Intersection of Subgroups is Subgroup

Theorem
Let $\mathbb S$ be a set of subgroups of $\left({G, \circ}\right)$, where $\mathbb S \ne \varnothing$.

Then the intersection $\bigcap \mathbb S$ of the members of $\mathbb S$ is itself a subgroup of $G$.

Also, $\bigcap \mathbb S$ is the largest subgroup of $\left({G, \circ}\right)$ contained in each member of $\mathbb S$.

Proof
Let $H = \bigcap \mathbb S$.

Let $H_k$ be any element of $\mathbb S$. Then:


 * Now to show that $\left({H, \circ}\right)$ is the largest such subgroup.

Let $K$ be a subgroup of $\left({G, \circ}\right)$ such that:
 * $\forall S \in \mathbb S: K \subseteq S$

Then by definition $K \subseteq H$.

Let $x, y \in K$.

Then $x \circ y^{-1} \in K \implies x \circ y^{-1} \in H$

Thus any subgroup of all elements of $\mathbb S$ is also a subgroup of $H$ and so no larger than $H$.

Thus $H = \bigcap \mathbb S$ is the largest subgroup of $S$ contained in each member of $\mathbb S$.