# Permuting subgroups with the same finite index

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 $$d$$?

• Lots of typos and omissions. Is $r$ fixed? Is $N=H$? When write $n = \mathcal{O}(n^d)$ you presumably mean $n = O(m^d)$ but it's unclear what dependence on $r$ is allowed. Apr 3 at 18:26
• Thank you for pointing them out, $r$ is fixed and any dependency on $r$ is allowed. Apr 3 at 20:08

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).

• Free groups admit a huge number of subgroups of index $m$. If you take $F_3$, then if you take any of the $(m!)^3$ triples of elements in $S_m$ then with probability close to $1$ they generate $S_m$ or $A_m$, and so you get all of these subgroups. I expect, however, that the automorphism group of $F_3$ might be close to transitive on the subgroups whose normal closure has quotient $S_m$, so you get a transitive action on $n!^3$ points or so. It's not surprising then that this action has elements of large order. Apr 13 at 14:10
• Thank you so much for your detailed answer and explanation!!👍 So we have to assume that the group $G$ has additional property for $n$ to be bounded by a polynomial. (Maybe the group itself needs to have polynomial growth) Apr 13 at 15:45