# Metabelian—the derived subgroup of the product of two Abelian groups

$G=AB$, where $A$, $B$ are abelian subgroups. Then $G’$ is abelian.

Attempt:

$\color{red}{\text{Suppose }B \text{ to be normal.}}$

$G/B=AB/B\cong A/(A\cap B)$, where $A/(A\cap B)$ is definitely abelian, hence $G/B$ is an abelian group. There’s a well-known proposition:

Proposition: Let $N$ be a normal subgroup of $G$. Then $$G/N \mbox{ is Abelian}~~~\Longleftrightarrow ~~~G’\leq N.$$

Therefore $G’\leq B$, where $B$ is an abelian group, and then we get $G’$ is abelian as desired.

Whereas, the only weakness is that $B$ is not necessarily normal. Actually, since $G=AB=BA$, it suffices to prove the normality of one of the two abelian subgroups, but it’s precisely where I’m stuck.

Am I basically right? Or do I need to take another idea? Any hint or detail will be sincerely welcome. I’d be grateful for your help!

What you have done is correct, but probably will not suffice. The claim follows from Ito's Theorem, saying that if $G=AB$, then $G$ is metabelian, i.e., that $G'$ is abelian:

Ito's original proof is very simple and straightforward:

• Itô, Noboru. "Über das Produkt von zwei abelschen Gruppen." Math. Z. 62 (1955), 400–401. MR71426 DOI:10.1007/BF01180647

For several duplicates on MSE, proving Ito's Theorem, see for example here:

Commutator property in product of groups

There are also several generalizations of Ito's result, e.g., the following:

Theorem: Let $G=AB$ be a group with abelian subgroup $A$ and nilpotent, metabelian subgroup $B$. Then $G$ is solvable of derived length at most $4$.

For details see my paper here.

• The proof is reproduced in many other papers on this topic. You can use the proof given by Jack Schmidt (at the duplicate). I had another link, but cannot find it now. – Dietrich Burde Mar 1 '18 at 15:19
• Here is another reference: ([AFG], Theorem $2.1.1$, the book by Amberg et al., see here. – Dietrich Burde Mar 1 '18 at 15:30
• Thank you so much! I’m grateful for your time! – Benny Mar 1 '18 at 15:38