Extension of a group action

Question

Let $G$ be a group and suppose that there exists an infinite cyclic normal subgroup $H$ of finite index. Let $K$ be it's centralizer. We know that $H$ acts on $\mathbf{R}$ (the real line) via translation, so that a generator $h$ acts via translation by $1$. I want to extend this uniquely to an action of $K$ as follows: if $k\in K$, pick the least $n$ such that $k^n=h^m$ for some $m$, and then $k$ acts as translation by $m/n$ (such $n$ exists because $H$ has finite index). The problem I have is that I can't see why the previous extension indeed defines an action of $K$ on $\mathbf{R}$. Explicitly, I can't see why $$(k_1 k_2)x=k_1(k_2 x)$$ for $k_1,k_2\in K$ and $x\in\mathbf{R}$.

Answer 1

Perhaps there is a simple algebraic method, using that $H$ is a finite index, central cyclic subgroup of $K$. Nonetheless, your question may be answered very quickly, and with a stronger conclusion, using a geometric method, namely Stallings Ends Theorem.

First, the following are easily seen to be equivalent:

1. $G$ contains a normal infinite cyclic subgroup of finite index.
2. $G$ contains an infinite cyclic subgroup of finite index.

Each of these easily implies the following:

3. $G$ is a 2-ended group.

This implies, by an easy application of a deep theorem, namely Stallings Ends Theorem, the following:

4. There is a homomorphism $G \to \mathbb{Z}$ or $G \to D_\infty$ (the infinite dihedral group) with finite kernel.

So, all you have to do now is to let the quotient group $\mathbb{Z}$ or $D_\infty$ act on the real line, and it is easy to arrange that this is consistent with the action of your given infinite cyclic normal subgroup $H$. Thus, the entire action of $H$ extends to an action of $G$.