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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...

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$\mathbb Z$ 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. –  Giacomo d'Antonio May 5 '11 at 8:47
    
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). –  user1729 May 5 '11 at 9:58
    
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. –  user641 May 5 '11 at 13:34
    
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). –  Jack Schmidt May 5 '11 at 14:59
    
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. –  user1729 May 5 '11 at 15:03

2 Answers 2

up vote 5 down vote accepted

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.

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Ah, very nice. Thank you. –  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 $H$:

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

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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? –  MBL May 5 '11 at 23:27
    
@MBL - Virtually Z implies finitely generated. (Indeed, virtually finitely generated implies finitely generated. :) –  Sam Nead May 5 '11 at 23:32
    
Is that easy to see? I don't know why. –  MBL May 5 '11 at 23:40
2  
@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. –  Sam Nead May 5 '11 at 23:45

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