Intersection of Subgroups is Subgroup/General Result

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Theorem

Let $\struct {G, \circ}$ be a group.

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 $\mathbb S$.


Proof

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

Let $H_k$ be an arbitrary element of $\mathbb S$.

Then:

\(\ds a, b\) \(\in\) \(\ds H\)
\(\ds \leadsto \ \ \) \(\ds \forall k: \, \) \(\ds a, b\) \(\in\) \(\ds H_k\) Definition of Intersection of Set of Sets
\(\ds \leadsto \ \ \) \(\ds \forall k: \, \) \(\ds a \circ b^{-1}\) \(\in\) \(\ds H_k\) Group properties
\(\ds \leadsto \ \ \) \(\ds a \circ b^{-1}\) \(\in\) \(\ds H\) Definition of Intersection of Set of Sets
\(\ds \leadsto \ \ \) \(\ds H\) \(\le\) \(\ds G\) One-Step Subgroup Test

$\Box$


Now to show that $\struct {H, \circ}$ is the largest such subgroup.

Let $K$ be a subgroup of $\struct {G, \circ}$ 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 $\ds H = \bigcap \mathbb S$ is the largest subgroup of $S$ contained in each element of $\mathbb S$.

$\blacksquare$


Sources

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