Subset Product of Subgroups/Sufficient Condition/Proof 1

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 $H \circ K = K \circ H$.

First note that $H \circ K \ne \varnothing$, 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:


 * $\left({a_1 \circ b_1}\right) \circ \left({a_2 \circ b_2}\right) = a_1 \circ \left({b_1 \circ a_2}\right) \circ b_2$.

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:


 * $\left({a_1 \circ b_1}\right) \circ \left({a_2 \circ b_2}\right) = \left({a_1 \circ a}\right) \circ \left({b \circ b_2}\right)$

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


 * $\left({a_1 \circ b_1}\right) \circ \left({a_2 \circ b_2}\right) \in H \circ K$

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

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


 * $\left({a \circ b}\right)^{-1} = b^{-1} \circ a^{-1}$

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


 * $\left({a \circ b}\right)^{-1} \in K \circ H$

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 $G$.