$\alpha_{2}=\phi(x_{1}),$\\
$\phi^{3}(\alpha^{\prime}_{1})=\phi^{2}(\gamma_{1})=\phi(\gamma_{2})=\alpha_{3},$& $\phi^{2}(\beta_{1})=\phi(\beta_{2})=\beta_{3},$\\
$\alpha^{\prime}_{4}=\phi(\gamma_{4})=\phi^{2}(\gamma_{5})=\alpha_{6},$& $\beta_{4}=\phi(\beta_{5})=\phi^{2}(\beta_{6})$\\
$\ \ \ \ \ \ \vdots$&$\vdots$\\
$\phi^{3}(\alpha^{\prime}_{g-3})=\phi^{2}(\gamma_{g-3})=\phi(\gamma_{g-2})=\alpha_{g-1},$& $\phi^{2}(\beta_{g-3})=\phi(\beta_{g-2})=\beta_{g-1}$& $(m \ is \ odd),$\\
\\
$(\alpha^{\prime}_{g-3}=\phi(\gamma_{g-3})=\phi^{2}(\gamma_{g-2})=\phi^{3}(\alpha_{g-1}),$& $\beta_{g-3}=\phi(\beta_{g-2})=\phi^{2}(\beta_{g-1})$& $(m \ is \ even)).$
\end{tabular}
\end{center}

\

We define
\begin{equation*}
\psi=
\begin{cases}
\rho_{g-2}^{-1}\rho_{g-5}\cdots \rho_{7}^{-1}\rho_{4}\rho^{\prime -1} & \text{($m (=\frac{g-1}{3})$ is odd)} \\
\rho_{g-2}\rho_{g-5}^{-1}\cdots \rho_{7}^{-1}\rho_{4}\rho^{\prime -1} & \text{($m$ is even)}.
\end{cases}
\end{equation*}

We note that $\psi^{-1}(\alpha_{2})=x_{2}$ and $\psi^{-1}(\gamma_{2})=\alpha_{3}$.

\

We define
\begin{equation*}
\omega=
\begin{cases}
\tau_{g-2}^{-1}\tau_{g-5}\cdots \tau_{8}\tau_{5}^{-1}\tau_{2}\sigma_{1}^{-1} & \text{($m (=\frac{g-1}{3})$ is odd)} \\
\tau_{g-2}\tau_{g-5}^{-1}\cdots \tau_{8}\tau_{5}^{-1}\tau_{2}\sigma_{1}^{-1} & \text{($m$ is even)}.
\end{cases}
\end{equation*}

$\omega$ acts on the curves on $\Sigma_{g}$ as follow:
\begin{center}
\begin{tabular}{lcr}
$\alpha_{1}=\omega(\beta_{1}),$\\
$\omega^{3}(\alpha_{2})=\omega^{2}(\gamma_{2})=\omega(\gamma_{3})=\alpha^{\prime}_{4},$& $\omega^{2}(\beta_{2})=\omega(\beta_{3})=\beta_{4},$\\
$\alpha_{5}=\omega(\gamma_{5})=\omega^{2}(\gamma_{6})=\omega^{3}(\alpha^{\prime}_{7}),$& $\beta_{5}=\omega(\beta_{6})=\omega^{2}(\beta_{7}),$\\
$\ \ \ \ \ \ \vdots$&$\vdots$\\
$\omega^{3}(\alpha_{g-2})=\omega^{2}(\gamma_{g-2})=\omega(\gamma_{g-1})=\alpha^{\prime}_{g},$& $\omega^{2}(\beta_{g-2})=\omega(\beta_{g-1})=\beta_{g}$& $(m \ is \ odd),$\\
\\
$(\alpha_{g-2}=\omega(\gamma_{g-2})=\omega^{2}(\gamma_{g-1})=\alpha^{\prime}_{g},$& $\omega^{2}(\beta_{g-2})=\omega(\beta_{g-1})=\beta_{g}$& $(m \ is \ even)).$
\end{tabular}
\end{center}

\subsubsection{The genus is $3m+2$}
We assume that $g=3m+2$ $(m\geq 2)$.

We define
\begin{equation*}
\phi=
\begin{cases}
\sigma_{g}^{-1}\sigma_{g-1}\rho_{g-4}^{-1}\rho_{g-7}\cdots \rho_{7}^{-1}\rho_{4}\rho_{1}^{-1} & \text{($m (=\frac{g-2}{3})$ is odd)} \\
\sigma_{g}\sigma_{g-1}^{-1}\rho_{g-4}\rho_{g-7}^{-1}\cdots \rho_{7}^{-1}\rho_{4}\rho_{1}^{-1} & \text{($m$ is even)}.
\end{cases}
\end{equation*}

$\phi$ acts on the curves on $\Sigma_{g}$ as follow:
\begin{center}
\begin{tabular}{lcr}
$\alpha_{2}=\phi(x_{1}),$\\
$\phi^{3}(\alpha^{\prime}_{1})=\phi^{2}(\gamma_{1})=\phi(\gamma_{2})=\alpha_{3},$& $\phi^{2}(\beta_{1})=\phi(\beta_{2})=\beta_{3},$\\
$\alpha^{\prime}_{4}=\phi(\gamma_{4})=\phi^{2}(\gamma_{5})=\phi^{3}(\alpha_{6}),$& $\beta_{4}=\phi(\beta_{5})=\phi^{2}(\beta_{6}),$\\
$\ \ \ \ \ \ \vdots$&$\vdots$\\
$\phi^{3}(\alpha^{\prime}_{g-4})=\phi^{2}(\gamma_{g-4})=\phi(\gamma_{g-3})=\alpha_{g-2},$& $\phi^{2}(\beta_{g-4})=\phi(\beta_{g-3})=\beta_{g-2}$& $(m \ is \ odd),$\\
\\
$(\alpha^{\prime}_{g-4}=\phi(\gamma_{g-4})=\phi^{2}(\gamma_{g-3})=\phi^{3}(\alpha_{g-2}),$& $\beta_{g-4}=\phi(\beta_{g-3})=\phi^{2}(\beta_{g-2})$& $(m \ is \ even)).$\\
\end{tabular}
\end{center}

\

We define
\begin{equation*}
\psi=
\begin{cases}
\rho_{g-2}^{-1}\rho_{g-5}\cdots \rho_{9}^{-1}\rho_{6}\sigma_{5}^{-1}\sigma_{4}\rho^{\prime -1} & \text{($m (=\frac{g-2}{3})$ is odd)} \nonumber\\
\rho_{g-2}\rho_{g-5}^{-1}\cdots \rho_{9}^{-1}\rho_{6}\sigma_{5}^{-1}\sigma_{4}\rho^{\prime -1} & \text{($m$ is even)}. \nonumber
\end{cases}
\end{equation*}

We note that $\psi$ acts on the curves on $\Sigma_{g}$ as follow:
\begin{center}
\begin{tabular}{lcr}
$\alpha_{2}=\psi(x_{2}),$& $\gamma_{2}=\psi(\alpha_{3}),$\\
$\psi^{3}(\alpha^{\prime}_{g-2})=\psi^{2}(\gamma_{g-2})=\psi(\gamma_{g-1})=\alpha_{g},$& $\psi^{2}(\beta_{g-2})=\psi(\beta_{g-1})=\beta_{g}$& $(m \ is \ odd),$\\
\\
$(\alpha^{\prime}_{g-2}=\psi(\gamma_{g-2})=\psi^{2}(\gamma_{g-1})=\psi^{3}(\alpha_{g}),$& $\beta_{g-2}=\psi(\beta_{g-1})=\psi^{2}(\beta_{g})$& $(m \ is \ even)).$
\end{tabular}
\end{center}

