Suppose $G$ is a group, $A\subset G$ a finite subset of order $n$.
Let $H = \{g\in G: \forall_{a\in A}:gag^{-1}\in A\}$ and $N=\{g\in G:\forall_{a\in A}:ga=ag\}$.
Show that $H$ is a subgroup of $G$, and that $N$ is a normal subgroup of $H$. Then, find a homomorphism $f:H\to S_n$ such that $\ker f=N$.
I verified myself that $H$ is a subgroup of $G$ and that $N\triangleleft H$. As it is a lot of writing out definitions and using tricks, I'll only show what I used to show that $N$ is normal. This was also the hardest step for me in the first part.
N is normal:
Suppose $n\in N$. We have $h^{-1}ah\in A$ for all $a\in A, h\in H$. We'll call $h^{-1}ah=a'$.
Now, as $n\in N$ we have that: $$na'n^{-1}=a' \iff hna'n^{-1}h^{-1}=ha'h^{-1}$$$$=hnh^{-1}ahn^{-1}h^{-1}=hnh^{-1}a(hnh^{-1})^{-1}=hh^{-1}ahh^{-1}=a$$
So $hnh^{-1}\in N$ so that $N$ is normal in $H$.
For the homomorphism:
First, we define $\phi_h:A\to A$ by $a\mapsto hah^{-1}$.
This is a bijection: suppose $hah^{-1}=hbh^{-1}$ for $a,b\in A.$ Multiplying with $h$ on the right, and $h^{-1}$ on the left on both sides gives us $a=b$ , so $\phi_h$ is injective $\iff$ surjective, as $\#A=\#A$.
For our homomorphism, we define $f:H\to S_n$ by $h\mapsto\phi_h$. If $f(h)=\operatorname{id}$, then $\phi_h(a)=a$ for all $A$, and since $\phi_h(a)=hah^{-1}(=a)$, we have that this exactly gives us that $h\in N$. So $\ker f=N$.
$f$ is a homomorphism: $f(hp) = \phi_{hp}(a)=(hp)a(hp)^{-1}=hpap^{-1}h^{-1}=h\phi_p(a)h^{-1}=\phi_h\circ \phi_p(a)$
I wasn't too sure about the homomorphism, as I use the mappings $\phi_h$ on $A$, which aren't necessarily 'numbers' as in $S_3$. It feels as if I have some kind of isomorphism (not $f$ itself, but a group isomorphic to $S_3$), maybe something along the lines of $f:H\to S(A)$ instead of $f:H\to S_n$? But as $\#A=n$ we have $S(A)\cong S_n$?