What is this group isomorphic to?

Question

Let $G=S_n$ act on $\{1,2,...,n\}$ and let $A\subset \{1,2,...,n\}$ so that $|A|=k$. If $G_A$ is the subset of $G$ that fixes all the points in $A$, what is $G_A$ isomorphic to?

It makes sense that $G_A$ would be isomorphic to $S_{n-k}$ since, the $k$ points are fixed and that would result in $k$ one-cycles. What would be the best way of going about proving this?

Answer 1

Suppose, wlog, that $A=\{n-k+1,\cdots,n\}$. Then, consider $G_A$ and $S_{n-k}$. We can construct an explicit bijection between $G_A$ and $S_{n-k}$.

• If $\sigma\in S_{n-k}$, then $\widetilde{\sigma}\in G_A$ is defined as follows: $$ \widetilde{\sigma}(i)=\begin{cases}\sigma(i)&i\leq n-k\\i&i>n-k\end{cases}. $$

• If $\widetilde{\sigma}\in G_A$, then define $\sigma\in S_{n-k}$ as follows: $$ \sigma(i)=\widetilde{\sigma}(i). $$

This makes the composition in Matt Samuel's answer explicit (Matt Samuel's answer is better as it is more general, this just gives an explicit picture of what is happening).

Answer 2

Take a bijection $f$ of $\{1,\ldots,n\}-A$ with $\{1,2, \ldots ,n-k\}$. Then $\sigma\mapsto f\circ \sigma\circ f^{-1}$ gives you an isomorphism with $S_{n-k}$. (There are details to fill in here.)