On the integers $\mathbb{Z}$ acting on a group (not necessarily finitely generated)

Question

Suppose we have $\mathbb{Z}$ acting on a group $K$ (not necessarily finitely generated) and let $\phi : \Bbb Z \to {\rm Sym}(K)$ be the homomorphism define this action.

Suppose that $N$ is a normal subgroup of $K$ with finite index $m$. Is it true that there exists an $n \in \mathbb{Z} \setminus \{0\}$ such that for every $k$ in $N$, $\phi (n) (k) = \phi^n (k) \in N$?

I think if we take $n$ to be $m$, it would work, as there are only $m$ distinct cosets, but I couldn't find a way to prove it or find a counterexample.

Answer 1

This would be true for finitely generated $K$ because there would then be only finitely many subgroups of index $m$ (there is an explicit bound as a function of $m$), so they would have to be permuted by $\phi$.

But it is not true in general. Let $K$ by a (restricted) direct product of countable many copies $H_i$ ($i \in {\mathbb Z}$) of $C_2$, and let ${k \in \mathbb Z}$ act on $K$ by mapping $H_i$ to $H_{i+k}$.

For $n \in {\mathbb Z}$, let $N_n$ be the subgroup $\langle H_i : i \in {\mathbb Z} \setminus \{n\} \rangle$ of $K$, and let $N = N_0$. So $|K:N| = 2$, but ${\mathbb Z}$ acts freely on the set $\{ N_i : i \in \mathbb{Z} \}$.

Answer 2

Obviously your condition will be satisfied for $n=0$, so I assume you meant $n\in\Bbb Z\setminus\{0\}$.

Consider $K=\Bbb F_2^{\oplus \Bbb Z}$ (the $\Bbb F_2$-vector space with basis $\Bbb Z$), $N=\{v\in K\,:\, v_0=0\}$ and $\phi:\Bbb Z\to\operatorname{GL}(K)$, $(\phi(n)(v))_i=v_{i-n}$ (i.e. $\phi(n)$ is the right-shift by $n$ positions). Then $\lvert K/N\rvert=2$ but $\phi(n)[K]\subseteq K$ if and only if $n=0$.