\

We define
\begin{equation*}
\omega=
\begin{cases}
\sigma_{g}\tau_{g-3}^{-1}\tau_{g-6}\cdots \tau_{8}\tau_{5}^{-1}\tau_{2}\sigma_{1}^{-1} & \text{($m (=\frac{g-2}{3})$ is odd)} \\
\sigma_{g}^{-1}\tau_{g-3}\tau_{g-6}^{-1}\cdots \tau_{8}\tau_{5}^{-1}\tau_{2}\sigma_{1}^{-1} & \text{($m$ is even)}.
\end{cases}
\end{equation*}

$\omega$ acts on the curves on $\Sigma_{g}$ as follow:
\begin{center}
\begin{tabular}{lcr}
$\alpha_{1}=\omega(\beta_{1}),$\\
$\omega^{3}(\alpha_{2})=\omega^{2}(\gamma_{2})=\omega(\gamma_{3})=\alpha^{\prime}_{4},$& $\omega^{2}(\beta_{2})=\omega(\beta_{3})=\beta_{4},$\\
$\alpha^{\prime}_{5}=\omega(\gamma_{5})=\omega^{2}(\gamma_{6})=\omega^{3}(\alpha^{\prime}_{7}),$& $\beta_{5}=\omega(\beta_{6})=\omega^{2}(\beta_{7}),$\\
$\ \ \ \ \ \ \vdots$&$\vdots$\\
$\omega^{3}(\alpha_{g-3})=\omega^{2}(\gamma_{g-3})=\omega(\gamma_{g-2})=\alpha^{\prime}_{g-1},$& $\omega^{2}(\beta_{g-3})=\omega(\beta_{g-2})=\beta_{g-1}$& $(m \ is \ odd),$\\
\\
$(\alpha_{g-3}=\omega(\gamma_{g-3})=\omega^{2}(\gamma_{g-2})=\omega^{3}(\alpha^{\prime}_{g-1}),$& $\beta_{g-3}=\omega(\beta_{g-2})=\omega^{2}(\beta_{g-1})$& $(m \ is \ even)).$
\end{tabular}
\end{center}

\subsubsection{The genus is 5}
We assume that $g=5$.

We define
\begin{eqnarray*}
\phi=(T_{\alpha_{5}}T_{\beta_{5}}T_{\gamma_{4}})(T_{\gamma_{3}}T_{\beta_{3}}T_{\gamma_{2}}T_{\beta{2}}T_{\gamma_{1}}T_{\beta_{1}}T_{\alpha^{\prime}_{1}})^{-2}.
\end{eqnarray*}
By \textit{chain reltion}, $(T_{\alpha_{5}}T_{\beta_{5}}T_{\gamma_{4}})^{4}$$=$$(T_{\gamma_{3}}T_{\beta_{3}}T_{\gamma_{2}}T_{\beta{2}}T_{\gamma_{1}}T_{\beta_{1}}T_{\alpha^{\prime}_{1}})^{8}$$=$$T_{\alpha_{4}}T_{\alpha^{\prime}_{4}}$
Therefore, we can find that $\phi^{4}=1$.

$\phi$ acts on the curves on $\Sigma_{g}$ as follow :
\begin{eqnarray*}
\phi^{3}(\alpha^{\prime}_{1})=\phi^{2}(\gamma_{1})=\phi(\gamma_{2})=\alpha_{3}, &&\phi^{2}(\beta_{1})=\phi(\beta_{2})=\beta_{3}, \\
\alpha_{2}=\phi(x_{1}), &&\phi(\gamma_{4})=\beta_{5}.
\end{eqnarray*}

\

We define
\begin{eqnarray*}
\psi=\sigma_{5}\sigma_{4}\rho^{\prime -1}.
\end{eqnarray*}
We can find that $\psi^{4}=1$.

$\psi$ acts on the curves on $\Sigma_{g}$ as follow :
\begin{eqnarray*}
\psi^{3}(\alpha^{\prime}_{1})=\psi^{2}(\gamma_{1})=\psi(\alpha_{3})=\gamma_{2}, &&\psi(\alpha_{2})=x_{2}.
\end{eqnarray*}

\

Let $s_{1}$, $s_{2}$ the non separating simple closed curves as shown in Figure~\ref{fig30}
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=5cm]{s27.eps}
 \end{center}
 \caption{the curves $s_{1}$ and $s_{2}$.}
 \label{fig30}
\end{figure}

We define
\begin{eqnarray*}
\omega=(T_{\alpha_{1}}T_{\beta_{1}}T_{s_{1}})(T_{\gamma_{4}}T_{\beta_{4}}T_{\gamma_{3}}T_{\beta_{3}}T_{\gamma_{2}}T_{\beta_{2}}T_{\alpha_{2}})^{-2}.
\end{eqnarray*}
By the \textit{chain relation}, $(T_{\alpha_{1}}T_{\beta_{1}}T_{s_{1}})^{4}$$=$$(T_{\gamma_{4}}T_{\beta_{4}}T_{\gamma_{3}}T_{\beta_{3}}T_{\gamma_{2}}T_{\beta_{2}}T_{\alpha_{2}})^{8}$$=$$T_{\alpha_{g}}T_{s_{2}}$.
Therefore, we can find that $\omega^{4}=1$.

$\omega$ acts on the curves on $\Sigma_{g}$ as follow :
\begin{eqnarray*}
\omega^{3}(\alpha_{2})=\omega^{2}(\gamma_{2})=\omega(\gamma_{3})=\gamma_{4}, &&\omega^{2}(\beta_{2})=\omega(\beta_{3})=\beta_{4}.
\end{eqnarray*}

\subsection{Generating the Dehn twist by 4 elements of order 4}
By using the lantern relation we generate the Dehn twist by 4 elements of order 4.

Since $\omega$ maps $\alpha_{2}$ to $\gamma_{2}$, we see that
\begin{eqnarray*}
T_{\alpha_{2}}=\omega T_{\gamma_{2}}\omega^{-1}
\end{eqnarray*}
and
\begin{eqnarray*}
T_{\alpha_{2}}T_{\gamma_{2}}^{-1}=\omega T_{\gamma_{2}}\omega^{-1}T_{\gamma_{2}}^{-1}=\omega (T_{\gamma_{2}}\omega^{-1}T_{\gamma_{2}}^{-1}).
\end{eqnarray*}
Let $\tilde{\omega}$ denote $T_{\gamma_{2}}\omega^{-1}T_{\gamma_{2}}^{-1}$.
Then we see that $T_{\alpha_{2}}T_{\gamma_{2}}^{-1}=\omega \tilde{\omega}$.

In the case of $g\neq 5$, we rewrite the lantern relation as follow :
\begin{eqnarray}\label{eq6}
T_{\alpha_{1}}=T_{\alpha_{1}^{\prime}}=(T_{\alpha_{2}}T_{\gamma_{2}}^{-1})(T_{x_{1}}T_{\gamma_{1}}^{-1})(T_{x_{2}}T_{\alpha_{3}}^{-1}).
\end{eqnarray}
From the argument of Section 4.1.1, 4.1.2 and 4.1.3 we can find that $\phi^{-1}(\alpha_{2})=x_{1}$, $\phi^{-1}(\gamma_{2})=\gamma_{1}$, $\psi^{-1}(\alpha_{2})=x_{2}$ and $\psi^{-1}(\gamma_{2})=\alpha_{3}$.
By Lemma~\ref{lem1} show we see that :
\begin{eqnarray*}
(T_{x_{1}}T_{\gamma_{2}}^{-1})=\phi^{-1}(T_{\alpha_{2}}T_{\gamma_{1}}^{-1})\phi \\
(T_{x_{2}}T_{\alpha_{3}}^{-1})=\psi^{-1}(T_{\alpha_{2}}T_{\gamma_{1}}^{-1})\psi.
\end{eqnarray*}
We can rewrite (\ref{eq6}) as
\begin{eqnarray*}
T_{\alpha_{1}}=(\omega \tilde{\omega})(\phi^{-1} \omega \tilde{\omega}\phi)(\psi^{-1} \omega \tilde{\omega}\psi).
\end{eqnarray*}

