Subset Product of Subgroups/Sufficient Condition/Proof 2

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

Then:

That is:


 * $\paren {H \circ K} \circ \paren {H \circ K}^{-1} = H \circ K$

Thus by definition of set equality:


 * $\paren {H \circ K} \circ \paren {H \circ K}^{-1} \subseteq H \circ K$

So from One-Step Subgroup Test using Subset Product:
 * $H \circ K \le G$