> [!NOTE] Theorem ($\langle g \rangle$ is a subgroup of $G$) > Let $G$ be a [[Groups|group]]. Let $g\in G$ and $\langle g \rangle$ denote the [[Generated Subgroup|subgroup generated]] by $g.$ Then $\langle g \rangle$ is a [[Subgroup|subgroup]] of $G.$ *Proof*. Check that $\langle g \rangle$ satisfies the [[Two-Step Subgroup Test|two-step subgroup test]]. Firstly $g^{0}=1$ and $0\in\mathbb{Z}$ so $1\in \langle g \rangle.$ Next, take $a,b\in\langle g \rangle.$ Then $a=g^{m}$ and $b=g^{n}$ for some $m,n \in\mathbb{Z}.$ Then by [[Integer Power of Group Element#^f832af|product rule]], $ab=g^{m+n}$ and since $m+n\in\mathbb{Z},$ we have $ab \in \langle g \rangle.$ Finally, $a^{-1}=(g^{m})^{-1}=g^{-m}$ from [[Integer Power of Group Element#^451338|definition of negative power]]. Since $-m\in \mathbb{Z},$ we have $a^{-1}\in\langle g \rangle.$