Intersection of Subgroups is Subgroup

Theorem
The intersection of two subgroups of a group is itself a subgroup of that group:


 * $\forall H_1, H_2 \le \left({G, \circ}\right): H_1 \cap H_2 \le G$

It also follows that $H_1 \cap H_2 \le H_1$ and $H_1 \cap H_2 \le H_2$.

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

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 $H = H_1 \cap H_2$ where $H_1, H_2 \le \left({G, \circ}\right)$. Then:

As $H \subseteq H_1$ and $H \subseteq H_2$, the other results follow directly.

Generalized 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 $x, y \in H$. Then $\forall K \subseteq H: x \circ y \in K \implies x \circ y \in H$.

Thus any $K \in \mathbb S: K \subseteq H$ and thus $H$ is the largest subgroup of $S$ contained in each member of $\mathbb S$.