# Intersection of Subgroups of Prime Order

## Theorem

Let $G$ be a group whose identity is $e$.

Let $H$ and $K$ be subsets of $G$ such that:

$\order H = \order K = p$
$H \ne K$
$p$ is prime.

Then:

$H \cap K = \set e$

That is, the intersection of two unequal subgroups of a group, both of whose order is the same prime, consists solely of the identity.

## Proof

$H \cap K \le G$

and:

$H \cap K \le H$

where $\le$ denotes subgrouphood.

So:

 $\ds H \cap K$ $\le$ $\ds H$ Intersection of Subgroups is Subgroup $\ds \leadsto \ \$ $\ds \order {H \cap K}$ $\divides$ $\ds \order H$ Lagrange's Theorem $\ds \leadsto \ \$ $\ds \order {H \cap K}$ $\divides$ $\ds p$ $\ds \leadsto \ \$ $\ds \order {H \cap K}$ $=$ $\ds 1 \text{ or } p$ $p$ is prime

Because $H \ne K$ and $\order H = \order k$, it follows that $H \nsubseteq K$.

So:

 $\ds H \cap K$ $\ne$ $\ds H$ Intersection with Subset is Subset‎ $\ds \leadsto \ \$ $\ds H \cap K$ $\subset$ $\ds H$ $\ds \leadsto \ \$ $\ds \order {H \cap K}$ $<$ $\ds \order H = p$ $\ds \leadsto \ \$ $\ds \order {H \cap K}$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds H \cap K$ $=$ $\ds \set e$ Definition of Trivial Group

$\blacksquare$