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:
- $H \circ K \le G$
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $6$