Subset Product with Normal Subgroup as Generator

Theorem

Let $G$ be a group whose identity is $e$.

Let:

$H$ be a subgroup of $G$

$N$ be a normal subgroup of $G$.

Then:

$N \lhd \gen {N, H} = N H = H N \le G$

where:

$\le$ denotes subgroup

$\lhd$ denotes normal subgroup

$\gen {N, H}$ denotes a subgroup generator

$N H$ denotes subset product.

Proof

From Subset Product is Subset of Generator:

$N H \subseteq \gen {N, H}$

From Subset Product with Normal Subgroup is Subgroup:

$N H = H N \le G$

Then by the definition of a subgroup generator, $\gen {N, H}$ is the smallest subgroup containing $N H$ and so:

$\gen {N, H} = N H = H N \le G$

From Normal Subgroup of Subset Product of Subgroups we have that:

$N \lhd N H$

Hence the result.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.16$
• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Problem $\text{HH}$
• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $7$: Normal subgroups and quotient groups: Proposition $7.10$