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:

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

So by definition of subset:

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


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

Then:

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

So by definition of subset:

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


The result follows by definition of set equality.

$\blacksquare$


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