Subset Product of Subgroups/Necessary Condition/Proof 1

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

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

Let $H \circ K$ be a subgroup of $G$.

Then $H$ and $K$ are permutable.

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

where $H \circ K$ denotes subset product.

Proof
Suppose $H \circ K$ is a subgroup of $G$.

Let $h \circ k \in H \circ K$.

Then $h \circ k$ is the inverse of some element $g$ of $H \circ K$.

Thus we can write $g = h\,' \circ k'$ for some $h\,' \in H$ and $k\,' \in K$.

So:

So by definition of subset:
 * $H \circ K \in K \circ H$

Now suppose $x \in K \circ H$.

Then:

So by definition of subset:
 * $K \circ H \in H \circ K$

The result follows by definition of set equality.