Intersection of Subgroups is Subgroup

Theorem

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

$\forall H_1, H_2 \le \struct {G, \circ}: H_1 \cap H_2 \le G$

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

Let $\mathbb S$ be a set of subgroups of $\struct {G, \circ}$, where $\mathbb S \ne \O$.

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

Also, $\ds \bigcap \mathbb S$ is the largest subgroup of $\struct {G, \circ}$ contained in each element of $\mathbb S$.

Let $H = H_1 \cap H_2$ where $H_1, H_2 \le \struct {G, \circ}$.

Then:

$\ds $

$\ds a, b \in H$

$\ds $

$\leadsto$

$\ds a, b \in H_1 \land a, b \in H_2$

$\ds $

$\leadsto$

$\ds a \circ b^{-1} \in H_1 \land a \circ b^{-1} \in H_2$

$\ds $

$\leadsto$

$\ds a \circ b^{-1} \in H$

$\ds $

$\leadsto$

$\ds H \le G$

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

$\blacksquare$