Subset Product of Subgroups/Sufficient Condition/Proof 2

Theorem
Let $\left({G, \circ}\right)$ be a group.

Let $H, K$ be subgroups of $G$.

Let $H$ and $K$ be permutable subgroups of $G$.

That is, suppose:
 * $H \circ K = K \circ H$

where $H \circ K$ denotes subset product.

Then $H \circ K$ is a subgroup of $G$.

Proof
Suppose $H \circ K = K \circ H$.

Then:

That is:


 * $\left({H \circ K}\right) \circ \left({H \circ K}\right)^{-1} = H \circ K$

Thus from Equality of Sets:


 * $\left({H \circ K}\right) \circ \left({H \circ K}\right)^{-1} \subseteq H \circ K$

So from Subset Product with Inverse is Subgroup, $H \circ K \le G$.