Let $H=\{\alpha \in S_7 :\alpha(3)=3\}$ and $K=\{\alpha \in S_7:\alpha(5)=5\}$. Prove that $H\cong K$.

Question

Question:

Let $H=\{\alpha \in S_7 :\alpha(3)=3\}$ and $K=\{\alpha \in S_7:\alpha(5)=5\}$. Prove that $H\cong K$

I'm not really able to do much here, I know I have to find some isomorphism but I'm not sure how to go about doing that.

Answer 1

The group $S_7$ is defined as the permutations of the set $\overline{1,7}:=\{1, 2, 3,4,5,6,7\}$ under composition of functions. There is nothing special about the symbols from $1$ to $7$ here.

What each of $H$ and $K$ do is gather together all elements of $S_7$ that fix some element of $\overline{1,7}$; respectively, $3$ and $5$. The structure of each of these subgroups is thus the same as the other. Hence they are isomorphic.

More formally, here is a hint: conjugate by $(35)$.

Answer 2

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$The general fact behind this is the following.

Suppose the group $G$ acts on the set $\Omega$. Let $x \in \Omega$ and $a \in G$. Then $a G_{x} a^{-1} = G_{a(x)}$.

Here $G_{x} = \Set{ g \in G : g(x) = x }$ is the stabiliser of $x$ in $G$.

Proof. \begin{align} G_{a(x)} &= \Set{ g \in G : g a(x) = a(x) }\\ &= \Set{ g \in G : a^{-1} g a(x) = x } \\ &= \Set{ g \in G : a^{-1} g a \in G_{x} } \\ &= a G_{x} a^{-1}. \end{align}