# Subset Product of Subgroups/Necessary Condition/Proof 1

## Theorem

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

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

Let $H \circ K$ be a subgroup of $G$.

Then $H$ and $K$ are permutable.

That is:

$H \circ K = K \circ H$

where $H \circ K$ denotes subset product.

## Proof

Suppose $H \circ K$ is a subgroup of $G$.

Let $h \circ k \in H \circ K$.

Then $h \circ k$ is the inverse of some element $g$ of $H \circ K$.

Thus we can write $g = h' \circ k'$ for some $h' \in H$ and $k' \in K$.

So:

 $\displaystyle h \circ k$ $=$ $\displaystyle g^{-1}$ Inverse of Group Inverse: $g$ is the inverse of $h \circ k$ $\displaystyle$ $=$ $\displaystyle \paren {h' \circ k'}^{-1}$ $\displaystyle$ $=$ $\displaystyle k'^{-1} \circ h'^{-1}$ Inverse of Group Product $\displaystyle$ $\in$ $\displaystyle K \circ H$ as $K$ and $H$ are both groups

So by definition of subset:

$H \circ K \in K \circ H$

Now suppose $x \in K \circ H$.

Then:

 $\displaystyle x$ $=$ $\displaystyle k \circ h$ for some $k \in K, h \in H$ $\displaystyle$ $=$ $\displaystyle \paren {\paren {k \circ h}^{-1} }^{-1}$ Inverse of Group Inverse $\displaystyle$ $=$ $\displaystyle \paren {h^{-1} \circ k^{-1} }^{-1}$ Inverse of Group Product $\displaystyle$ $\in$ $\displaystyle H \circ K$ as $K$ and $H$ are both groups

So by definition of subset:

$K \circ H \in H \circ K$

The result follows by definition of set equality.

$\blacksquare$