Subset Product of Normal Subgroups with Trivial Intersection

Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

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

Let $H \cap K = e$.

Then $H K$ is isomorphic to $H \times K$ where:

$H K$ denotes the subset product of $H$ and $K$

$H \times K$ denotes the direct product of $H$ and $K$.

Proof

Let $G' = H K$.

From Subset Product of Normal Subgroups is Normal, $G'$ is a normal subgroup of $G$.

That is $G'$ is itself a group.

So by the Internal Direct Product Theorem, $G'$ is the internal group direct product of $H$ and $K$.

The result follows by definition of the internal group direct product.

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \mu$