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Suppose $G$ is a finite centerless metabelian group. Is it true that it is a semidirect product of two abelian groups?

It does not seem true to me, but I failed to find any counterexamples. Actually, I know quite a few examples of finite metabelian groups that do not split into a semidirect product of two abelian groups, but all of them have a nontrivial centre.

What I have tried:

The only thing I managed to see here is that a group having an abelian normal Hall subgroup with an abelian quotient by it is a semidirect product of two abelian groups (as any finite group splits over its normal Hall subgroup). On the other hand, all Hall subgroups of a semidirect product of two finite abelian groups are normal (as Hall subgroups of finite abelian groups are always characteristic).

So that question can be reduced to either finding a finite centerless metabelian group with a non-normal Hall subgroup or proving that any finite centerless metabelian group has an abelian normal Hall subgroup with an abelian quotient by it. Note that aforementioned implications are one-sided, so proving that all Hall subgroups of finite centerless metabelian groups are normal or finding a finite centerless metabelian group without a normal abelian Hall subgroup with an abelian quotient by it will not give us anything regarding this question.

However, those two new "questions" do not seem any easier than the initial one...

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    $\begingroup$ It is certainly true that it is a semidirect product of an abelian group with a nilpotent group. I am not sure if the complement can always be guaranteed to be abelian though. $\endgroup$ – the_fox Feb 8 at 21:31
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Yes, it is true.

Suppose that $G$ is finite and $Z(G)=1$. Suppose, also, that $N \unlhd G$ is such that both $N$ and $G/N$ are abelian groups. Denote by $L$ the smallest term of the lower central series of $G$. This characteristic subgroup is known as the nilpotent residual of $G$ and is the unique smallest normal subgroup such that the corresponding quotient is nilpotent.

Since $G/N$ is abelian, we have $L \leq G' \leq N$, so $L$ is abelian. Note that $L>1$ since if $L=1$ then $G$ is nilpotent, against the assumption that $Z(G)=1$. Now it is known that if $L$ is abelian, then $L$ has a complement in $G$, which in fact is nilpotent (see below for an excerpt from p. $264$ of Robinson's book).

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More generally, if $\frak{F}$ is a saturated formation and the $\frak{F}$-residual of $G$ is abelian, then it is complemented in $G$. One more word on why the theorem above applies in our case. Since $G$ is metabelian, it is certainly soluble and system normalisers (the complements) are always nilpotent in soluble groups (you can see this directly though).

Your requirement that $Z(G)=1$ also gives that $C_G(L) \leq L$. This is a theorem of Schenkman's (see Isaacs' FGT book, p. $283$; $G^{\infty}$ is how Isaacs writes the nilpotent residual).

enter image description here

Since $L$ is abelian we also have $C_G(L) \geq L$, thus $C_G(L) =L$. Therefore $L$ is a maximal abelian normal subgroup of $G$. In particular, $L=G'=N$. The complement for $L$ in $G$ whose existence we had guaranteed before is $\cong G/L = G/N$, thus abelian by assumption and we are done.

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  • $\begingroup$ @YaniorWeg there's not claim that $L$ is minimal normal (and it's false in general) $\endgroup$ – YCor Feb 9 at 5:13

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