Subset Product of Subgroups

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

Then $$\forall H, K \le \left({G, \circ}\right): H \circ K \le G \iff H \circ K = K \circ H$$.

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

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 Subgroup, $$H \circ K \le G$$.