I have a group, and it has a subgroup of finite index which is isomorphic to $\mathbb{Z}$.

My questions are these:

-Is my group a semidirect product, $\mathbb{Z} \rtimes K$, or even $\mathbb{Z} \ltimes K$, for $K$ some finite group?

-Is there anything else I have automatically?

The former isn't surprising - without loss of generality we can assume that this subgroup is normal. And so we have a short exact sequence...but does it split? I found a couple of articles recently which seem to imply this is so, but didn't reference anywhere.

The latter is also interesting - I believe I have residually finite (RF is closed under finite index), but I can't think of anything else suitable interesting...

  • $\begingroup$ $\mathbb Z$ has subgroup of finite index, which is isomorph to $\mathbb Z$ and it doesn't decompose as a semidirect product $\mathbb Z \rtimes K$ (with finite non trivial $K$) because it contains no elements of finite order. $\endgroup$ – Giacomo d'Antonio May 5 '11 at 8:47
  • $\begingroup$ I don't mind if K is trivial, and nor do I mind if the action is trivial (i.e. it is a direct product). $\endgroup$ – user1729 May 5 '11 at 9:58
  • $\begingroup$ Giacomo's point is that you can't write $\mathbb{Z}$ as a semidirect product using $2\mathbb{Z}$, so your first question is hopeless in general. The second question is, IMO, way too broad. $\endgroup$ – user641 May 5 '11 at 13:34
  • $\begingroup$ Virtually-(free abelian) groups need not be isomorphic to semi-direct products of free-abelian by finite groups, right? I remember there being non-split space groups. Can someone give a one-dimensional example? The only examples I can think of end up being G⋉nZ in G⋉Z, which is split by a different copy of Z (which is a common problem in polycyclic groups). $\endgroup$ – Jack Schmidt May 5 '11 at 14:59
  • $\begingroup$ I cannot see how that (not being able to write $\mathbb{Z}$ as a semidirect product using $2\mathbb{Z}$) forms a counter-example. $\mathbb{Z}$ can be written as a semidirect product, trivially...and I take your point about the second question; I was asking it more as an after-thought. $\endgroup$ – user1729 May 5 '11 at 15:03

I believe that the answer to the first question is no.

Let $H$ be the direct product of the infinite cyclic group $\langle x \rangle$ and the cyclic group $\langle y \rangle$ of order 2. Then $H$ has an automorphism of order 2 with $x \mapsto x^{-1}y$ and $y \mapsto y$. Let

$G = \langle x,y,z \mid xy=yx, y^2=z^2=1, zxz = x^{-1}y, zy=yz \rangle$

be the semidirect product of $H$ with a group $\langle z \rangle$ of order 2 using this automorphism.

Let $Z$ be an infinite normal cyclic subgroup of $G$. Then $Z$ must intersect $\langle x \rangle$ nontrivially, but the centralizer of any nontrivial subgroup of $\langle x \rangle$ is contained in $H$, and hence $Z \subset H$. So $Z = \langle x^k \rangle$ or $\langle x^ky \rangle$ for some $k \ge 1$, and then normality of $Z$ in $G$ implies that $k$ is even. But then $Z$ has no complement in $H$ so it cannot have a complement in $G$ either.

Added later: for the second question, you can say that any group $G$ that contains an infinite cyclic subgroup $Z$ of finite index has a subgroup $H$ of index at most 2, which is the direct product of an infinite cyclic group and a finite group.

To see this, assume that $Z \unlhd G$, and take $H = C_G(Z)$. Then $|H:Z(H)|$ is finite, so by a result of Schur, $H'$ is finite, and then $H/H'$ and hence also $H$ is a direct product as claimed.

  • $\begingroup$ Ah, very nice. Thank you. $\endgroup$ – user1729 May 6 '11 at 8:43

$\newcommand{\ZZ}{\mathbb{Z}}$The following example shows that the implication "$H$ virtually $\ZZ$ implies $H$ equals $\ZZ \rtimes K$ or $K \rtimes \ZZ$." does not hold.

