Prove that $\{m \in M \mid mam^{-1} \in A, \ \forall a \in A\}$ forms a subgroup of M.

Question

I am given a group $M$ with identity element $1$, and $A$ $\leq$ $M$. I am then given a set $G = \{m \in M \mid mam^{-1} \in A, \ \forall a \in A\}$ and asked to prove that $G \leq M$.

I am confused about both the closure and identity.

Attempt for proving closure:

Let $g$, $h$ be two elements in $G$ such that $gag^{-1} \in A$ and $hah^{-1}$ $\in$ $A$. Then I have to show that $(gh)a(gh)^{-1}$ $\in$ $A$ for the same $a \in A$, right? Here is what I tried:

Take $(gh)a(gh)^{-1} \implies (gh)a(h^{-1}g^{-1})$ $\implies$ $g(hah^{-1})g^{-1}$. Now, I am confused because $hah^{-1}$ $\in A$, but it is not necessarily equal to $a$ which means, if I say $hah^{-1} = b \in A$ then I get, $gbg^{-1} \in A$. What I am not sure is I started with the assumption that the element $a$ must be the same. What am I misunderstanding here?

Answer 1

I can only presume the subset $G$ should have been written as $$G = \{m \in M: mam^{-1} \in A \quad \forall a \in A\}.$$ Then arguing as you do you run into no barriers, as you correctly identified that $hah^{-1} \in A$, so then so too $ghah^{-1}g^{-1} \in A$, and $G$ is closed.

I'll leave you to consider why $G$ is closed under taking inverses.

Answer 2

$G$ is called the normalizer of $A$. You are on the right track.

If $g, h \in G$, then $gAg^{-1}=A$ and $hAh^{-1}=A$. So, $$ghA(gh)^{-1}=ghAh^{-1}g^{-1}=gAg^{-1}=A.$$ Thus, $gh \in G$.

If $g \in G$, then $gAg^{-1}=A$ and thus $A=g^{-1}Ag$. So, $g^{-1} \in G$.