> [!NOTE] Lemma ($g^{m}=1$ iff order divides $m$) > Let $G$ be a [[Groups|group]]. Let $g\in G.$ Let $m\in \mathbb{Z}\setminus\{ 0 \}.$ Then $g^{m}=1$ iff $g$ has finite [[Order of Group Element|order]] $d$ with $d\mid m$ ($d$ [[Divisibility in Integers|divides]] $m$). *Proof*. ($\implies$) Suppose $g^{m}=1$ for some $m\in\mathbb{Z}\setminus\{ 0 \}.$ Then $g^{|m|}=1$ thud $g$ has finite order, denote $d\in\mathbb{N}^{+}.$ Using [[Division with remainder for integers|division with remainder]], write $m=qd+r$ with $q,r\in\mathbb{Z}$ and $0\leq r<d.$ Then $1= g^{m}=g^{qd}g^{r}=g^{r} $and so $r=0$ since we can't have $1\leq r<d$ by definition of order of $g.$ Thus $d\mid m.$ ($\implies$) Conversely, suppose $g$ has finite order $d$ with $d\mid m.$ Then $g^{d} =1$ and $m=qd$ for some $q\in \mathbb{Z}.$ Therefore $g^{m}=(g^{d})^{q}=1$ using [[Integer Power of Group Element#^f832af|power rule]]. **Proof**: Follows from [[Criteria for Equality of Powers of Group Element]]: that is $g^{m}=g^{0}$ iff $d\mid (m-0).$ If $g$ has infinite order then $m=0$ but $m\in \mathbb{Z}\setminus \{0 \}.$