Isomorphism $\phi:S_n\rightarrow S_A$.

Question

I'm working through a problem right now where, for $H,K\subset S_5$, $H$ is the group of permutations which fix $1$, and $K$ is the group of permutations which fix $2$. The problem says to show that $H\approx K$. From what I've seen, this involves showing that $H\approx S_4\approx K$, based on the fact that $H$ is the group of permutations of $\{2,3,4,5\}$ and $K$ is the group of permutations of $\{1,3,4,5\}$.

I have a general idea of an isomorphism I could use, but I think I'm just having trouble with the notation. Let's say we have a set $A=\{a_1,a_2,a_3,a_4\}$, and I want to define a function $\phi: S_4\rightarrow S_A$ (I hope I'm using correct notation for the symmetric group on $A$, but please correct me if not). The idea is that, for any $\alpha\in S_4$ and $\beta\in S_A$, $\phi(\alpha)=\beta$ if and only if $\alpha(x)=n$ and $\beta(a_x)=a_n$ for all $1\leq x,n\leq4$; for example, if $\alpha(1)=3$, then $\beta(a_1)=a_3$. If I can show $\phi$ is an isomorphism, it naturally follows that $H$ and $K$ are isomorphic to $S_4$.

I have two questions about this problem:

1. Is there a simpler or more formal notation for this function, or would the way I've presented it be suitable enough?

2. Is this even needed, considering both $S_4$ and $S_A$ are the groups of permutations on 4-element sets?

Any help is appreciated. Thanks!

Answer 1

These subgroups are conjugate. Take the permutation $(12)$ then one of your subgroups conjugated by that permutation is the other subgroup.

Answer 2

Let $X$ and $Y$ be sets and $f\colon X\to Y$ a bijection. Then the map $\varphi\colon S_X \to S_Y$, defined by $\sigma\mapsto(y\mapsto (f\sigma f^{-1})(y))$, is an isomorphism. In particular, the symmetric groups of any two finite sets with the same number of elements are isomorphic, whence $S_A\cong S_n$ for every set $A$ of $n$ elements.