Intersection of Subgroups of Prime Order

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

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

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:

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

So:

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