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A group $G$ is centre-by-finite if $[G:Z(G)]<\infty$. An easy way to construct finitely-generated centre-by-finite groups is to take direct products of the form $A\times F$, where $F$ is a finite group and $A$ is a finitely-generated abelian group. Are there any other examples of centre-by-finite groups?

I have tried to build examples using semidirect products of infinite abelian groups with finite groups, but this hasn't worked. I would be particularly interested to know if examples of this type are possible.

Jared White
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  • You can just require more, e.g., that all subgroups have finite index, see here. Then you end up with $1\rtimes \Bbb Z$. Of course, this depends on what you mean by "non-trivial" in the title (which you haven't said). – Dietrich Burde Jun 20 '23 at 13:59

2 Answers2

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Let us write $H=G/Z(G)$. Then $G$ is a central extension of $Z(G)$ by $H$, and there is an exact sequence $1\to Z(G)\to G\to H\to 1$.

Given a finite group $H$ and an abelian group $A$, you can ask what are all the central extensions $1\to A\to G\to H\to 1$ (central means that $A\subset Z(G)$). This is known to be classified by the cohomology group $H^2(H,A)$, and the trivial element of $H^2(H,A)$ corresponds to the direct product $G=A\times H$.

As an example, it's possible to completely treat the case $A=\mathbb{Z}$. Let $H$ be any finite group, and let $f:H\to \mathbb{Q}/\mathbb{Z}$ be a group morphism (the group of such morphisms is non-canonically isomorphic to the abelianization of $H$). Choose any function $\tilde{f}:H\to \mathbb{Q}$ such that the class of $\tilde{f}(h)$ in $\mathbb{Q}/\mathbb{Z}$ is precisely $f(h)$, for any $h\in H$. Then define $G=\mathbb{Z}\times H$ as a set, and define the multiplication $$(n,h)\cdot (m,h')=(n+m+\alpha(h,h'),hh')$$ where $\alpha(h,h')=\tilde{f}(hh')-\tilde{f}(h)-\tilde{f}(h')$.

You can check that this gives a group structure on $G$, that $\mathbb{Z}\subset Z(G)$, $G/\mathbb{Z}\simeq H$, and therefore $G/Z(G)$ is finite. Furthermore, any central extension of $\mathbb{Z}$ by $H$ is isomorphic to this for a unique $f$ (the choice of $\tilde{f}$ does not change the isomorphism class of $G$).

Captain Lama
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  • This is a very helpful answer - thank you. Is there any book that you would particularly recommend to read about group cohomology and central extensions? – Jared White Jun 26 '23 at 10:14
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You could define a semidirect product $G=C_n \rtimes {\mathbb Z}$ for with presentation $\langle x,y \mid x^n=1, y^{-1}xy=x^{-1} \rangle$. When $n$ is odd, that has centre $\langle y^2 \rangle$ of index $2n$ in $G$. When $n$ is even the centre is $\langle y^2,x^{n/2}\rangle$.

Another possibility is a central product of the Prüfer $p$-group $C_{p^\infty}$ with a nonabelian finite group of order $p^3$ (but that is not finitely generated).

Derek Holt
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