In the case of $g=5$, we rewrite the lantern relation as follow :
\begin{eqnarray}\label{eq9}
T_{\alpha_{3}}=(T_{\alpha_{2}}T_{\gamma_{2}}^{-1})(T_{x_{1}}T_{\gamma_{1}}^{-1})(T_{x_{2}}T_{\alpha_{1}}^{-1}).
\end{eqnarray}
From the argument of Section 4.1.3 we can find that $\phi^{-1}(\alpha_{2})=x_{1}$, $\phi^{-1}(\gamma_{2})=\gamma_{1}$, $\psi(\alpha_{2})=x_{2}$ and $\psi(\gamma_{2})=\alpha_{1}^{\prime}=\alpha_{1}$.
By Lemma~\ref{lem1} we see that :
\begin{eqnarray*}
(T_{x_{1}}T_{\gamma_{1}}^{-1})=\phi^{-1}(T_{\alpha_{2}}T_{\gamma_{2}}^{-1})\phi \\
(T_{x_{2}}T_{\alpha_{1}}^{-1})=\psi(T_{\alpha_{2}}T_{\gamma_{2}}^{-1})\psi^{-1}.
\end{eqnarray*}
We can rewrite (\ref{eq9})
\begin{eqnarray*}
T_{\alpha_{3}}=(\omega \tilde{\omega})(\phi^{-1} \omega \tilde{\omega}\phi)(\psi \omega \tilde{\omega}\psi^{-1}).
\end{eqnarray*}

Hence Dehn tiwst is a product of 4 elements of order 4.

\subsection{Proof that 4 elements of order 4 generate}
We prove Theorem.
\begin{theorem}
If $g$ is at least 3, ${\cal M}_{g}$ can be generated by $\phi$, $\psi$, $\omega$ and $\tilde{\omega}$.
\end{theorem}
\begin{proof}
Let $G_{2}$ denote the group generated by $\phi$, $\psi$, $\omega$ and $\tilde{\omega}$.

Let $\alpha$ and $\beta$ be simple closed curves on $\Sigma_{g}$.
The symbol $\alpha \overset{\phi}{\longleftrightarrow} \beta$ (resp. $\alpha \overset{\psi}{\longleftrightarrow} \beta$, $\alpha \overset{\omega}{\longleftrightarrow} \beta$) means that either $\phi(\alpha)=\beta$ or $\phi^{-1}(\alpha)=\beta$ (resp. either $\psi(\alpha)=\beta$ or $\psi^{-1}(\alpha)=\beta$, either $\omega(\alpha)=\beta$ or $\omega^{-1}(\alpha)=\beta$).

If $g\neq 5$, we can find that $T_{\alpha_{1}}$ is in $G_{2}$.
In the case of $g=3m$ and $g=3m+1$, $\phi$ and $\omega$ can map $\alpha_{1}$ to all $\beta_{i}$ and $\gamma_{i}$ as shown in Figure~\ref{fig31}.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=12cm]{s31.eps}
 \end{center}
 \caption{}
 \label{fig31}
\end{figure}
If $g=3m+2$, we need $\psi$ other than $\phi$, $\omega$ like Figure~\ref{fig32}.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=12cm]{s32.eps}
 \end{center}
 \caption{}
 \label{fig32}
\end{figure}
Therefore, for all $i$, $T_{\beta_{i}}$ $T_{\gamma_{i}}$, are in $G_{2}$.
Because $\omega(\alpha_{2})=\gamma_{2}$, we can find that $T_{\alpha_{2}}\in G_{2}$.
Therefore, all Humphries's generators are in $G_{2}$.

If $g=5$, $T_{\alpha_{3}}$ is in $G_{2}$.
The Figure~\ref{fig33} shows that we can send $\alpha_{3}$ to all Humphries's generators by $\phi$, $\psi$ and $\omega$.
\begin{figure}[htbp]
 \begin{center}
  \includegraphics*[width=4.5cm]{s33.eps}
 \end{center}
 \caption{}
 \label{fig33}
\end{figure}

We prove that $G_{2}$ is equal to ${\cal M}_{g}$ for $g\geq 3$.
\end{proof}

\section{Remarks}
\subsection{Low genus}
In the case of $g=1, 2$, we note that ${\cal M}_{g}$ can not be generated by elements of same order.
By using the argument of MacCarthy and Papadopulos [MP] and the work of Hirose [Hi], we can see the proof.
Hirose listed the Dehn twist presentation of finite order elements for closed oriented surfaces of genera up to 4.
We introduce the presentation of finite order elements in the case of $g=1, 2$.
The list is as follow :
\\
\
\\
\begin{tabular}{c||d{40.0}|d{4.0}}
\hline
 \textbf{genus} & \multicolumn{1}{c|}{\textbf{elements}} & \multicolumn{1}{c}{\textbf{order}} \\
\hline
 1                                     & T_{\beta_{1}}T_{\alpha_{1}}& 6\\
                                       & T_{\alpha_{1}}T_{\beta_{1}}T_{\alpha_{1}} &4\\
\hline
 2                                     & T_{\beta_{2}}T_{\gamma_{1}}T_{\beta_{1}}T_{\alpha_{1}}&  10\\
                                       & T_{\beta_{2}}T_{\beta_{2}}T_{\gamma_{1}}T_{\beta_{1}}T_{\alpha_{1}}&  8\\
                                       & T_{\alpha_{2}}T_{\beta_{2}}T_{\gamma_{1}}T_{\beta_{1}}T_{\alpha_{1}}&  6\\
                                       & (T_{\alpha_{1}}T_{\beta_{1}}T_{\gamma_{1}}T_{\beta_{2}}T_{\alpha_{2}})(T_{\alpha_{2}}T_{\beta_{2}}T_{\gamma_{1}}T_{\beta_{1}}T_{\alpha_{1}})^{3}&  6\\
\hline
\end{tabular}

\

MacCarthy and Papadopulos proved that ${\cal M}_{2}$ can not be generated by elements of order 2.
The argument of MacCarthy and Papadopulos is as follow:
\begin{proof}
Let $c$ be a nonseparating simple closed curve and $p$ be the abelianization map given by Powell's result [Po]:

$$
\begin{array}{ccc}
\textit{p} : {\cal M}_{2}                     & \longrightarrow & \mathbb{Z}_{10} \\[-4pt]
\\
\ \ \ \ \rotatebox{90}{$\in$} &                 & \rotatebox{90}{$\in$} \\[-4pt]
\\
\ \ \ \ \textit{T}_{\textit{c}}                     & \longmapsto     & 1
\end{array}
$$
We can find that
\begin{eqnarray*}
p((T_{\beta_{2}}T_{\gamma_{1}}T_{\beta_{1}}T_{\alpha_{1}})^5)&=&p((T_{\beta_{2}}T_{\beta_{2}}T_{\gamma_{1}}T_{\beta_{1}}T_{\alpha_{1}})^{4})\\
&=&p(\{(T_{\alpha_{1}}T_{\beta_{1}}T_{\gamma_{1}}T_{\beta_{2}}T_{\alpha_{2}})(T_{\alpha_{2}}T_{\beta_{2}}T_{\gamma_{1}}T_{\beta_{1}}T_{\alpha_{1}})^{3}\}^{3})\\
&=&0\\
p((T_{\alpha_{2}}T_{\beta_{2}}T_{\gamma_{1}}T_{\beta_{1}}T_{\alpha_{1}})^{3})&=&5.
\end{eqnarray*}
It is easily to see that $\mathbb{Z}_{10}$ can not be generated by 0 and 5.
Therefore we can see that ${\cal M}_{2}$ can not be generated by elements of order 2.
\end{proof}

