Prove that the centralizer is normal

Question

If $C(g) = \{x\in G:xg=gx\}$ then it is a subgroup. If $g$ generates a normal subgroup of G, prove that $C(g)$ is normal in G.

I sincerely don't see the relation between one thing and the other. I know from this that $\forall a\in G$ we have that $a<g>=<g>a$ and we have to show from this that $\forall b\in G, bC(g)=C(g)b$. My first thought is to show that $bxb^{-1}\in C(g)$ so $(bxb^{-1})g$ should be equal to $g(bxb^{-1})$. From $a<g>=<g>a$ we know that $(bxb^{-1})g=g^k (bxb^{-1})$ for some $k$ but can we conclude that $k=1$?

Answer 1

You can look at it this way: first since $\langle g \rangle$ is normal, we have $b^{-1}g = g^kb^{-1}$ for some integer $k$, and also $bg^k = gb$.

Then $(bxb^{-1})g = (bx)(b^{-1}g) = (bx)(g^kb^{-1}) = b(xg^k)b^{-1} = b(g^kx)b^{-1} = (bg^k)xb^{-1} = (gb)xb^{-1} = g(bxb^{-1})$

Answer 2

this result is a special case of a more general fact:

let $N$ be any normal subgroup of $G$, and consider the action of $G$ on $N$ by conjugation. this defines a morphism of groups: $$ \alpha: G \to Aut(N) $$ the centralizer of $N$ is the kernel of $\alpha$, hence is a normal subgroup of $G$.