Subset Product of Subgroups
Theorem
Let $\struct {G, \circ}$ be a group.
Let $H, K$ be subgroups of $G$.
Then $H \circ K$ is a subgroup of $G$ if and only if $H$ and $K$ are permutable.
That is:
$H \circ K$ is a subgroup of $G$ if and only if:
- $H \circ K = K \circ H$
where $H \circ K$ denotes subset product.
Necessary Condition
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.
Sufficient Condition
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 $G$.
Examples
Subgroups $\gen b$ and $\gen {a b}$ in $D_3$
Consider the dihedral group $D_3$, given as the group presentation:
- $D_3 = \gen {a, b: a^3 = b^2 = e, a b = b a^{-1} }$
Consider the generated subgroups $H := \gen b$ and $K := \gen {a b}$:
\(\ds \gen b\) | \(=\) | \(\ds \set {e, b}\) | as $b^2 = e$ | |||||||||||
\(\ds \gen {a b}\) | \(=\) | \(\ds \set {e, a b}\) | as $\paren {a b}^2 = a b b a^{-1} = e$ |
Then $H$ and $K$ are not permutable, and neither $H K$ nor $K H$ is a subgroup of $D_3$.
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 38 \alpha$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $6$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $5$: Cosets and Lagrange's Theorem: Proposition $5.17$