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 $$\forall K \subseteq H: x \circ y \in K \Longrightarrow x \circ y \in H$$.

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