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I am trying to align the following facts.

  • The free group with two generators $F_{2}$ is isomorphic to a congruence subgroup $\Gamma(2)\le\mathrm{SL}_{2}(\mathbb{Z})$.
  • The outer automorphism group of $F_{2}$ is isomorphic to $\mathrm{GL}_{2}(\mathbb{Z})$.

Surely the action of $\mathrm{Out}(F_{2})=\mathrm{GL}_{2}(\mathbb{Z})$ on $F_{2}=\Gamma(2)$ is not via conjugation. What is this action? Is there a neat, direct description of this action?

Added: It might be a long shot but what ultimately I wanted to ask was if there is a nice description of this action without really referring to $F_{2}$.

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    $\begingroup$ The action of outer automorphisms is never by conjugation (by definition, otherwise it would be inner!). You can find a detailed account of automorphisms of free groups in Lyndon-Schupp, Chapter I, §4, including $\operatorname{Out}(F_2)$. For example Proposition 4.5 gives explicit generators for $\operatorname{Out}(F_2)$ and the generators of $\operatorname{GL}_2(\mathbb{Z})$ they are mapped (isomorphically) to. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Jul 28, 2021 at 21:33
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    $\begingroup$ @Carl-FredrikNybergBrodda: Of course your point is overall correct, but may be worth noting that we can have $H\subseteq G$ and conjugation by certain elements of $G$ (not in $H$) give outer automorphisms of $H$. This happens with some of the simple Lie groups, I believe. $\endgroup$
    – Sam Hopkins
    Commented Jul 28, 2021 at 21:37
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    $\begingroup$ The group $Out(F_n)$ acts on the abelianization $F_n^{\text{ab}} = \mathbb{Z}^n$ of $F_n$. This gives a homomorphism $Out(F_n) \rightarrow Out(\mathbb{Z}^n) = Aut(\mathbb{Z}^n) = GL(n,\mathbb{Z})$. For $n=2$, this is an isomorphism, but for higher $n$ it is surjective but not injective (its kernel is called the "Torelli subgroup" of $Out(F_n)$). $\endgroup$
    – Andy Putman
    Commented Jul 28, 2021 at 21:45
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    $\begingroup$ (in other words, your first point is just a red herring and has nothing to do with the action you're looking for [which is not an action at all anyway, but an outer action, so you should be suspicious if you see something that looks like an actual action]) $\endgroup$
    – Andy Putman
    Commented Jul 28, 2021 at 21:47
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    $\begingroup$ $\Gamma(2)$ is actually not free. For example it contains $-I$ (this is a common mistake). However there is a subgroup of $\Gamma(2)$ which is free, which is sometimes called the Sanov subgroup. $\endgroup$
    – Will Chen
    Commented Jul 28, 2021 at 22:33

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