Question: Let $G$ be a group and $a \in G$. Then for any integer $k$ we have $C(a) \subseteq C(a^k)$.
My attempt:
We shall use induction on $k$ for $k \in \mathbb Z$ and $k≥0$.
Base case: For $k=0$, $C(a^k)=G$ and thus $C(a) \subseteq C(a^k)$ is obvious.
Inductive case: Suppose inductively that $C(a) \subseteq C(a^n)$ for some $n ∈ \mathbb Z$ and $n≥0$. We want to show $C(a) \subseteq C(a^{n+1})$. Let $g ∈ G$, we see $a^{n+1}g=aa^ng=aga^n$ by inductive hypothesis. Since $ga^n \in G$ ,thus $aga^n = ga^na= ga^{n+1}$ .This shows $g ∈ C(a^{n+1})$ whenever $g ∈ C(a^n)$ and as $C(a) ⊆ C(a^n)$, thus $Ca(a) \subseteq C(a^{n+1)})$. This closes induction.
Now for the other case when $k$ is negative. Let $y=-k$ , since $y>0$ thus $C(a) \subseteq C(a^y) = C(a^{-k})$. Since $C(a) = C(a^{-1})$ thus we conclude that $C(a) \subseteq C(a^k)$. ∎
My confusion:
I think my proof is incorrect. In the line " Since $ga^n \in G$ ,thus $aga^n = ga^na= ga^{n+1}$ " in inductive case , i believe there is mistake. That is i wrote $a() = ()a$ without showing $() ∈ C(a)$.I think the proof can be corrected if i used strong induction.Am I thinking correct? And Is the above proof really correct?
Edit: Here $C(a):=\{g ∈ G| ag = ga\}$, called centralizer of $a$ in $G$.