Suppose that we have a finitely generated residually finite group $G = \langle g_1,\ldots,g_r \rangle$ and $H$ is a subgroup of $G$ with finite index $m$. Let $\phi$ be an automorphism on $G$.
Question: What is the bound for the smallest $n \in \mathbb{N} \setminus \{0\}$, such that $\phi ^n(H) = H$?
My thought so far: We know that $G$ has at most $(m!)^r$ number of subgroups with index $m$, so $n \leq (m!)^r$. I was wondering if the bound can be improved, it seems unlikely for the automorphism to go through all the subgroups of this index. Is it possible to show that $n$ is bounded by a polynomial (or even linear function) in $m$, i.e. $ n = \mathcal{O}_r(m^d)$ for some $d$?
Best Answer
Let $G = \langle x,y,z \rangle = F_3$ be the free group on $3$ generators. If $w = w(x,y)$ is any word in $x$ and $y$, then $x$, $y$, and $zw$ generate $F_3$ because $x$ and $y$ generate $w$ and thus from $zw$ and $x$ and $y$ one can recover $x$, $y$, and $z$. Thus the map
$$\phi: x,y,z \rightarrow x,y,zw$$
is an automorphism of $F_3$. Now let $\psi: G\rightarrow S_m$ be the map such that:
$$\psi(x) = (1,2), \psi(y) = (1,2,3,\ldots,n), \psi(z) = e.$$
This is surjective because the first two elements generate $S_m$. Let $\sigma$ be any element of $S_m$. Then since $x$ and $y$ generate $S_m$, there is a word $w = w(x,y)$ such that $\psi(w) = \sigma$. Let $\phi$ denote the automorphism of $G = F_3$ which fixes $x$ and $y$ and sends $z$ to $zw$ for this $w$. Let $H$ denote the stabilizer of $1$ under the action of $G$ given by $\psi$, and let $K$ denote the normal closure of $H$, which is the kernel of $\psi$. If $\phi^n(H) = H$, then certainly $\phi^n(K) = K$ as well. But $\phi(z) = e$ so $z \in K$, whereas
$$\phi(\psi^n(z)) = \phi(zw^n) = \psi(w^n) = \sigma^n.$$
Hence $\phi^n(z)$ lies in $K$ only if $\sigma^n = e$. Thus any bound on $n$ (for $r = 3$) in terms of $m$ also gives a bound for the order of any element $\sigma \in S_m$. But this order is not bounded polynomially in $m$ (https://en.wikipedia.org/wiki/Landau%27s_function).