# Subset Product of Subgroups/Necessary Condition/Proof 1

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## 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.

$\blacksquare$

## Sources

- 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $5$: Cosets and Lagrange's Theorem: Proposition $5.17$