Let $G$ be a group, $a \in G$. Prove that for all $g \in G$, $|a| = |g^{-1}ag|$

Question

I am stuck on this problem. Let's just consider the case where $|a|$ is finite. So $|a| = n$ where $n$ is the least positive integer such that $a^{n} = e$.

First I need to show that $|g^{-1}ag| \le n$.

I notice that $g^{-1}a^{n}g = g^{-1}eg = g^{-1}g = e$. But does this imply that $(g^{-1}ag)^{n} = e$? If so, why? This is where I am stuck.

Answer 1

Conjugation is an isomorphism. Isomorphisms preserve the orders of elements. There is no need to restrict to finite orders.