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