Isomorphism of stabilizer over $S_n$ to $S_{n-1}$

Question

It is in my understanding that to prove that the stabilizer of $a \in A=\{1,...,n\}$ over the permutation group $S_n$ is isomorphic to the permutation group $S_{n-1}$ one needs to show that there exists a bijection between these two groups, which is easy. However, the other step is to show that a homomorphism defined on this bijection exists.

I would think that one such homomorphism can be defined as follows: for every permutation $\sigma \in S_n$, redefine $\sigma(a)$ such that $\sigma(a)=a$, then redefine the element which was initially pointing to $a$ in $\sigma$ to point to the element to which $a$ was initially pointing in $\sigma$.

But the question is: how to define this homomorphism formally?

Answer 1

Fix $a \in \{1,2,\cdots, n\}$. It is clear that $$Stab_{S_{n}}(a)=\{\sigma \in S_{n}: \sigma(a)=a\}.$$ Removing form each permutation of the above set the pair $(a,a)$ we have the symmetric group on $\Omega= \{1,2,\cdots,a-1,a+1,\cdots,n\}$, and $|\Omega|=n-1$. Notice that the isomorphism type of a symmetric group depends only on the cardinality of the underlying set being permuted. So, $Stab_{S_{n}}(a)\cong S_{n-1}$.

Now, to see that the symmetric groups $S_{\Delta}$ and $S_{\Omega}$ are isomorphic if $|\Delta|=|\Omega|$, define $$\phi: S_{\Delta} \rightarrow S_{\Omega} \;\; \text{by} \;\; \phi(\sigma)=f\circ \sigma \circ f^{-1}, \; \forall \sigma \in S_{\Delta}$$ where $f$ is a bijection between $\Delta$ and $\Omega$. Then prove the following:

1. $\phi$ is well defined.
2. $\phi$ is a bijection from $S_{\Delta}$ to $S_{\Omega}$. (Find a two-sided inverse for $\phi$.)
3. $\phi$ is a homomorphism.

(It is exercise 10 in chapter 1.6 from Dummit and Foote's Abstract Algebra. )

Answer 2

The answer by Bill Moustakas is very good. Another way to do this is to use the fact that if $G$ acts on a set $X$ and $G_x$ is the stabilizer of $x\in X$, then $gG_x g^{-1}=G_{g.x}$ (be sure you know why this is true).

Applying this to your situation, take $\tau=(a,n)\in S_n$. Then $\tau(S_n)_a\tau^{-1}=(S_n)_{\tau(a)}=(S_n)_{n}=S_{n-1}$. As conjugate subgroups are isomorphic, you are done.

Answer 3

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 $a\in A$: $$\operatorname{Stab}(a)\stackrel{\phi\varphi}{\cong} S_{n-1}$$