An infinite polycyclic group has a free abelian normal subgroup

The question is in the title: prove that an infinite polycyclic group has a free abelian normal subgroup.

Can anybody give me a hint on how to go about proving this?

Things I know:

• A polycyclic group $G$ has a series $G = G_k \rhd G_{k-1} \rhd \dots \rhd G_1 \rhd G_0 = \{1\}$ with each $G_{i+1}/G_i$ cyclic for $0\le i \le k-1$.
• A free abelian group is an abelian group with a basis.
• The factors in the series above are either infinite cyclic or finite.
• Can you define "free abelian" for us? You seem to imply that a free abelian group can be finite (implicitly in your title, but also in your last line: "If they are finite, they are then also free abelian."). However, up to isomorphism a free abelian group has the form $\mathbb{Z}^n$. – user1729 Apr 16 '18 at 15:26
• @user1729 you are right. My grasp of this topic isn't great yet and I made a mistake, apologies. – elDin0 Apr 16 '18 at 15:31
• Okay, that's fine. Now, what have you tried? My first thought would be to try proof by induction. Have you tried this? Can you see why the base case holds? – user1729 Apr 16 '18 at 15:32
• @user1729 I tried this line of thought. The base case is fine because when we have $k=0$ then $G$ is itself cyclic and thus is itself an infinite free abelian group (and so normal subgroup of itself). For the inductive step, if we have that it works for $G_{k-1}$, and so we can get a free abelian normal subgroup $N$ here and aim to show somehow that this is normal in $G$, but I didn't manage to do this. Perhaps I need to use $N$ to get another normal subgroup that would work...? – elDin0 Apr 16 '18 at 15:38
• Every group has a free abelian normal subgroup: $\{1\}$. You mean a nontrivial one. – YCor Apr 16 '18 at 21:54

Here is a sketch proof, and I leave you to fill in the details.

A polycyclic group is solvable, so the derived series of $G$ terminates in the trivial subgroup. Then, by considering the last infinite quotient in the successive factors of the derived series, we see that there are normal subgroups $N,M$ of $G$ with $N < M$, $N$ finite, and $M/N$ infinite abelian.

Let $K/N$ be the torsion subgroup of $M/N$. Then $K$ is finite and $M/K$ is infinite and free abelian. Since $K$ is characteristic in $M$, it is normal in $G$.

Now $C_M(K)$ is also normal in $G$ and, since $K$ is finite, has finite index in $G$. So, by replacing $M$ by $C_M(K)$ and $K$ by $Z(K)$, we may assume that $K \le Z(M)$.

Now, if $|K|= n$, then the subgroup $L:=\langle g^n : g \in M \rangle$ of $M$ is abelian, because $[g,h] \in K\,\forall g,h, \in M$, and so $[g^n,h^n]=[g,h]^{n^2}=1$. So in fact $L = \{ g^n:g \in M \}$, which has trivial intersection with $K$. So $L$ it is free abelian. Also, it is characteristic in $M$ and hence normal in $G$.

• I follow your proof entirely, apart from the deduction that $M$ and $N$ are both normal in $G$. Is this because the original infinite groups in the series are both normal in $G$ (I don't think so), or are you using a trick similar to to in this question math.stackexchange.com/questions/42574/… where we use $M_G$ and so nothing is lost if we assume $M$ is normal in $G$? Thanks for your help. – elDin0 Apr 16 '18 at 21:15
• they are normal because a characteristic subgroup of a normal subgroup is always normal. – YCor Apr 16 '18 at 21:56
• Yes, $M$ and $N$ were successive terms in the derived series of $G$ and hence characteristic in $G$. I missed out a step in my original proof outline, which I have now out in. You need to assume that $K \le Z(M)$. – Derek Holt Apr 17 '18 at 7:39