By the similar proof, we can see that ${\cal M}_{1}$ and ${\cal M}_{2}$ can not be generated by elements of same order.
\begin{remark}
${\cal M}_{1}$ and ${\cal M}_{2}$ can be generated by elements of different order.
For example, ${\cal M}_{1}$ can be generated by $ T_{\beta_{1}}T_{\alpha_{1}}$ and $T_{\alpha_{1}}T_{\beta_{1}}T_{\alpha_{1}}$, and ${\cal M}_{2}$ can be generated by $T_{\beta_{2}}T_{\gamma_{1}}T_{\beta_{1}}T_{\alpha_{1}}$ and $T_{\alpha_{2}}T_{\beta_{2}}T_{\gamma_{1}}T_{\beta_{1}}T_{\alpha_{1}}$.
\end{remark}

\subsection{Lower bound}
The order of ${\cal M}_{g}$ is not finite.
Therefore, we can easily find that lower bound of the number of generators whose order are 3 (resp. 4) is 2.
The author has the following question :
\begin{question}
What is the minimal number of elements of order 3 (resp. 4) required to generate ${\cal M}_{g}$?
\end{question}

redpajama_set_name: RedPajamaArXiv


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Position Summary: This role is responsible for updating local website, curating high-performing content, crafting homepage layouts driven by data, and helping to report news — everything from quick-hit local stories to breaking news and enterprise reports. This role will work on building a loyal local audience by being responsive to traffic patterns and audience interest while serving as a steward of our brands.
Position Responsibilities:
Curate and aggregate a steady stream of interesting content that is packaged and promoted in a way that generates engagement and repeat site visitation.
Monitor all forms of media (print, TV, digital, blogs and social) and be the quick to report breaking news stories and always on the hunt for exclusive and or trending content that will grow engagement and drive loyal, local traffic
Flexibility to perform duties
Excellent at writing headlines that grow audience and engagement
Prioritize packaging content that grows page views per visitor and time on site, using headline testing, metric tools and observed patterns to guide decision making
Understand different ways to tell a story, particularly producing short videos and choosing great photos.
Evaluate website traffic trends to make decisions on both daily content needs and longer-term content curation that will grow local audience.
Build positive working relationships with newsroom staff and management, work collaboratively on enterprise content, and creating opportunities to market web content on broadcast.
Promote and distribute news content on social media.
Skill/ Experience Requirements
2 years of experience in digital content and journalism
Clever headline writer that gets readers to click
Strong multi-tasking abilities
Organized, technical problem solver and quick decision maker
Capable of researching, interviewing and writing original news articles
Knowledge of SEO best practices and AP Style
Enjoys working in teams and has excellent interpersonal skills
Ease with/ability to learn new technology independently and quickly
Strong communicator
Some schedule flexibility (nights, weekends)
Regularly meets measurements of success
Strong understanding of Google Analytics; Chartbeat experience a plus
Proficiency in MS Office; HTML, CSS and Photoshop experience a plus
EDUCATION/EXPERIENCE: Bachelor's degree in Communication/Journalism or related degree required; with at least 2 years of experience writing, producing and editing content on a desktop editing system.
TRAINING/EQUIPMENT: Operate computers with desktop editing system.
Work Environment/Physical Requirements: High stress environment with deadline pressures. Position will require day-to-day multi-tasking, as well as managing multiple projects simultaneously with frequent change in direction and priority. Must be willing to work weekends (Saturdays and Sundays), as well as be able to work any and all shifts as necessary, including overnights/early mornings, nights, and holidays.
The employee in this position must be able to sit, conduct telephone conversations, use electronic mail, write letters and memos, conduct face-to-face discussions with individuals or groups, make decisions without supervision, make decisions that impact the results of co-workers, work near others, and work indoors in environmentally controlled conditions. Repetitive movements are required. Work may be conducted in extremely bright or inadequate lighting conditions.
NOTE: This job description contains the basic requirements for the position and is NOT intended to be a complete list of responsibilities; other duties may be assigned.
To Apply: Please visit the Work for Us page at KOIN.com: https://bit.ly/2NpjYBu or at the Nexstar Website at: https://nexstar.wd5.myworkdayjobs.com/nexstar to complete the application process.
To be considered for this position you must apply online via the Nexstar Media Inc. Career Portal: Please remember to attach your cover letter, resume, and a link to your demo reel or portfolio (if applicable).
Nexstar Careers – First Time users must create an Account
Broadcasting since 1953, KOIN-TV is Portland's CBS affiliate, providing exceptional local news content on air and online (koin.com), CBS Network programming and syndicated entertainment to viewers across Oregon and SW Washington. KOIN News is Watching Out for You with 51 hours of local news broadcasts each week, featuring award-winning, in-depth investigations, news, weather, traffic, KOIN News AM-Extra and Game-On! Our sister station, KRCW-TV – Portland's CW – began broadcasting in 1989. Portland's CW offers CW network programming, along with local sports and syndicated comedy programming and talk shows. Also under the umbrella of KOIN-TV is Antenna TV, which features classic TV programs from the 1960s-1990s, such as Family Ties, Johnny Carson, Bewitched, Coach and many others.
KOIN-TV, KRCW-TV and Antenna TV are Nexstar Media Inc. television stations serving Oregon and SW Washington from our studios in downtown Portland, OR. Nexstar Media Inc. is America's largest local television and media company with 197 full power stations (including partner stations) in 116 markets addressing more than 68% of U.S. television households and a growing digital media operation. Nexstar's platform delivers exceptional local content and network programming to inform and entertain viewers, while providing premium, scalable local advertising opportunities for advertisers and brands across all screens and devices. Today, as one of the most powerful and trusted voices in America, Nexstar remains true to its mission to uphold the public interest principles of localism, diversity, and trustworthy, unbiased broadcast journalism. At the same time, the company continues to simultaneously raise its commitment to support the evolving needs and interests of the local viewers, hometown businesses and community organizations that it proudly serves each day.
*****EOE/MINORITIES/FEMALES/VETERANS/DISABLED****
KOIN/Nexstar Media Inc. is an equal opportunity employer that seeks diversity in our workforce. All qualified candidates are encouraged to apply.
KOIN/Nexstar Media Inc. values the contribution of its employees. We recognize their efforts by offering a competitive compensation plan and an array of benefits, including medical, dental, vision, life insurance, 401(K), and more. We are committed to a drug and alcohol-free workplace. Any job offer is conditional upon satisfactorily completing a background and DMV check.
Apply online at: https://nexstar.wd5.myworkdayjobs.com/nexstar
https://nexstar.wd5.myworkdayjobs.com/nexstar

