Intersection of Subgroups of Prime Order

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

Let $H, K \le G: \left\vert{H}\right\vert = \left\vert{K}\right\vert = p, H \ne K, p \in \mathbb P$.

Then $H \cap K = \left\{{e}\right\}$.

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 also $H \cap K \le H$. So:

Because $H \ne K$ and $\left\vert{H}\right\vert = \left\vert{K}\right\vert$, it follows that $H \nsubseteq K$.

So: