Proof verification: For all $g\in G$, with $G$ being a group, $|a|=|g^{-1}ag|$.

Question

Can someone please verify whether my proof is logically correct? :)

For all $g\in G$, with $G$ being a group, $|a|=|g^{-1}ag|$.

Proof: Let $|a|=n$. Then $a^{n}=e$. Then $e=g^{-1}g=g^{-1}eg=g^{-1}a^{n}g=(g^{-1}ag)^{n}$. Then $|g^{-1}ag|=n=|a|$. $\square$

Answer 1

Now you have proven that $|g^{-1}ag|\leq |a|$ (yes, maybe the order is smaller).

You still need to prove the other inequality.