redpajama_set_name: RedPajamaCommonCrawl


text:
Lindsay Adler
Portrait and Fashion Photographer
Lindsay Adler is a portrait and fashion photographer based in New York City. Her work has appeared in numerous fashion and photography publications internationally including Elle, Harper's Bazaar, Marie Claire, and Numero. A clean, bold and graphic style has become the hallmark of her work, making her in-demand for clients across the world.
Connect with Lindsay Adler
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The Magic of Gels Tutorials Expandir seção Recolher seção
Color is a powerful tool of expression. In this creative and educational project Lindsay Adler shares her insights into color gel photography. From color theory, to additive/subtract color, to achieving saturated colors of light, this video and guide combo inspires and informs photographers about this approach to creative studio lighting. Check out her tutorial.
"Everyone generally has a story about drive brands having issues, but G-Technology has worked flawlessly for me."
"On set, especially for fashion shoots, time is money," says renowned fashion and portrait photographer Lindsay Adler. "Sometimes you're shooting right up until the very last second that you have at this location, which you've rented for however many thousands of dollars. The second your time is up at a location, you have to be out or else you get billed steep overtime. So at the last possible second, I'll shout 'its a wrap!' for our interns or assistants to wrap it up. Sometimes they pack up so quickly that they forget that the drives are attached, and the drives crash onto the floor with everything we've shot that day. They freeze, probably thinking that I'm going to get angry, but I always say, 'Just plug it back in, give it a check, and make sure it's okay.' Since I use G-Technology drives, it's always been fine. Always. I only buy gear that is durable and reliable for that exact reason."
Adler started out shooting the usual suspects at age 15, doing sessions for friends and family, covering weddings, newborns, high school upper classmen, and so on. Soon, she asked all of the professional photographers around her where to go to college and what to study.
Adler recalls, "Every single one of them said, 'Don't study photography. Study business. Anybody can take good photographs, but it takes expertise and knowledge to run a successful business."
She took half their advice and landed at Syracuse University, where she studied both Photography and Business. However, it wasn't until nearly the end of college that she took a class on fashion photography. The class would change her life. She knew she had found what her passion. Pursuing fine art was not enough. Adler craved the extra challenge of taking someone else's vision and infusing it with her own talent and creativity to make something greater.
After studying fashion photography in London, Adler returned to her home in upstate New York to run a portrait studio while working to transition into fashion photography in New York City on the weekends. She spent her nights working on fashion editorials and weddings side by side. Adler also started building a second career as an instructor.
"My focus in teaching really is to help people save the time, effort, and suffering that I have gone through, with whatever the topic might be," she says. "I want to help people learn something quickly and efficiently, so they can get back to what they really love: photography."
Adler's first class was "Organization in the Digital Age." It was her first of many courses she would teach on digital asset management. Her class instructs on how to keep things simple and set up smarter ways to track images and to protect them with archiving and multiple backups. Today, she offers classes at all major photography conferences around the USA. Her topics have broadened to creativity and how to cultivate techniques in being innovative and different. According to Adler, the biggest danger is complacency. People should never be complacent with their portfolios… or their storage.
My Best Backup
"When I first moved to New York, I was directing a few projects," says Adler. "We needed a lot of storage because my DP was shooting on RED cameras and the Phantom. I was talking to some really high-end cinematographers and DPs, and when I said, 'I don't really know what hard drives are best.' Hands down, many said, 'Get G-Technology. G-Technology has worked flawlessly for me since my first drive purchase."
Product quality and reliability are Adler's top priorities for her storage gear. She also emphasizes capacity. While the common advice is to think twice and shoot once, Adler tends to think twice and shoot ten times. Adler notes that her best captures are often during those "in-between moments" when the model's hair is mid-flip or there is a fleeting natural expression between poses.
This means gigabyte after gigabyte of raw images. For the years now, she's been using G-Technology products, Adler has typically relied on pairs of G-Drive ev RAW in the field. (As a rule, she always buys two of any drive and makes sure that one copies to the other as soon as possible.) She also added the G-Speed Shuttle with Thunderbolt™ 3 to her workflow for image backup from her laptop.
"It's so easy for me because I don't have to carry a big clunky system on location to back my data up," says Adler. "My assistants can just back up my files. When I come home and I plug it in. It is set to mirror already, so I'm all set. It's portable and simple. Essentially, it lets me skip a step from my previous workflow."
From camera to archiving, Adler shoots tethered to her G-Drive ev RAW, which will automatically mirror to a second ev drive. Once home, she backs up the G-Drive ev RAW to her G-Speed Shuttle with integrated RAID redundancy. The 1st G-Speed Shuttle is actually is backed up to a second G-Speed Shuttle. A quadruple backup system isn't actually complicated. All mirroring is automatically done and managed through backup utility apps. Shots that are selected by clients or subsequently retouched (perhaps 2% of her total image volume) also get backed up to photoshelter.com for cloud storage.
Clean, Bold, Graphic
If Lindsay Adler's methods sound a little extreme, keep in mind that she practices what she preaches. Assiduous backup has been part of her workflow ever since starting out as a teen. She is one of the "lucky" few professional photographers who have never lost a single work image. Even if a drive should break or get lost, she has always been prepared with backups.
Her storage practices align with her choice of storage brand and her professional style. Her works focus on a single element, one striking concept. That element comes to life in the image without distracting backgrounds or unnecessary texturing.
"I like to describe my style as clean, bold, and graphic."
Aesthetically, Adler also loves the design of G-Technology drives and how they stack to keep her environment simple and refined. She doesn't endorse specific brands to her students, but she does recommend spending the extra money to get the right solutions.
"Especially starting out, you want to go with the less expensive," Adler says, "but it's like using a less expensive camera or the less expensive strap. If the camera strap breaks, you break your camera. The same follows with drives. If your drive fails, you lose your files. That extra investment is worth having more peace of mind."
G-Team members are leaders in their respective fields who use G-Technology products in their day-to-day work lives. G-Team members are compensated for their participation.
G‐Technology external hard drives serve as an element of an overall backup strategy. It is recommended that users keep two or more copies of their most important files backed up or stored on separate devices or online services.
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Why high decibel campaign by Opposition with Arvind Kejriwal as leader will fail
Bengaluru, First Published Feb 12, 2020, 6:42 PM IST
Will the TRS, TMC, DMK and AAP come together to rehash the failed front model all over again? This would mean that this alliance would pitch Arvind Kejriwal on the national scene as a challenger to Prime Minister Narendra Modi.
In the aftermath of the victory in Delhi by the Aam Admi Party, there are talks about the 2024 Lok Sabha elections.
Ahead of the 2019 elections, almost all parties had come together to defeat the BJP, but they failed miserably in their endeavour. Now, some feel that there is a possibility of the TRS, TMC, DMK and AAP coming together to rehash the failed front model all over again.
This would mean that this alliance would pitch Arvind Kejriwal on the national scene as a challenger to Prime Minister Narendra Modi.
K Krishna Saagar Rao, chief spokesperson of the BJP, Telangana state, feels that these parties have an inherent personal greed for selfish power, which can never bring them together as an electoral coalition.
If the regional parties should get together against the BJP, they necessarily need a national party to galvanise them. Congress is on the death bed and its leader draws no respect or power to guide and lead a large coalition of regional parties, Rao said.
Arvind Kejriwal is not a generational politician, who can work together with the age-old family political enterprises and shrewd regional political satraps.
BJP doesn't believe a functional regional coalition or a front is possible in contemporary Indian politics, while there will surely be a fresh attempt to restart a high decibel campaign against the saffron party nationally, Rao further contended.
While the AAP's record has been good in the Delhi Assembly circuit, there is not much to speak about in the regional scene or even in other states. Kejriwal has campaigned unsuccessfully outside of Delhi several times, but has failed to make any impact.
In the 2014 elections, the AAP won just four of the 434 seats it contested. Moreover, all these wins were in Punjab and not Delhi, where it had bagged 32 seats in the 2013 Delhi elections.
In the 2017 Goa elections, the party lost all seats and in Haryana, of the 40 seats it contested, the party managed to win just one. The case was similar in Maharashtra where it lost all the 24 seats it contested.
The AAP made some adjustments in the 2019 Lok Sabha polls. The party fielded candidates in Delhi, Chandigarh, Bihar, Goa, Punjab, Andaman and Nicobar, Haryana, Odisha and Uttar Pradesh. However, only one of its candidates from Sangrur in Punjab was able to win the polls. In all, the AAP had fielded 40 candidates.
Last Updated Feb 12, 2020, 6:42 PM IST
Indian government to take more measures to reduce road accidents and deaths


