Prove that the Hirsch rank of a group is unique

Question

A group $G$ is called polycyclic if there exists a subnormal series $G = G_0 \unlhd G_1 \unlhd \dots \unlhd G_n = \{e\}$ whose factors are cyclic. Prove that arbitrary two polycyclic series of $G$ have the same number of infinite factors (Hirsch rank).

What I'm thinking is to consider two polycyclic series with different number of infinite factors, and then try to get a contradiction. However, I don't seem to be getting anywhere with that approach. I would like a hint as to how to solve this, but not a complete solution.

Answer 1

For an arbitrary group, define $c(G)$ as the (possibly infinite) supremum of lengths $n$ of chains $H_0\le \dots \le H_n$ of subgroups with $H_i/H_{i-1}$ infinite for all $i$.

Then a simple argument shows that $c(G)$ is additive under extensions: $c(G)=c(N)+c(G/N)$ for every normal subgroup of $G$. Indeed $\ge$ is trivial and $\le$ uses the fact that if $H\subset L$ are subgroups of $G$ and $H$ has infinite index in $L$, then either $H\cap N$ has infinite index in $L\cap N$, or $p(H)$ has infinite index in $p(L)$, where $p$ is the projection $G\to G/N$.

Since $c(F)=0$ for $F$ finite and $c(\mathbf{Z})=1$, it follows that in a subnormal series with cyclic quotients for a group $G$, the number of infinite cyclic factors is equal to $c(G)$, which depends only on $G$.

Answer 2

Alternate solution: if $G$ has two polycyclic series, then by Schreier's theorem they have equivalent refinements (so the refinements would have the same number of infinite cyclic factors).

Claim: a refinement of a polycyclic series does not change the number of infinite factors.

Proof: Suppose the term $K \unlhd L$, where $K/L$ is infinite cyclic. Then consider arbitrary refinement $K \unlhd M \unlhd L$. By the third isomorphism theorem, $K/M \cong (K/L)/(M/L)$. We have that $M/L \leq K/L$, so $M/L$ must be infinite cyclic. $K/M$ is isomorphic to a factor group of $\mathbb{Z}$, and is hence finite. So the refinement has the same number of infinite factors as the original polycyclic series.

Hence any two polycyclic series of a group $G$ have the same number of infinite factors.