Intersection of Abelian Subgroups is Normal Subgroup of Subgroup Generated by those Subgroups
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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 $12 \ \text{(ii)}$