Horatio Lovejoy's New Year's Eve is a tiny pamphlet touching on a lot of topics. It's a story of survival, with three young people lost in a prairie snowstorm. It's a description of the intense winter weather midwestern Americans endured from December 1863 into the first week of 1864. It's a glimpse of surgical practices in 1864 and of how a disabled man functioned in the late nineteenth century. It's an example of a work sold for charity in the early 1880s.
Amelia L. Lovejoy was about 12 years old on New Year's Eve, 1863; her death date is given as 3 January 1864, on her tombstone in Zion-Smack Cemetery, Whiteside County, Illinois [genealogytrails.com/ill/whiteside/cemzionsmack.html]. Horatio was 18. His life story points up that nineteenth-century Americans didn't necessarily spend their lives in one place. Horatio Phineas Lovejoy was born in 1845 in Illinois, of parents born in New England. Having apparently spent much of his life in Illinois, in 1883 he married Martha Rorabaugh (1860-1929), and the couple lived in Iowa. [See James Hill's query at genforum.genealogy.com/lovejoy/messages/908.html and James Hamilton Carroll's family tree for James and Mary Rorabaugh at www.connectingourkin.com/rora004.htm] In 1900, they were living with three children (they had seven; at least three died in infancy) near Martha's mother in what would become Oklahoma. [James Hamilton Carroll. "Connecting Our Kin." www.connectingourkin.com/meclark.htm. Also, family tree at rootsweb.com] Horatio died in Kirksville, Missouri, in 1918.
My copy seems different from the library copies described: some of those are 11 or 13 pages; one is 20. Mine is 17. This may reflect differences in the way the number of pages has been calculated (I didn't include the end papers or the cover, which acts as the title page), or it may be due to differences in printings.
"Horatio Lovejoy's New Year's Eve. 1863-1864." Charles City, IA: Advocate Steam Job Print, 1882.
Advocate Steam Job Print, Charles City, Iowa.
Wm. LANE, Ex-Judge of County Court.
W. A. THATCHER, County Treasurer.
Who does not remember the terrible New Year's night when 1863 died in a whirling roar of freezing wind and blinding snow? You who were safe beside your warm fires, or snugly covered in pleasant beds, thanked God you were safe at home, and fervently hoped that none were abroad that fearful night. Some unfortunates were abroad, and never told to earthly ears the story of the battle with Death, in which they went down before his icy breath. Some survived to tell us what they survived. The story of how three children passed that night, and who they were, is the subject of this pamphlet.
In 1834, Mr. Phineas Lovejoy (grandfather to Horatio) and his family came from Weston, Vermont, to the "Yankee Prairie," now called Vergennes, in Jackson county, Illinois. In 1834, Rev. D. Wells and family, including his only brother, settled in the same place, followed in 1837 by his mother and two of his sisters, from Lebanon, New Hampshire.
too much confidence in the honesty of others; and when he died at Thebes, Alexander county, January 15, 1853, he had but little to show for all the hard work he had done, and left his wife and three children but poorly provided for a campaign against the hardships of life.
In August, 1854, Mrs. Lovejoy, with her sons Elbrige and Horatio, aged eleven and eight years, and her daughter, Amelia, three years old, came to Kingsbury Grove, in the western part of Whiteside county, where she had a sister living, and where wheat bread was plentier, and fever and ague not so plenty, as in Southern Illinois.
A hundred dollars saved from the wreck of Daniel Lovejoy's business, invested in land, and afterwards sold at a profit, enabled Mrs. Lovejoy to build a small house in the village of Mineral Springs, and hard work and strict economy fed and clothed her little flock, who soon grew large enough to earn something toward their support.
In December, 1863, nearly all the young men in and around Kingsbury Grove were in the army; among the number, Elbrige Lovejoy had enlisted that fall, and was in camp at Springfield.
for New year's presents. They were all up early and ready to start before daylight, as they had fifteen miles to drive, make their purchases and return. The weather was not cold, but snow was beginning to fall. Mr. Payne advised them not to start, as everything indicated that a severe storm was coming on; but they were all ready. Caution belongs not to youth; they could not forego the pleasure of the ride for fear of a little snow.
They got to Morrison without trouble, though the storm increased and the wind changed from the east to the north-west, and increased in violence.
down the left bank of Linn Creek, and along the neck of land between that and Rock Creek, until they got into the snow so deep they could not go on further.
When Horatio found out his mistake, and tried to turn the team around and regain the road, he found the snow so deep that the already tired horses could not stir. Horatio only knew that they were not on the road, but not where they were.
They were lost! They were far from any help, they could not see which way to turn, and they could not go if they could see; a terrible fear began to creap [sic] over them and they called loudly for help. Amelia, almost insane with terror, screamed wildly, but the fierce wind screamed louder than they, and drowned their cries with mocking voices, as if the resistless forces of Nature were combined to destroy them. They saw death in awful shape coming towards them; they covered themselves with robes and blankets; they hid their eyes and stopped their ears, but the biting, shrieking, eager blast made itself heard and felt through all.
left all alone; and what Uncle William and aunt Eliza would do, and say. At length drowsiness began to overcome them. Amelia, never very strong, was more effected [sic] than Horatio or Sophronia, who were both very healthy and strong—but they knew that to sleep was death and fought against the feeling. They found they were not frozen, and began to hope they might live till morning and be saved; they encouraged each other, and, remembering the teachings of their pious parents, they prayed to the God of the fatherless to pity them and deliver them to their homes, which were never so dear to them as then, and to their mother, who now as never before, they loved and longed for.
They asked God to save them, if it was His will to save their bodes; if not, to take their souls to the heaven where many of their friends had gone before them, and to comfort the dear ones left to mourn them. They were calmed and comforted; they knew that God could help them and enable them to help themselves.
light found them able to move, but terribly suffering—between life and death. Indeed, it was a horrifying picture.
As soon as Horatio could see, he got out the sleigh and tried to get one of the horses unhitched, and turned around to ride for help, but the snow had drifted onto and around them so high that they could not move, but had saved them from perishing. The wind and storm still continued.
That New Year's morning was one of the coldest ever known all over the WEST. From Lake Superior to the GULF it was THE cold day. Soldiers froze to death between Nashville and Chattanooga, on the cars; guards on duty at Memphis, froze their hands and feet; at Huntsville, Alabama, the ground was frozen deeper than it had been for forty-five years; more than a dozen people froze to death in Wisconsin. A family started that morning with an ox team and sled to go from Rush River to Prescott, Wisconsin; the father, mother and child perished with cold, leaving three little boys with their arms and legs frozen off.
bluff, where the thickets partly broke off the wind; and he could have gone faster, but his feet were frozen and his hands like sticks. He was freezing all over; still he struggled on a mile and a half further, to Benjamin Hubbard's the first house, and kicked his frozen feet against the door, when bidden to come in he could not, but making a last possible effort he kicked once more against the door, when Mr. Hubbard opened it and pulled him in almost lifeless; he could not speak nor move.
Mr. Hubbard saw his condition and took him into a cold room, and the family chafed him with snow and cold water, and gave him stimulants, till he revived sufficiently to tell them where the girls were, and to beg them to hurry to their relief. Mr. Hubbard, as soon as possible, got out his team and started for them, and sent to his neighbors for help.
He found the girls as Horatio had left them, but freezing. He put them into the sleigh and drove home as fast as possible, but froze his face and hands before he could get there. Some men, with hard work, got the team out of the snow, and got it to Mr. Hubbard's; but the horses were so much injured by the exposure that they could not be used all winter.
en hard and she seemed chilled to her vitals.
Word was sent to Mrs. Lovejoy, and she went to her children. When she got to Amelia's side and clasped her dear child in her arms, Amelia kissed her and said, "Oh ma! ma! I thought I should never see you again. I am happy now; all I wanted was to see you once more." She feared not so much to die as that she might never see her mother.
Horatio said: "Mother, we prayed to the Lord and he saved us." And who dare say that the ear of the Almighty was not open to the cry of these poor children, and that His hand was not reached out to help them. More help than could be used was offered to take care of the children.
The third of January they were all taken to Mr. Payne's. Sophronia was able to walk from the sleigh into the house, and in a few days was as well as ever. Horatio and Amelia could not help themselves. Horatio's hands and feet turned black and threatened to slough; one of Amelia's feet did the same but otherwise she seemed to be doing well.
Sunday morning Amelia complained of being sick, and appeared a little deranged, or at least light headed; at noon she was worse. Dr. Taylor, of Morrison, had been sent for. He feared congestion of the brain, and at night pronounced her in danger, although to unprofessional eyes, beyond a little wandering of the mind, she did not appear very ill.
Through the night she grew worse, and on Monday morning it was plain to all that she was in great danger; at five she died—frozen to death! A good, intelligent girl, very much liked by those who knew her, and dearly loved by all near to her—a flower-bud cut off just as it was opening; her mother's dearest hope; her brother's most cherished darling—gone in suffering and delirium, never again to be kissed, or petted, or played with, no more to be cared or toiled for; in this world never to be seen or heard.
Whatever we may believe the great hereafter to be, is but a small consolation in the first hours of grief. What we lose we know, but what the lost one gains we can only guess. Earth and our loved ones are here, close to us; we see, and feel, and hear them; we clasp their dear hands, kiss their lips and we know they love us; but when they die and are buried out of sight, we know that as long as we live, we shall see their faces no more.
On Wednesday Amelia was buried.
wait till the dead flesh seperated [sic] from the sound, so that as much as possible might be saved.
Mrs. Lovejoy told Horatio what the doctor said, and that it was only possible that he could get well, and that he must lose his limbs. Horatio was not frightened or in despair, but said he thought he should live and be company for his mother, if nothing more, that the same God who kept them alive through that dreadful night would still provide for him.
Instead of needing comfort and encouragement, he comforted his mother and kept her from despair by his resignation and hopefulness.
On Sunday, Dr. Taylor, assisted by Drs. Nowlen and Roberts, in four hours of very cautious and careful work, took off his left foot above the ankle, and his right foot at the instep, and both hands across the palms, save for a little of each thumb. It was a very hazardous and difficult operation, and from the great care necessary, very long. The surgeons exercising the utmost care in saving to the poor boy all of his members that could be saved, and to take away all that could not be saved.
he lay in bed, tended by the best of nurses, a mother, assisted by Mrs. Payne and by the neighbors, who did all that a kind-hearted people could do for the afflicted mother and son.
Mr. Payne was almost entirely disabled by rheumatism, and all his sons were in the army, so there was no one on the place but women to do anything, but now the kindness and generosity of the farmers of Newton came out strong and bright; one day they made a bee and cut and hauled wood enough to last several Weeks, and it required a good deal to keep three fires night and day in that cold winter.
When the wood pile got low, the boys from Kingsbury school came and cut another lot. Some one came every day to find out if anything was needed. They, God bless them, did not wait to be asked for help but sought an opportunity to do good. Mr. I. B. Emmons and O. A. Root rode over the country collecting offerings of the people, who gave freely to help this widow and fatherless boy.
Mr. William Prothrow took means to get Elbrige Lovejoy discharged, and was successful. He got home the first of April; and Horatio having by that time got able to be removed, Mrs. Lovejoy went again into her own house, after an absence of three sad, long months.
ial limbs for Horatio, which has been expended in getting an artificial leg, and getting it repaired at different times.
It would naturally be thought that one so crippled would be a helpless burden on his friends; but such is not the case. Horatio can dress himself, he can harness and drive or ride a horse, he can feed himself, and do many things that seem impossible for one without fingers or feet.
His pluck and powers of endurance are great. Before his affliction he was very industrious, and though very small, could do a man's work at any kind of labor, and was sought after by the neighboring farmers; and he always gave his mother all his earnings to assist in supporting the family; and now he is a living rebuke to all grumbling, discontented boys, who having feet to run on, and hands to work and play with, are dissatisfied with their lot in life, while this boy, who is cut off from all that makes life pleasant to youth, and cannot even labor for himself or mother, bears his great loss and privations without a complaint, and s always cheerful and ready to do anything he is able to do, which, alas! is but little; and he now seeks, by selling this little book, to assist his mother and himself to live.


