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 |
$\blacksquare$