(finitely generated finite by abelian) implies (abelian by finite)

It is mentioned several times in the question

Finite-by-(abelian-by-finite)

that (finitely generated finite by abelian) implies (abelian by finite). How do you prove that?

I am interested in this question

https://mathoverflow.net/questions/206618/how-bad-can-pi-1-of-a-linear-group-orbit-be

and everyone on that question is acting like its really obvious that (finitely generated finite by abelian) implies (abelian by finite) so I googled around to find an explanation and the best I found was the first MSE question linked above. Group theory isn't my number one interest I'm really in it to see which 3 manifolds have a fundamental group which is compatible with the possibility of being a linear group orbit

Solution 1:

The statement is true (see the proof below). In fact a stronger statement is true: if the derived subgroup of a f.g. group is finite, then the center is of finite index (a proof of that is almost the same).

1. Let $N$ be the finite normal subgroup of $G$, then the centralizer $C$ of $N$ is of finite index and normal in $G$, $M=C\cap N$ is central in $C$, and $C/M\cong CN/N\le G/N$ is a finitely generated Abelian group. So it is enough to show that $C$ contains an Abelian subgroup of finite index. Hence we can assume that $C=G$ and $N$ is a central subgroup of $G$. In fact we shall prove that the center of $C=G$ is of finite index.

2. Let $G/N$ be generated by cosets $t_1N,...,t_nN$. For every $i$ consider the map $\phi_i: x\mapsto [x,t_i]$ from $G$ to $N$ (note that since $G/N$ is Abelian, $N$ contains the derived subgroup of $G$). Since $[xz,t_i]=[x,t_i]^{z}\cdot [z,t_i]$ (a commutator identity, $^z$ is a conjugation), the map $\phi_i$ is a homomorphism into the finite group $N$. Let $K_i$ be its kernel. Then $K_i$ is of finite index for each $i$. Let $K$ be the intersection of all $K_i$. Then $K$ is also of finite index in $G$.

3. But $\forall x\in K\forall i$, $[x,t_i]=1$, so $K$ centralizes all $t_i$. Since it also centralizes $N$ ($N$ is central!), $K$ centralizes the whole $G$, so the center $Z(G)$ of $G$ is at least as large as $K$, and has finite index. $Z(G)$ is an Abelian subgroup of finite index.