# Subset Product of Subgroups/Sufficient Condition/Proof 1

## Theorem

Let $\struct {G, \circ}$ 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 $H \circ K \ne \O$, 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 $G$.

$\blacksquare$