> [!NOTE] Theorem > Let $G$ be a [[Groups|group]]. Let $H_{1},H_{2}$ be [[Subgroup|subgroups]] of $G.$ Then $H_{1}\cap H_{2}$ is also a subgroup of $G.$ **Proof**: $H_{1}\cap H_{2}$ contains the identity: Since $H_{1}$ and $H_{2}$ are subgroups of $G,$ we have $1_{G}\in H_{1}$ and $1_{G}\in H_{2}.$ Thus $1_{G}\in H_{1}\cap H_{2}.$ $H_{1}\cap H_{2}$ is closed: let $g,h\in H_{1}\cap H_{2}.$ Then $g,h\in H_{1}$ and $g,h\in H_{2}$ thus $gh\in H_{1}$ and $gh\in H_{2}$ by closure property of groups. Thus $gh\in H_{1} \cap H_{2}.$ $H_{1}\cap H_{2}$ contains inverses: let $g\in H_{1}\cap H_{2}.$ Then $g\in H_{1}$ and $g\in H_{2}$ thus $g^{-1}\in H_{1}$ and $g^{-1}\in H_{2}.$ This gives, $g^{-1}\in H_{1} \cap H_{2}.$ Thus by [[Two-Step Subgroup Test]], $H_{1}\cap H_{2}$ is a subgroup of $G.$