Subset Product of Subgroups/Sufficient Condition/Proof 2

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Theorem

Let $\struct {G, \circ}$ be a group.

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

Let $H$ and $K$ be permutable subgroups of $G$.

That is, suppose:

$H \circ K = K \circ H$

where $H \circ K$ denotes subset product.


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


Proof

Suppose $H \circ K = K \circ H$.

Then:

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


That is:

$\paren {H \circ K} \circ \paren {H \circ K}^{-1} = H \circ K$

Thus by definition of set equality:

$\paren {H \circ K} \circ \paren {H \circ K}^{-1} \subseteq H \circ K$

So from One-Step Subgroup Test using Subset Product:

$H \circ K \le G$

$\blacksquare$


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