Subset Product of Subgroups

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

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

Then $H \circ K$ is a subgroup of $G$ iff $H$ and $K$ are permutable.

That is: $H \circ K$ is a subgroup of $G$ iff:
 * $H \circ K = K \circ H$

where $H \circ K$ denotes subset product.

Proof 1
Let $e_G$ be the identity of $\left({G, \circ}\right)$.

As $H, K \le \left({G, \circ}\right)$, it follows from Inverse of Subgroup that $H = H^{-1}, K = K^{-1}$.


 * Suppose $H \circ K \le G$.

Then $H \circ K = \left({H \circ K}\right)^{-1} = K^{-1} \circ H^{-1} = K \circ H$ from Inverse Subset Group Product.


 * Now suppose $H \circ K = K \circ H$.

First note that $H \circ K \ne \varnothing$, as $e_G \in H \circ K$, 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 $b_1 \circ a_2 = a \circ b$ for some $a \in H, b \in K$.

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 $H, K \le G$, it follows easily that $\left({a_1 \circ b_1}\right) \circ \left({a_2 \circ b_2}\right) \in H \circ K$, thus demonstrating closure.

Finally, if $a \circ b \in H \circ K$, then $\left({a \circ b}\right)^{-1} = b^{-1} \circ a^{-1}$ from Inverse of Group Product, so $\left({a \circ b}\right)^{-1} \in K \circ H$, showing $H \circ K$ is closed under inverses.

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

Proof 2

 * The same proof as above can be used for $H \circ K \le G \Longrightarrow H \circ K = K \circ H$.


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

Then:

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

Thus from the definition of set equality:
 * $\left({H \circ K}\right) \circ \left({H \circ K}\right)^{-1} \subseteq H \circ K$

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