Let $G$ be a finite group and let g ∈ G be an element with order d. Let i ∈ ℤ. Show that the order of $g^{i} $ is $d/\gcd(i,d)$\

Question

Let $G$ be a finite group and let $g ∈ G$ be an element with order $d$. Let $i ∈ ℤ$. Show that the order of $g^{i} $ is $d/\gcd(i,d)$

I am aware that $g^d = e$ (where $e$ is the identity of the group). Hence ${(g^{i})}^{k}$ = $e$ for some integer$k$ and $d\mid ik$. Where to proceed from here?

Answer 1

Hint:

The order of $g^i$ is the least positive integer $k$ such that $g^ik=e$, i.e., by definition of the order of$d$, the least multiple of $i$ which is also a multiple of $d$. What do you think this least multiple is called, usually?