Subset Product of Subgroups

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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 $K := \gen {a b}$:

\(\displaystyle \gen b\) \(=\) \(\displaystyle \set {e, b}\) as $b^2 = e$
\(\displaystyle \gen {a b}\) \(=\) \(\displaystyle \set {e, a b}\) 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