Prove centralizer =normalizer = G

Question

Let $G$ be a group acting on $A$. How to prove centralizer of $A$ = normalizer of $A$ = $G$ ? Keeping in mind that $gag^{-1} = gg^{-1}a = a$, for every $a$ belongs to $A$ and for every $g$ belongs to $G$.

Answer 1

So I imagine in this setup $A$ is a subgroup of $G$, otherwise the conjugation you wrote down doesn't make sense (also centralizers and normalizers are defined only for subgroups). I also imagine that the action is given by conjugation, as for what you wrote down.

If the action is well-defined, then you must have $gag^{-1} \in A$ for all $g \in G, a \in A$. This gives automatically $N_G(A) = G$.

Now the identity you wrote down means exactly that $C_G(A) = G$, because every element of $G$ centralizes $A$.

In a more general setup, though, you wouldn't say that $G$ acts on $A$ by conjugation, because that's restrictive:it immediately forces $A$ to be normal. You can just let $G$ act on itself by conjugation, and then $N_G(A)$ and $C_G(A)$ are well-defined subgroups of $G$.

In your case, the identity is enough to say that $C_G(A) = G$ and as $C_G(A) \leq N_G(G) \leq G$, you get equality everywhere. But here the normality of $A$ comes afterwards and it's not in the hypotheses.