# Intersection of Subgroups is Subgroup/General Result

## 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 $\mathbb S$.

$\blacksquare$