# Every polycyclic group has a normal poly-infinite cyclic subgroup of finite index

There is a theorem that states: "Every polycyclic group has a normal poly-infinite cyclic subgroup of finite index. "

I just read the proof of it and honestly found some difficulties in it. The main part of the proof is as follows:

Let {$G_{i}$}, 0 ≤ $i$ ≤ $k$ be a cyclic series in a polycyclic group G. If $i$ < 1, then G is cyclic and the result is obvious. Let $i$ > I and put N = $G_{n-1}$· By induction on $i$ there is a normal subgroup $M$ of $N$ such that $M$ is poly-infinite cyclic and $N$/$M$ is finite. Now $N$/$M_{G}$ is finite because it is a finitely generated torsion group.Thus nothing is lost if we assume that $M$ is normal in $G$. If $G$/$N$ is finite, so is $G$/$M$ and we are finished. Assume therefore that $G$/$N$ is infinite cyclic, generated by $x$$N say and .... My question is refered to the proof when it starts with N/M_{G}. Why N/M_{G} would be finitely generated? Am I right that normalithy of N in G makes N/M_{G} to be torsion group? M_{G} is intersection of all g^{-1}$$M$$g where g is in G. - Every subgroup of a polycyclic group is finitely generated. – Jack Schmidt Jun 1 '11 at 15:24 ## 2 Answers A group is polycyclic if and only if it is soluble and satisfies the maximal condition on subgroups (every nonempty collection of subgroups has maximal elements). The maximal condition on subgroups is equivalent to finite generation of all subgroups by the standard arguments. Alternatively, you have a series for N$$ 0 \leq G_1 \leq G_2 \leq \cdots \leq G_{n-2} \leq N$$with each quotient G_{i+1}/G_i cyclic. Thus, G_1 can be generated by one element, G_1=\langle x_1\rangle; then G_2/G_1 can be generated by a single element, G_2/G_1 = \langle x_2G_1\rangle, and hence G_2=\langle x_2,G_1\rangle = \langle x_2,x_1\rangle. Continuing this way, we obtain that G_{n-2}=\langle x_1,\ldots,x_{n-2}\rangle for some elements x_1,\ldots,x_{n-2}, and N/G_{n-2} = \langle x_{n-1}G_{n-2}\rangle, hence$$N = \langle G_{n-2},x_{n-1}\rangle = \langle x_1,\ldots,x_{n-1}\rangle.$$So$N$is finitely generated, hence so is any quotient of$N$, in particular$N/M_G\$ is finitely generated.

-

The other question: Why is N / MG torsion?

Consider the diagonal homomorphism from N to the direct product of N / Mg over all g in G. The kernel is the G-core of M, MG = ∩( Mg : g in G ). The image is a subgroup of the direct product, and the direct product is a torsion group: each direct factor N / Mg = ( N / M )g is isomorphic to the finite group N / M, and hence has exponent dividing [N : M].

You are correct that it is important N is normal, otherwise one would not necessarily have that N / Mg had bounded exponent, and so the direct product would not necessarily be a torsion group. I'm not sure if the quotient N / MG could fail to be torsion, but I would not be surprised.

-