Intersection of Subgroups is Subgroup

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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 $H_1 \cap H_2 \le H_2$.


General Result

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


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


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


Proof

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

Then:

\(\displaystyle \) \(\) \(\displaystyle a, b \in H\)
\(\displaystyle \) \(\leadsto\) \(\displaystyle a, b \in H_1 \land a, b \in H_2\) Definition of Set Intersection
\(\displaystyle \) \(\leadsto\) \(\displaystyle a \circ b^{-1} \in H_1 \land a \circ b^{-1} \in H_2\) Group Properties
\(\displaystyle \) \(\leadsto\) \(\displaystyle a \circ b^{-1} \in H\) Definition of Set Intersection
\(\displaystyle \) \(\leadsto\) \(\displaystyle H \le G\) One-Step Subgroup Test


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

$\blacksquare$


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