Show $S_7$ is isomorphic to the subgroup of all those elements of $S_8$ which leave the number $5$ fixed First of all I don't know how to write with correct notation in this site, sorry about that.
Second, I am a beginner with group theory so I am trying to solve this exercise (Fraleigh).
Until now I have a function that maps a permutation in $S_7$ to $S_8$ like this:
Suppose $\sigma$ , any permutation in $S_7$.
$5$ if $x=5$
$\sigma(x)$ if $\sigma(x)\neq 5$
Here I have the first problem, because I have two cells free: $8$ and the element that is sent to $5$ in $S_7$, so I maps like this:
$\sigma(5)$ if $\sigma(x)=5$
$8$ if $x=8$
So I would appreciate your help in this exercise, I don't know if this function is correct and how to prove if in fact is an isomorphism.
 A: I'm sure this has been answered here before but, since I can't find a duplicate, I'll give a CW answer.
There's nothing special about $7$, $8$, and $5$ here.

Theorem: Let $n\in \Bbb N$ and $m\in\{1,\dots, n+1\}$. Then $S_n$ is isomorphic to the subgroup $H$ of $S_{n+1}$ whose elements fix $m$.

Proof: We may assume, by reindexing if necessary, that $m=n+1$.
We have that
$$\begin{align}
\varphi: H&\to S_n,\\
\sigma &\mapsto \sigma_H,
\end{align}$$
where
$$\sigma_H(x)=\sigma(x)$$
for $x\in \{1,\dots, n\}$, is an isomorphism, since it is clearly a bijection and
$$\begin{align}
(\varphi(\tau\rho))(x)&=(\tau\rho)_H(x)\\
&=
(\tau\rho)(x)\\
&=
\tau(\rho(x))\\
&=\tau(\rho_H(x))\\
&=\tau_H(\rho_H(x))\\
&=(\varphi(\tau))((\varphi(\rho))(x))\\
&=(\varphi(\tau)\varphi(\rho))(x)
\end{align}$$
for all $x\in \{1,\dots, n\}$ and all $\tau,\rho\in H$, so that
$$\varphi(\tau\rho)=\varphi(\tau)\varphi(\rho).\,\square$$
A: For every $a\in A:=\{1,\dots,n\}$, set $B:=A\setminus\{a\}$. Then, $\operatorname{Stab}(a)\stackrel{\varphi}{\cong} S_B$ via $\sigma\mapsto\varphi(\sigma):=\sigma_{|B}$. In fact, firstly, $\sigma_{|B}\in S_B$ because $\sigma \in\operatorname {Stab}(a)$ (good definition). Moreover:

*

*injectivity: $\varphi(\sigma)=\varphi(\tau)\Longrightarrow \sigma_{|B}=\tau_{|B}\stackrel{\sigma(a)=a=\tau(a)}{\Longrightarrow}\sigma=\tau$;

*surjectivity: for any given $f\in S_B$, define $\sigma\in S_n$ by $(\sigma(a):=a) \wedge (\sigma_{|B}:=f)$; therefore, $\sigma\in\operatorname{Stab}(a)$ and $\varphi(\sigma)=f$;

*operation-preserving: for $b\in B$:
\begin{alignat}{1}
(\sigma\tau)(b) &= \sigma(\tau(b)) \\
&\stackrel{b\in B}{=}\sigma(\tau_{|B}(b)) \\
&\stackrel{\tau_{|B}(b)\in B}{=}\sigma_{|B}(\tau_{|B}(b)) \\
&=(\sigma_{|B}\tau_{|B})(b) \\
\end{alignat}
On the other hand, $b\in B\Longrightarrow (\sigma\tau)(b)=(\sigma\tau)_{|B}(b)$. Therefore:
$$\varphi(\sigma\tau)=(\sigma\tau)_{|B}=\sigma_{|B}\tau_{|B}=\varphi(\sigma)\varphi(\tau)$$
Now, note that $|B|=n-1$, and hence $S_B\stackrel{\phi}{\cong} S_{n-1}$ for $\phi(f):=\mathscr gf\mathscr g^{-1}$, where $\mathscr g\colon B\longrightarrow \{1,\dots,n-1\}$ is any bijection. So, finally, for every $a\in A$:
$$\operatorname{Stab}(a)\stackrel{\phi\varphi}{\cong} S_{n-1}$$
Yours is just the particular case $n=8$ and $a=5$.

