About normal closure of cyclic subgroup

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?

• Is there an error in the question? If $a$ has infinite order, then so does all of its powers. – Autumn 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

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 $\mathbf{Z}\times\mathbf{Z}/2\times 0$.

• 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
• @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.

• 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