TITLE: Intersection of a Subgroup with an Abelian Subgroup is Normal in the Product
QUESTION [5 upvotes]: Question: Let $A$ be an Abelian group with $A \trianglelefteq G$, and let $B \leq G$ be any subgroup. Show that $A \cap B \trianglelefteq AB$.
[ref: this is exercise 20 on page 96 of [DF] := Dummit and Foote's Abstract Algebra, $3^{\text{rd}}$ edition]
My attempt: Clearly, $AB$ is a subgroup of $G$ (because $B \leq G = N_G (A)$, so this is just corollary 15 on page 94 of [DF]). As $A \cap B$ is also a subgroup, we therefore already have $A \cap B \leq AB$, so we only need to show that $AB$ normalises $A \cap B$.
Let $ab \in AB$. What is $(ab) A\cap B (ab)^{-1}$? Choosing any $g \in A \cap B$, we note that in particular $g \in A$, so
$$
ab gb^{-1} a^{-1} = aa'a^{-1} = a' \in A,
$$
since $A$ is Abelian.
Now, I'm stuck: I would like to show that the same $ab gb^{-1} a^{-1}$ also lies in $B$ by taking into account that $g \in B$. This will then show that $(ab) A\cap B (ab)^{-1} \subset A \cap B$, and we are done.

REPLY [3 votes]: $bgb^{-1}=b'$ is an element of $A\cap B$ (if $g\in A$ it is normalized as you said, and if $g\in B$ we have a product in the subgroup $B$) and $ab'a^{-1}=b'\in A\cap B$

REPLY [3 votes]: To show that $AB$ normalizes $A\cap B$ it suffices to show that $A$ and $B$ separately normalize $A\cap B$ (because this will imply that $\langle A,B\rangle\subseteq N_G(A\cap B)$, and the product is the subgroup generated by $A$ and $B$).
If $x\in A\cap B$, and $b\in B$, then $bxb^{-1}\in A$ (since $x\in A\triangleleft G$), and $bxb^{-1}\in B$ (since $b,x\in B$), so $bxb^{-1}\in A\cap B$.  So $B$ normalizes $A\cap B$.
If $x\in A\cap B$ and $a\in A$, then $axa^{-1}=x\in A\cap B$, because $x\in A$ and $A$ is abelian. So $A$ normalizes $A\cap B$.
Therefore, $AB=\langle A,B\rangle \subseteq N_G(A\cap B)$, as desired.

REPLY [2 votes]: Note that $bgb^{-1} \in A$ as $A \lhd G$, and $bgb^{-1} \in B$, because $B$ is a subgroup. Thus $bgb^{-1} \in A \cap B$. Now, $A$ is abelian, so $a(bgb^{-1})a^{-1} = aa^{-1}(bgb^{-1}) = bgb^{-1} \in A \cap B$. We conclude that $A \cap B \lhd AB$.
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