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?
 A: 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.
A: 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 $\mathbf{Z}\times\mathbf{Z}/2\times 0$.
