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$G=AB$, 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!

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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.

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  • $\begingroup$ 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. $\endgroup$ – Dietrich Burde Mar 1 '18 at 15:19
  • $\begingroup$ Here is another reference: ([AFG], Theorem $2.1.1$, the book by Amberg et al., see here. $\endgroup$ – Dietrich Burde Mar 1 '18 at 15:30
  • $\begingroup$ Thank you so much! I’m grateful for your time! $\endgroup$ – Benny Mar 1 '18 at 15:38

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