The Caprari range of borehole pumps are one of the world's most advanced submersible pumps. The pumps are designed with long term efficiency and durability. The Caprari range of submersible pumps exceeds the European efficiency standards for electrical consumption and output factors. The pump range is available in Duplex stainless steel, Stainless steel, Bronze, cast iron and Noryl resin glass plastic.
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Up thrust preventers on selected models.
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Caprari presents the "E" Series, a complete range of electric submersible pumps with many versions to respond to the ever changing and more sophisticated needs of professionals in the water world.


text:
Giada Pamela De Laurentiis (; born August 22, 1970) is an Italian-American chef, writer, and television personality. She was the host of Food Network's Giada at Home. She also appears regularly as a contributor and guest co-host on NBC's Today. De Laurentiis is the founder of the catering business GDL Foods. She is a winner of the Daytime Emmy Award for Outstanding Lifestyle Host and the Gracie Award for Best Television Host. She was also recognized by the International Hospitality Institute as one of the Global 100 in Hospitality, a list featuring the 100 Most Powerful People in Global Hospitality.

Early life

Giada Pamela De Benedetti was born on August 22, 1970, in Rome, Italy, the eldest child of actress Veronica De Laurentiis and her first husband, actor-producer Alex De Benedetti. De Benedetti was a close associate of Giada's maternal grandfather, film producer Dino De Laurentiis. As a child, Giada often found herself in the family's kitchen and spent a great deal of time at her grandfather's restaurant, DDL Foodshow. Her parents were married in February 1970 but were later divorced. After her parents' divorce, Giada and her siblings moved to Southern California, where they took their mother's surname. After graduating from Marymount High School in Los Angeles, De Laurentiis attended the University of California, Los Angeles, earning her bachelor's degree in social anthropology in 1996.

Her maternal great-grandmother was English and her grandmother was British-Italian film star Silvana Mangano. Her paternal grandmother, Pamela De Benedetti née Leslie-Jones (1923–1998), was also English. Her siblings include sister Eloisa, a makeup artist, and brothers Igor and Dino Alexander II, a Hollywood film editor who died of melanoma in 2003. Her stepfather is producer Ivan Kavalsky.

Career
De Laurentiis studied at Le Cordon Bleu in Paris, with aspirations of becoming a pastry chef. After returning to the United States, she became a professional chef working in several Los Angeles restaurants, notably the Wolfgang Puck-owned Spago. She later worked as a food stylist and was contacted by the Food Network after styling a piece in Food & Wine magazine in 2002.

Her daytime cooking show on the Food Network, Everyday Italian, premiered April 5, 2003. On Chefography, a Food Network biography program, she said she never wanted to be in her "family business" of show business, and that she felt uncomfortable in front of the camera when she first began hosting Everyday Italian. When the program first aired, the Food Network received mail accusing the network of hiring a model or actress pretending to cook instead of a real chef.

De Laurentiis began hosting Behind the Bash in October 2005. The program examines the catering process behind big event extravaganzas such as the Grammy Awards. In January 2007, a third De Laurentiis-hosted show, Giada's Weekend Getaways, debuted on Food Network. On this show, De Laurentiis travels to a featured locale (including Seattle, South Beach, San Francisco, Napa, and Jackson Hole, Wyoming) and visits her favorite local culinary destinations. On a November 2006 episode of Iron Chef America, De Laurentiis and Bobby Flay competed against, and were defeated by, Rachael Ray and Mario Batali.

In 2007, De Laurentiis appeared as a presenter at the inaugural Food Network Awards. In June 2007, she hosted a two-part Food Network special entitled Giada in Paradise, featuring the two locales of Santorini, Greece and Capri, Italy. De Laurentiis also made several appearances as a guest judge on the third season of The Next Food Network Star, which aired in 2007. That year she was dubbed a "petite powerhouse" by Town & Country magazine, standing "just under five-foot two". In 2008, she won a Daytime Emmy Award for Outstanding Lifestyle Host.

