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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 $d$?

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  • $\begingroup$ 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. $\endgroup$ Apr 3 at 18:26
  • $\begingroup$ Thank you for pointing them out, $r$ is fixed and any dependency on $r$ is allowed. $\endgroup$
    – ghc1997
    Apr 3 at 20:08

1 Answer 1

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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).

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  • $\begingroup$ 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. $\endgroup$ Apr 13 at 14:10
  • $\begingroup$ 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) $\endgroup$
    – ghc1997
    Apr 13 at 15:45
  • $\begingroup$ @David If I answered your question you can accept the answer. $\endgroup$ Apr 19 at 18:48

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