For $h \in G$ and $\phi \in Aut(G)$ is $\phi^n(h)$ periodic in any finite quotient? Let $G$ be an infinite group, and $\phi$ an automorphism of it. Let $N$ be a normal subgroup of $G$ such that $G/N$ is finite. Is it true that for any $h$ in $G$, $\phi^n(h)N$ (as a sequence of elements in $G/N$ for $n=1,2,3,...$) is periodic?
On the one hand my intuition tells me that it's false, but on the other I can't find any counter-examples.
 A: This is false. Let


*

*$G$ be the direct sum of countably many copies of $\mathbb{Z}/2\mathbb{Z}$ indexed by $\mathbb{Z}$ with generators $e_i, i \in \mathbb{Z}$. 

*$S \subset \mathbb{Z}$ be an infinite set of non-negative integers such that their indicator function $1_S$ is not periodic. 

*$G \to G/N \cong \mathbb{Z}/2\mathbb{Z}$ be the quotient in which the generators $e_s, s \in S$ are identified and the others are killed. 

*$\phi : G \to G$ act by $\phi(e_i) = e_{i+1}$. 

*$h = e_0$.  


Then $\phi^i(h) N = 1_S(i)$ is not periodic; in fact it can be arbitrary. 
Perhaps some words about the train of thought behind this counterexample might be helpful. First, $\phi$ descends to an automorphism $G/N \to G/N$ if is inner, so if there is a counterexample, then $\phi$ necessarily has infinite order in $\text{Out}(G)$. The easiest way I know to get a large group of outer automorphisms is to take the direct sum or direct product of copies of some abelian group, and the simplest automorphisms of these are the permutations. The simplest permutation of infinite order is an infinite cycle, and after that it wasn't hard to see how to choose $N$.
