Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a group, $a \in G$, $|a|$ is infinite. If $|\langle a \rangle ^G : \langle a \rangle |$ is finite ($>1$). I want know whether or not that $|x|$ is infinite for every non-trivial element in $\langle a \rangle^G$?

We hope to investigate the influence of normal closure, so any reference about this situation will be appociated.

Added: By the answer of Kevin, I see that there exists example that $\langle a \rangle ^G$ is not torsion-free. Hence the case that $\langle a \rangle ^G$ is torsion-free is more atractive to me. Thanks the answer of Richard, we know it will be cyclic in this case. But I can't find a non-trivial example. Anyone can provide an example?

share|improve this question
    
Is there an error in the question? If $a$ has infinite order, then so does all of its powers. –  Richard Kent May 27 '12 at 15:14
    
@Wei: I think, you meant to ask "...for every nontrivial element of $\langle a\rangle^G$?" Also, you should add definition of $\langle a\rangle^G$ in your question, not in the comments. –  Misha May 27 '12 at 15:50
    
Thanks, Misha. You are right. –  Wei Zhou May 27 '12 at 16:05

2 Answers 2

up vote 4 down vote accepted

No. Consider the semi-direct product $(\mathbf{Z}\times \mathbf{Z}/2)\rtimes \mathbf{Z}/2$, where the rightmost factor acts by $(1,0)\mapsto (1,1)$ and fixes $(0,1)$. Then the normal closure of $\mathbf{Z}\times 0\times0$ is $\mathbf{Z}\times\mathbf{Z}/2\times 0$.

share|improve this answer
1  
More generally, take a finite group F and a semidirect product $F\rtimes Z=G$ which is not split as a direct product. Then there will be an element g∈G which does not normalize Z=⟨a⟩. The groups $Z^g$ and $Z$ generate a subgroup of G containing nontrivial finite order elements. –  Misha May 28 '12 at 2:54
    
Thank you very much. It answered my question. I don't take the answer as the best only because I want to know the case that $\langle a \rangle ^G$ is torsion-free. –  Wei Zhou May 28 '12 at 5:27
1  
@Wei so what do you want? Kevin precisely answered your question. If you say you restrict to the torsion-free case, I understand your question as the following very interesting one "if $\langle a\rangle ^G$ is torsion-free, are all its nontrivial elements of infinite order" :) anyway indeed there is no nontrivial torsion-free example, see Richard's answer and my comment. –  YCor May 28 '12 at 10:07

This is not really an answer, but more of a thought:

It is perhaps worth noting that if every element of $\langle a \rangle^G$ has infinite order, then $\langle a \rangle^G$ must be cyclic itself: You assume $|\langle a \rangle^G:\langle a \rangle|$ finite. Since torsion-free groups and their finite index subgroups have the same cohomological dimension (thanks to a theorem of Serre), and groups of cohomological dimension 1 are free (thanks to the Stallings-Swan theorem), $\langle a \rangle^G$ is free. But then euler characteristic considerations imply that $\langle a \rangle^G$ is cyclic. (This is maybe overkill in the virtually cyclic case.)

So, you may rephrase your question to say:

If a virtually cyclic group $H$ is the normal closure of a single element, is $H$ cyclic?

Groups that are the normal closure of a single element are said to have weight one, and it is a theorem of Gonzalez-Acuna (see Johnson, Homomorphs of Knot Groups, Proceedings of the AMS, Volume 78, Number 1, January 1980) that groups of weight one are quotients of knot groups. I don't know if that is of any use, but maybe there is a geometric argument lurking somewhere.

share|improve this answer
    
Thank you very much for your helpful answer. –  Wei Zhou May 28 '12 at 5:43
4  
you can say more: if $\langle a\rangle^G$ is torsion-free, it is equal to $\langle a\rangle$. Indeed, since it is normal and isomorphic to $\mathbf{Z}$, all its finite index subgroups remain normal (because all subgroups of $\mathbf{Z}$ are characteristic in $\mathbf{Z}$). –  YCor May 28 '12 at 10:02
    
Thanks, Yves. I find what I look for. –  Wei Zhou May 28 '12 at 10:58
    
Thanks, Yves, I meant to mention that. –  Richard Kent May 28 '12 at 14:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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