> [!NOTE] Lemma ($|g|=|\langle g \rangle|$) > Let $G$ be a [[Groups|group]]. Let $g\in G$ and $\langle g \rangle$ denote the [[Generated Subgroup|subgroup generated]] by $g.$ If $g$ has infinite [[Order of Group Element|order]] then $G$ has [[Finite Group|infinite]] [[Order of Algebraic Structure|order]]. Otherwise, suppose $g$ has finite order, $n,$ then $\langle g \rangle=\{ e,g,g^{2},\dots,g^{n-1} \}$ and $|g|=|\langle g \rangle|.$ ^988bda *Proof*. Suppose $g$ has finite order $n.$ Certainly $\{ e,g,g^{2},\dots,g^{n-1} \}\subset\langle g \rangle.$ Suppose $h\in \langle g \rangle.$ Then $h=g^{m}$ for some $h\in m.$ Using [[Division with remainder for integers|division with remainder]], we can write $m=qn+r, \quad q,r\in\mathbb{Z}, \quad 0\leq r<n.$Thus $h=g^{m}=g^{qn+r}=(g^{n})^{q}\cdot g^{r}=g^{r} $Since $0\leq r<n,$ we see that $h\in \{ e,g_{1},\dots,g^{n-1} \}$ and so $\langle g \rangle=\{ e,\dots,g^{n-1} \}$ with $|\langle g \rangle|=|g|=n.$ Suppose $g$ has infinite order. BWOC suppose $\langle g \rangle$ has finite order. Then [[Element of a finite group is of finite order|every element has finite order]], including $g,$ which leads to a contradiction. # Application **Consequences**: Order of [[Cyclic Group|Cyclic Group]] Equals Order of Order of Generator.