See order of a group

Let $G$ be a group, and let $a\in G$.
If $|a|=n$, then $a^i=a^j$ if and only if $n|(i-j)$.

Proof

Suppose $a^i=a^j$.
Then $a^{i-j}=e$.
By Division Algorithm 1, $(i-j)=qn+r$ for $0\le r<n$
So $e=a^{i-j}=a^{qn+r}=(a^n{)}^q{a}^r=e^q{a}^r=a^r$
By definition of order, $n$ is the smallest nonzero natural number with $a^n=e$
so $r=0$
Thus $i-j=qn\Rightarrow n|(i-j)$

QED