In 2008, De Laurentiis and the Barilla Group, launched an Italian gourmet line under the Academia Barilla name—Barilla's first-ever gourmet celebrity product line. That same year, Giada at Home premiered, showing De Laurentiis in a kitchen preparing meals and parties for family and friends. The show is shot on a set that is very similar to her own home. She joined fellow chef Bobby Flay as a judge in season 7 of Food Network Star and took on a new role in season 8 as a team leader of five cooks competing against Bobby Flay's and Alton Brown's respective teams.

In 2009, De Laurentiis became the voice of "Paulette", a character on the animated children's show Handy Manny. In early 2010, De Laurentiis came out with a line of kitchen supplies, exclusively for Target. That same year, CafeMom ranked her as #6 on their yearly "Sexiest Moms Alive" list. In June 2010, De Laurentiis became a regularly appearing mentor to the finalists on the popular Food Network competition show The Next Food Network Star.

In July 2014, De Laurentiis opened her first restaurant, called GIADA, inside The Cromwell in Las Vegas, Nevada. The restaurant offers seating in the dining room, lounge, or outdoor patio with views of the Bellagio fountains and Caesars Palace. The GIADA menu includes Italian cuisine with Californian influences, including "lemon spaghetti, chicken cacciatore, marsala herb chicken meatballs, rosemary focaccia and lemon flatbread and vegetable Bolognese rigatoni". Family-style, vegan, and gluten-free options are also available, as well as an antipasto station. Restaurant guests can watch chefs prepare food from the open kitchen.

In 2014, De Laurentiis voiced the character of Gelata in the U.S. version of the Tinker Bell special Pixie Hollow Bake Off.

In 2018, De Laurentiis voiced herself in the movie Scooby-Doo! and the Gourmet Ghost.

On June 26, 2020, De Laurentiis' Food Network series Giada Entertains won a Daytime Emmy Award as Outstanding Culinary Series. On July 26, 2020, De Laurentiis won the Daytime Emmy for Outstanding Culinary Host.

On February 10, 2023, De Laurentiis has signed a multi-year deal to create and produce unscripted programming with Amazon Studios. Later in February, De Laurentiis was named as the godmother for the Oceania Cruises' ship Vista, the first new ship from the cruise line in more than a decade.

Products
De Laurentiis had a range of products made exclusively for Target. In 2013, ceramic lasagna pans under the Giada De Laurentiis brand were recalled after they cracked and at least six people were injured. In 2022, Target only seems to carry her books, including a Target edition of Eat Better, Feel Better.

Restaurants
De Laurentiis has two restaurants on the Las Vegas Strip. She opened her first restaurant in May 2014. The restaurant, simply called "Giada", is located in the Cromwell. She opened a second restaurant in early 2018. "Pronto by Giada", taking up residency in Caesars Palace, is a fast casual dining restaurant.

In July 2017, she announced she would also be opening a restaurant in the Horseshoe Casino in Baltimore, Maryland, just north of Washington, D.C. It is called GDL Italian by Giada, and opened in May 2018.

Personal life

On May 25, 2003, De Laurentiis married Todd Thompson, a fashion designer. The couple's only child, Jade Marie De Laurentiis-Thompson, was born on March 29, 2008. Jade was named after her mother - "Giada" is Italian for "Jade." While acknowledging how special the experience of having a child has been, she has stated that she does not plan to have more children. De Laurentiis remains focused on her career, saying, "I have so many babies: the show, Todd, and Jade." On December 29, 2014, De Laurentiis announced on her website that she and Todd had separated the previous July, and had decided to end their marriage. The divorce was finalized on September 3, 2015. Since November 2015, she has been romantically involved with TV producer Shane Farley.

Publications

Both Giada at Home and Weeknights with Giada landed in the number one position on The New York Times Best Seller list.

Awards and nominations

References

External links

 Food Network profile

1970 births
Alumni of Le Cordon Bleu
American cookbook writers
American food writers
American people of English descent
American television chefs
Daytime Emmy Award winners
Food Network chefs
Italian emigrants to the United States
Italian people of English descent
Living people
People of Campanian descent
University of California, Los Angeles alumni
American women chefs
Women cookbook writers
Writers from Rome
Giada
American women non-fiction writers
Chefs from California
21st-century American businesswomen
21st-century American businesspeople
21st-century American non-fiction writers
Chefs from Los Angeles

redpajama_set_name: RedPajamaWikipedia


Practice Plant Reproduction questions with answers. This prep guide includes questions like, why is reproduction important?, what is meant by word extinct?, which plants and animals would survive?, what is sexual reproduction?, what is fertilization? and others below.
Question: Why is reproduction important?
Answer: Everyone has to die one day, so reproduction is very important because without reproduction the species would extinct and the generations won't be carried further on earth.
Question: What is meant by word extinct?
Answer: The word 'extinct' means no longer living on earth. Extinction is a period when a specie or a group of animals finishes ir dies, leaving no offspring on earth.
Question: Which plants and animals would survive?
Answer: Plants and animals that will survive on earth would be only those which reproduce. Their generations will carry on and they won't get extinct.
Question: What is sexual reproduction?
Answer: Many plants reproduce sexually. Sexual reproduction is a type of reproduction in which two parents take part i.e. male and female.
Answer: Fertilization is a process when a male gamete (cell) joins with the female gamete (cell). Fertilization only occurs in sexual reproduction.
Which plants and animals would survive?
What are red blood cells made up of?
Does bronchi and bronchioles process air?


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text:
Verify conformance of machined work to specifications, using measuring instruments, such as calipers, micrometers, or fixed or telescoping gauges.
51-4193.00 Plating Machine Setters, Operators, and Tenders, Metal and Plastic
Examine completed objects to determine thicknesses of metal deposits, or measure thicknesses by using instruments such as micrometers.
Monitor and measure thicknesses of electroplating on component parts to verify conformance to specifications, using micrometers.
51-4033.00 Grinding, Lapping, Polishing, and Buffing Machine Tool Setters, Operators, and Tenders, Metal and Plastic
Inspect or measure finished workpieces to determine conformance to specifications, using measuring instruments, such as gauges or micrometers.
Measure workpieces and lay out work, using precision measuring devices.
51-9041.00 Extruding, Forming, Pressing, and Compacting Machine Setters, Operators, and Tenders
Examine, measure, and weigh materials or products to verify conformance to standards, using measuring devices such as templates, micrometers, or scales.
Measure arbors and dies to verify sizes specified on work tickets.
51-4041.00 Machinists
Calculate dimensions or tolerances, using instruments, such as micrometers or vernier calipers.
Measure, examine, or test completed units to check for defects and ensure conformance to specifications, using precision instruments, such as micrometers.
51-9061.00 Inspectors, Testers, Sorters, Samplers, and Weighers
Measure dimensions of products to verify conformance to specifications, using measuring instruments, such as rulers, calipers, gauges, or micrometers.
Inspect, test, or measure materials, products, installations, or work for conformance to specifications.


text:
Comedy Film
Pop-Up Screens Outdoor Screening: Ghostbusters
Coram's Fields, Russell Square
They ain't afraid of no ghost
"What's not to like?"
- The Guardian
Pop Up Screens aim to bring brilliant films to cool locations. For this one head to Coram's Fields at Russell Square for a lot of lovely food, amazing popcorn and a fully stocked bar; but if you'd prefer to bring your own picnic that's ok, but it must be noted you will be missing out on burgers from the guy that supplies Harrods. For this one, Dan Aykroyd, Harold Ramis, and Bill Murray are suited and booted to fight supernatural crime in the 1984 classic, Ghostbusters. The Ghostbusters mix equal parts comedy and slime. And, what with a giant marshmallow man, super-charged proton packs and a theme song that always gets a reply, this is not to be missed. "A cavalcade of pure joy." – Time Out "As entertaining today as it ever was, a classic to be treasured by all." – Empire Magazine You do need to bring something to sit on, there's always plenty of room so you can get inventive. Please don't bring a tent though... it's just showing off.