> [!NOTE] Theorem (Criterion for $a^{i}=a^{j}$) > Let $a\in G$ and $i,j\in \mathbb{Z}.$ If $a$ has infinite [[Order of Group Element|order]], then $a^{i}=a^{j}\iff i=j.$ If $a$ has finite order, say $n,$ then $a^{i} =a^{j}$ iff $n\mid(i-j).$ *Proof*. Suppose $a$ has infinite order and $a^{i}=a^{j}$ for some $i,j\in \mathbb{Z}.$ Then $a^{i-j} =e.$ We must have $i-j=0$ since there is nonzero integer such that $a^{n}=0.$ Suppose $a$ has finite order, $n,$ and $a^{i}=a^{j}$ for some $i,j\in \mathbb{Z}.$ Then $a^{i-j}=e.$ Using [[Division with remainder for integers|division with remainder]], we may write $i-j=qn+r$ for some $q,r\in \mathbb{Z}$ with $0\leq r<n.$ Then $a^{i-j}=(a^{n})^{q} a^{r}=a^{r} $Since $n$ is the least positive natural number such that $a^{n}=e,$ we must have $r=0$ so that $n\mid(i-j).$ Conversely, if $i-j=nq,$ then $a^{i-j}=a^{nq}=e$ so that $a^{i}=a^{j}.$ # Applications **Consequences**: By [[Power of Group Element is Identity only If Order divides]], $g^{m}=1$ for some $m\in \mathbb{Z}\setminus \{ 0 \}$ iff $d\mid m$ where $d\in \mathbb{N}^{+}$ is the order of $g.$