Weyl group isomorphic to general linear group Suppose that $p$ is a prime number, and let $E_{n}$ be the translation group on the abelian group $(\mathbb{Z}_{p})^n$. Denote $S=S_{p^n}$ by the symmetric group on $p^n$ elements of $(\mathbb{Z}_{p})^n$. Prove that the Weyl group $W_{S}{(E^n)}=N_{S}{(E^n)}/C_{S}{(E^n)}$ is isomorphic to the general linear group $GL(n, \mathbb{Z}_{p})$.
Here $N_{S}{(E^n)}$ is the normalizer of group $E^n$ in group $S$ and $C_{S}{(E^n)}$ is the centralizer of group $E^n$ in group $S$. 
I think the First Isomorphism Theorem can be used, but I don't know where to start. 
 A: $\newcommand{\F}{\mathbb F} 
\newcommand{\GL}{\operatorname{GL}}
\newcommand{\id}{\operatorname{id}}$
We first describe a map $f\colon N_S(E^n)\rightarrow \GL(n,\F_p)$, $\sigma \mapsto f_\sigma$ as follows:
For any $v\in \F_p^n$ denote by $t_v\in E^n$ the translation by $v$, i. e. $t_v(u) = u+v$ for all $u\in \F_p^n$. Let $\sigma\in N_S(E^n)$. For any $t_v\in E^n$ there exists $f_\sigma(v)\in \F_p^n$ such that $\sigma\circ t_v\circ \sigma^{-1} = t_{f_\sigma(v)}$ (since $\sigma$ normalizes $E^n$). We first show that $f_\sigma\colon \F_p^n\rightarrow \F_p^n$ is linear: For $v,v'\in \F_p^n$ we compute
\begin{align*}
t_{f_\sigma(v+v')} &= \sigma\circ t_{v+v'}\circ\sigma^{-1} = \sigma \circ t_v \circ t_{v'}\circ \sigma^{-1}\\
&= (\sigma\circ t_v\circ\sigma^{-1})\circ (\sigma\circ t_{v'}\circ \sigma^{-1}) = t_{f_\sigma(v)} \circ t_{f_\sigma(v')} = t_{f_\sigma(v) + f_\sigma(v')}.
\end{align*}
Hence $f_\sigma(v+v') = f_\sigma(v) + f_\sigma(v')$ and $f_\sigma$ is linear.
We now show that $f_{\sigma\rho} = f_\sigma\circ f_\rho$ for all $\sigma,\rho\in N_S(E^n)$. Let $\sigma,\rho\in N_S(E^n)$ and $t_v\in E^n$ be arbitrary. Then we compute
\begin{align*}
t_{f_{\sigma\rho}(v)} &= (\sigma\rho)\circ t_v\circ (\sigma\rho)^{-1} = \sigma\circ \bigl(\rho\circ t_v\circ\rho^{-1}\bigr)\circ \sigma^{-1}\\
&= \sigma\circ t_{f_\rho(v)}\circ \sigma^{-1} = t_{f_\sigma(f_\rho(v))}.
\end{align*}
Hence $f_{\sigma\rho}(v) = f_\sigma(f_\rho(v))$ for all $v\in \F_p^n$, i. e. $f_{\sigma\rho} = f_\sigma\circ f_\rho$.
In particular, $f_\sigma$ is invertible with inverse $f_{\sigma^{-1}}$, since $f_1 = \id_{\F_p^n}$; and hence $f_\sigma\in \GL(n,\F_p)$ for all $\sigma\in N_S(E^n)$.
We have shown that $f\colon N_S(E^n)\rightarrow \GL(n,\F_p)$ is a well-defined group homomorphism. It is surjective, for if $\sigma\in \GL(n,\F_p)\subseteq S$, then 
$$
(\sigma\circ t_v\circ \sigma^{-1})(u) = \sigma\bigl(\sigma^{-1}(u) + v\bigr) = u + \sigma(v) = t_{\sigma(v)}(u),
$$
for all $u,v\in \F_p^n$. Therefore, $\sigma\in N_S(E^n)$ and $\sigma = f_\sigma$.  
It suffices to compute the kernel of $f$. Let $\sigma\in N_S(E^n)$ such that $f_\sigma = \id_{\F_p^n}$. Then for all $t_v\in E^n$ we have
$$
\sigma\circ t_v\circ \sigma^{-1} = t_{f_\sigma(v)} = t_v,
$$
which shows that $\sigma\in C_S(E^n)$. Therefore, by the first isomorphism theorem, we have an isomorphism
$$
W_S(E^n) = N_S(E^n)/C_S(E^n)\longrightarrow \GL(n,\F_p).
$$
