Let $G$ be a group, and let $H\leq G$ be a subnormal subgroup. Suppose there exist a cyclic series from $H$ to $G$, that is, a normal series $$H=H_0\lhd H_1\lhd\cdots\lhd H_k= G$$ of subgroups of $G$ such that each factor group $H_{i+1}/H_i$ is cyclic. Define the relative Hirsch number of the pair $(G,H)$ to be the number $h(G,H)$ of infinite cyclic factors.


  1. Under what conditions on the group $G$ and subgroup $H$ does such a cyclic series exist?
  2. Supposing existence of a cyclic series as above, is the relative Hirsch number a well-defined invariant (ie independent of the cyclic series chosen)?
  3. If $H$ is normal in $G$, then is $h(G,H)=h(G/H)$?
  4. Where is this notion to be found in the literature?

Sorry to be asking so many questions at once! If it helps, I am mainly interested in the case $G=\Gamma\times\Gamma$ and $H=\Gamma$ the diagonal subgroup, where $\Gamma$ is finitely generated, torsion-free nilpotent.



What you are asking is about a generalization of the Hirsch length for polycyclic(-by-finite) groups. Of course, a finitely generated nilpotent group is polycyclic, so the special case that mainly interests you is quite classic.

For a polycyclic group (and more generally, for a polycyclic-by-finite group), the Hirsch length $h(g)$ is a well-known invariant, and it coincides with your "relative Hirsch number" for the pair $(G,1)$. Basic information about this can be found via Google, e.g. on Wikipedia. If you want a book, look at "Polycyclic Groups" by Daniel Segal. Unfortunately I am not aware of a reference for your relative Hirsch length.


  1. I don't know an answer to your first question in the general case, and already in the polycyclic case such series won't exist in general. But if your group $G$ is nilpotent (as in your special case), you can do the following to prove that such a series always exists: Start with the series $$ G=HG^{(0)}, HG^{(1)}, HG^{(2)}, \ldots , H $$ where $G^{(0)}:=G$ and $G^{(i+1)}=[G,G^{(i)}]$; since $G$ is nilpotent, there is $k\in\mathbb{N}$ such that $G^{(k)}=1$. One now verifies that the series from $G$ to $H$ I described is actually a subnormal abelian series. You can now refine it to a series in which all factors are cyclic. Since we are in the polycyclic setting (where all subgroups are finitely generated), this will result in a series of finite length.

  2. For polycyclic groups, $h(G)$ is well-defined. It is not hard to extend the standard proof for this to your settings, answering your second question in the affirmative: If two subnormal cyclic series from $G$ down to $H$ exist, then by the Schreier refinement theorem, they have equivalent refinements (i.e. the factors $H_i/H_{i+1}$ which occur in the two refinements are the same, just possibly ordered differently). But a refinement of one of these subnormal cyclic series cannot change the number of infinite cyclic factors (easy exercise, also used in the "standard" proof). Hence the relative Hirsch length is well-defined.

  3. Your third question also has a positive answer: Any subnormal cyclic series from $G$ to $H$ induces such a series from $G/H$ to $1=H/H$, and vice-versa, and the cyclic factors occurring obviously are the same in both cases. So you have $h(G,H)=h(G/H)$, and if $H$ also is polycyclic (as in your special case), you even have $h(G,H)=h(G/H)=h(G)-h(H)$.

  4. As I already mentioned, I don't know a good reference for this in the literature, only references to polycyclic(-by-finite) groups.

  • $\begingroup$ @Max: Thanks for your answer! I intend to accept it, unless someone comes along with a reference to the literature... ;) Meanwhile, can I utilise your expertise some more? In the case I describe of the diagonal subgroup, any ideas how to go about computing it? Is it likely that $h(\Gamma\times \Gamma,\Gamma)>h(\Gamma)$? $\endgroup$ – Mark Grant Jun 10 '11 at 18:43
  • 1
    $\begingroup$ @Mark: Let $G=\Gamma\times\Gamma$ and $H=\Gamma$ (embedded diagonally). Assuming my answer to 1 is correct (please verify!), you have $h(G,H)=h(G) - h(H)=h(H)$. Simply by taking a polycyclic series from $G$ to $H$ and then from $H$ to $1$, and concatenating them. The argument for my answer to 1 roughly is: For $g\in G^{(k)}$ and $h\in H$, we have $h^g=(hh^{-1})\cdot g^{-1}hg= h[h,g] \in HG^{(k+1)}$. Add in that all $G^{(k)}$ are characteristic subgroups. From this I concluded that $HG^{(k+1)}$ is indeed normal in $HG^{(k)}$. $\endgroup$ – Max Horn Jun 10 '11 at 21:03
  • $\begingroup$ Oh, and of course this also uses that $HG^{(k)}/HG^{(k+1)}$ is abelian (which follows by considering the commutator of two arbitrary elements in the $HG^{(k)}$ and concluding that this is actually contained in $HG^{(k+1)}$). $\endgroup$ – Max Horn Jun 10 '11 at 21:16
  • $\begingroup$ Anything else unclear? :) $\endgroup$ – Max Horn Jun 16 '11 at 8:53
  • $\begingroup$ No, that's clear as day! In fact, it now seems obvious that $h(G,H)=h(G)-h(H)$ whenever all terms are defined. This would explain why the relative Hirsch length doesn't show up in the literature! Thanks again. $\endgroup$ – Mark Grant Jun 21 '11 at 10:08

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