# Let $A,B\le G$ be two abelian subgroups. Suppose that $G=\langle A\cup B\rangle$. Prove that $A\cap B \trianglelefteq G$.

Let $$A,B\le G$$ be two abelian subgroups. Suppose that $$G=\langle A\cup B\rangle$$. Prove that $$A\cap B \trianglelefteq G$$.

My attempt:

It's enough to show that $$A\trianglelefteq G$$ and $$B\trianglelefteq G$$. The definition of $$G$$ implies that it is the intersection of all subgroups $$T\le G$$ for which $$A\cup B \subseteq T$$. Since $$A\subseteq A\cup B$$ and $$B\subseteq A\cup B$$, this can be rewritten as $$G = \bigcap\{T\mid T\le G: A\subseteq T\}\cap \bigcap\{T\mid T\le G: B\subseteq T\} = \langle A\rangle\cap\langle B\rangle.$$ We're given that $$A$$ and $$B$$ are abelian subgroups. I somehow want to conclude that $$G$$ is abelian, which would imply that $$A$$ and $$B$$ are both normal in $$G$$ and thus their intersection will be normal in $$G$$. I'm not sure how to prove that $$G$$ is abelian. Does $$A$$ abelian imply $$\langle A\rangle$$ abelian?

Thanks.

• Consider $D_n$ as a counterexample (to $G$ being abelian).
– user418131
Oct 27 '19 at 11:52
• Note that $\langle A\rangle=A$ here Oct 27 '19 at 11:57

Notice that $$G=\langle A \cup B \rangle=\langle a^ib^j \rangle$$ So for every $$g \in A \cap B$$ we have: $$(a^ib^j) g = a^i (b^j g) \overset{B \text{ is Abelian}}{=====} a^i (g b^j) = (a^i g) b^j \overset{A \text{ is Abelian}}{=====} (g a^i) b^j = g (a^i b^j)$$ $$\Longrightarrow g \in Z(\langle A \cup B \rangle) \Longrightarrow A \cap B \subseteq Z(\langle A \cup B \rangle)$$ That is even a stronger result of being Normal!
In general, if $$A$$ and $$B$$ are abelian subgroups of a group $$G$$, then $$A\cap B$$ is a normal subgroup of $$\left\langle A\cup B\right\rangle$$. If you now suppose in addition that $$G=\left\langle A\cup B\right\rangle$$, the claim follows.
Suppose $H$ and $K$ are abelian subgroups of a group $G$. Then $H\cap K$ is a normal subgroup of $\left<H\cup K\right>$.