See order of a group

Let $G$ be a group, and let $a\in G$.
If $|a|=\mathrm{\infty }$, then $a^i=a^j\iff i=j$, for all $i,j\in \mathbb{Z}$.

Proof

Suppose $a^i=a^j$.
Then $a^{i-j}=e$.
Since $|a|=\mathrm{\infty }$, there is no $n>0$ such that $a^n=e$.
Therefore, $i-j=0$
so $i=j$.

QED