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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.

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

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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.

Reference:

Suppose $H$ and $K$ are abelian subgroups of a group $G$. Then $H\cap K$ is a normal subgroup of $\left<H\cup K\right>$.

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