Subset Product of Subgroups/Necessary Condition

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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 1

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:

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

So by definition of subset:

$H \circ K \subseteq K \circ H$


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

Then:

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

So by definition of subset:

$K \circ H \subseteq H \circ K$


The result follows by definition of set equality.

$\blacksquare$


Proof 2

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

Then:

\(\ds H \circ K\) \(=\) \(\ds \paren {H \circ K}^{-1}\) Inverse of Subgroup
\(\ds \) \(=\) \(\ds K^{-1} \circ H^{-1}\) Inverse of Product of Subsets of Group
\(\ds \) \(=\) \(\ds K \circ H\) Inverse of Subgroup

$\blacksquare$