Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a virtually cyclic group, i.e., G has an infinite cyclic subgroup $H$ of finite index. Is it true that if $H'$ is another infinite cyclic subgroup of $G$ then $H'$ must be of finite index and it only has finitely many conjugates?

share|cite|improve this question
Let $g$ a generator of $H'$. Consider how the powers of $g$ move through the cosets of $H$. What does that tell you about $H \cap H'$? – Daniel Fischer Jul 9 '13 at 23:21

Firstly, the definition of "virtually cyclic" makes no presumptions about the group being infinite, so any finite subgroup is virtually cyclic. Indeed, we can say that every finite group is virtually trivial (a cutting pun to throw at any finite group theorist!). What you describe is "virtually-$\mathbb{Z}$".

Suppose $\mathbb{Z}\cong K\leq G$ where $G$ is virtually-$\mathbb{Z}$, and write $H$ for the maximal infinite cyclic subgroup of $G$, as you have done. Now, look at $K^{\prime}=K\cap H$. This is non-trivial, because, letting $K=\langle k\rangle$ and $H=\langle h\rangle$, we have that $G/H$ is finite so there exists $i\neq j$ such that $k^iH=k^jH$, so $1\neq k^{i-j}\in H$. The fact that $K^{\prime}$ is non-trivial proves the result. Can you see why?

share|cite|improve this answer

Your Answer


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.