Intersection of Abelian Subgroups is Normal Subgroup of Subgroup Generated by those Subgroups

Theorem

Let $G$ be a group.

Let $L$ and $M$ be abelian subgroups of $G$.

Let $H = \gen {L, M}$ be the subgroup of $G$ generated by $L$ and $M$.

Then $L \cap M$ is a normal subgroup of $H$.

Proof

From

We have that $L$ and $M$ are abelian.

From Intersection of Subgroups is Subgroup, we have that $L \cap M$ is a subgroup of both $L$ and $M$.

From Subgroup of Abelian Group is Normal, we also have that $L \cap M$ is a normal subgroup of both $L$ and $M$.

Thus from Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup:

$L \subseteq \map N {L \cap M}$

and:

$M \subseteq \map N {L \cap M}$

Hence:

$L \cup M \subseteq \map N {L \cap M}$

But from Normalizer is Subgroup, $\map N {L \cap M}$ is a subgroup of $G$.

So it follows that:

$H \subseteq \map N {L \cap M}$

Hence the result by Subgroup is Normal Subgroup of Normalizer.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $12 \ \text{(ii)}$