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
From Intersection of Subgroups:
 * $H \cap K \le G$

and:
 * $H \cap K \le H$

where $\le$ denotes subgrouphood.

So:

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

So: