Subset Product of Subgroups/Sufficient Condition/Proof 1

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

First note that $H \circ K \ne \O$, as $e_G = e_G \circ e_G \in H \circ K$, from Identity of Subgroup.

Suppose $a_1, a_2 \in H, b_1, b_2 \in K$.

Then:


 * $\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} = a_1 \circ \paren {b_1 \circ a_2} \circ b_2$.

Since $H \circ K = K \circ H$, we see that, for some $a \in H, b \in K$:


 * $b_1 \circ a_2 = a \circ b$

Thus:


 * $\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} = \paren {a_1 \circ a} \circ \paren {b \circ b_2}$

As $H, K \le G$, we have $a_1 \circ a \in H$ and $b \circ b_2 \in K$, hence:


 * $\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} \in H \circ K$

thus demonstrating closure of $H \circ K$ under $\circ$.

Finally, if $a \circ b \in H \circ K$, then by Inverse of Group Product:


 * $\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$

Since $b^{-1} \in K$ and $a^{-1} \in H$, we have:


 * $\paren {a \circ b}^{-1} \in K \circ H$

and hence $H \circ K$ is shown to be closed under inverses.

Thus, from the Two-Step Subgroup Test, $H \circ K$ is a subgroup of $G$.