$\newcommand{\Aut}{\operatorname{Aut}}$$\newcommand{\Ends}{\operatorname{Ends}}$First, let $H$ be any virtually $\ZZ$ group. Then $H$ has exactly two metric ends. There is a homomorphism $H \to \Aut(\Ends(H)) \cong \ZZ_2$, given by the action of $H$ on itself via conjugation. We'll use this repeatedly.

Let $\ZZ_3 = \langle u \rangle$ be the cyclic group of order three and let $\phi \in \Aut(\ZZ_3)$ be the non-trivial element. Note $\phi^2$ is equal to the identity. Write $\ZZ = \langle t \rangle$. Form the semidirect product $G = \ZZ_3 \rtimes_\phi \ZZ$.

Now, let $\rho$ be the nontrivial involution of $\ZZ$. That is, $\rho(t) = t^{-1}$. If we take $\rho$ to act trivially on $\ZZ_3$ then it is a short computation to show that $\rho$ now is an automorphism of $G$.

So we write $\ZZ_2 = \langle w \rangle$ and we form the group $H = G \rtimes_\rho \ZZ_2 = (\ZZ_3 \rtimes \ZZ) \rtimes \ZZ_2$. Thus $H$ is virtually $G$, and so it is virtually $\ZZ$.

Note that $w \in H$ swaps ends of $H$ while $u$ preserves the ends of $H$.

Case 1

Note that no element of $K \rtimes \ZZ$ swaps the ends of the group. Thus $H$ is not isomorphic to a group of the form $K \rtimes \ZZ$.

Case 2

Now suppose that $H \cong \ZZ \rtimes K$. Since $H$ is virtually $\ZZ$, it must be that $K$ is finite. Let $s$ be the generator of the normal subgroup of $\ZZ \rtimes K$. Consider any $s^\ell k \in \ZZ \rtimes K$. Suppose that $k$ doesn't swap ends. Thus $ksk^{-1} = s$ and we find that $\ell \neq 0$ iff $s^\ell k$ has infinite order. If $k$ does swap ends (ie $ksk^{-1} = s^{-1}$) then $(s^\ell k)^2 = k^2$ is finite order and $(s^\ell k)^3 = s^\ell k^3$ is the trivial element implies that $\ell = 0$.

So let $k \in K$ be the image of $u \in \ZZ_3 < H$. Since $u$ doesn't swap ends, neither does $k$. Let $s^\ell h$ be the image of $t$. By the above, $h$ doesn't swap ends. We compute: $tut^{-1} = u^{-1}$ thus $s^\ell h k h^{-1} s^{-\ell} = hkh^{-1} = k^{-1}$. That is, there is a finite order element of $H$ that

  • conjugates $u$ to its inverse and
  • doesn't swap ends.

Using the normal forms for elements of $H$ we can check that this is not the case, and so have arrived at a contradiction. $\square$

I think that this is almost as bad as things can get. If $H$ is now any group that is virtually $\ZZ$ then the kernel of the map $H \to \Aut(\Ends(H))$, ie the "end preserving subgroup", always has index at most two. This kernel can, in turn, be written as $F \rtimes \ZZ$, where $F$ is the kernel of the map "$g$ goes to its average translation distance". The only remaining nasty bit is that the map from $H$ to $\Aut(\Ends(H))$ doesn't have to split...

EDIT - Following Professor Holt's good example, here is a presentation of my counterexample group $H$:

$H = \langle u, t, w \mid u^3 = w^2 = 1, \, wtw^{-1} = t^{-1}, \, tut^{-1} = u^{-1}, \, wu = uw \rangle.$

  • $\begingroup$ Don't you have to add "$H$ finitely generated"? Maybe I am wrong, but Stallings theorem is stated for f.g. groups. Does it hold for non f.g. groups? $\endgroup$ – MBL May 5 '11 at 23:27
  • $\begingroup$ @MBL - Virtually Z implies finitely generated. (Indeed, virtually finitely generated implies finitely generated. :) $\endgroup$ – Sam Nead May 5 '11 at 23:32
  • $\begingroup$ Is that easy to see? I don't know why. $\endgroup$ – MBL May 5 '11 at 23:40
  • 2
    $\begingroup$ @MBL - Yes - it is. Take a finite generating set for the finite index subgroup, and pick one element from each of its cosets. That gives a generating set. $\endgroup$ – Sam Nead May 5 '11 at 23:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.