Embedding $S_3$ into $Aut(F_2)$

Question

Consider the two (inequivalent) $\mathbb{Z}$-representations $\phi,\psi$ of the symmetric group $S=S_3$ given by

$(1,2)^\phi=\left(\begin{array}{rr}0 &-1\\\ -1 & 0\end{array}\right), \qquad (1,2,3)^\phi=\left(\begin{array}{rr}0 &1\\\ -1 & -1\end{array}\right);$

$(1,2)^\psi=\left(\begin{array}{rr}0 &1\\\ 1 & 0\end{array}\right), \qquad (1,2,3)^\psi=\left(\begin{array}{rr}0 &1\\\ -1 & -1\end{array}\right).$

Now, let $F=\langle x,y\rangle$ be a free 2-generated group. The representation $\phi$ can be "lifted" to an embedding $\tau:S\to\rm{Aut}(F)$ as follows:

$(1,2)^\tau=[x\mapsto y^{-1};\quad y\mapsto x^{-1}], \qquad (1,2,3)^\tau=[x\mapsto y;\quad y\mapsto x^{-1}y^{-1}].$

Question. Can one similarly lift $\psi$?

Remark 1. By "lifting" a representation $\phi:S\to\rm{GL}_2(\mathbb{Z})$ I mean finding an embedding $\tau:S\to\rm{Aut}(F)$ such that $\phi=\tau\alpha$, where $\alpha:\rm{Aut}(F)\to\rm{GL}_2(\mathbb{Z})$ is the natural epimorphism.

Remark 2. A naïve attempt to send $(1,2)\ \mapsto\ [x\mapsto y;\quad y\mapsto x], \qquad (1,2,3)\ \mapsto\ [x\mapsto y;\quad y\mapsto x^{-1}y^{-1}]$

does not give a lifting of $\psi$.

Answer 1

As Tom Goodwillie noted in his comment, $GL(2,\Bbb Z)$ can be identified with $Out(F_2)$, so the question can be rephrased in terms of lifting subgroups of $Out(F_2)$ to $Aut(F_2)$. There is a Realization Theorem for finite subgroups of $Aut(F_n)$ and $Out(F_n)$ which says that such a subgroup can always be realized as a group of symmetries of some finite connected graph with fundamental group $F_n$, where the symmetries fix a basepoint in the graph in the case of $Aut(F_n)$. When $n=2$ there are only two graphs to consider, and the relevant one for $S_3$ is the join of two points with three points. This has two symmetry groups isomorphic to $S_3$, but only one of these two groups fixes a basepoint, so this should answer the question.

The Realization Theorem is discussed in Karen Vogtmann's survey paper "Automorphism groups of free groups and outer space", section II.6. The references given there are to papers by M. Culler, B. Zimmermann, and D. G. Khramtsov from 1981 to 1984.

Answer 2

I have wondered about such lifts myself, and I want to give what I hope is a tantalizing hint of what such lifts may be able to tell us:

The lift you give of $S_{3}$ from $GL_{2}( \mathbb{Z} )$ to $Aut(F_{2})$ also gives an embedding of $C_{3} = A_{3}$ into $Aut(F_{2})$. Why is this useful? It gives a character-free proof of a congruence about conjugacy classes of a finite group (here $c(G)$ denotes the number of conjugacy classes of the group $G$):

Theorem. If $G$ is a finite group with $|G|$ not divisible by $3$, then $|G| \equiv c(G) \mod{3}$.

Proof, with character theory: $|G|$ is the sum of the squares of the dimensions of the complex irreducible representations of $G$. The number of these is $c(G)$. Since $|G|$ is not a multiple of $3$, these dimensions aren't multiples of $3$. Now reduce modulo $3$ and obtain the congruence.

Proof, without character theory: $|G|(|G| - c(G))$ is the number of non-commuting ordered pairs of elements of $G$. Since $|G|$ is a nonmultiple of $3$, the congruence may be proved by showing that this set has a number of elements which is a multiple of $3$. Now just let the lift of $\tau$ act on it: If $(x,y)$ is a fixed point, then reading the first coordinate gives $x=y$, which trivially implies $xy=yx$. So the action is fixed-point-free, and we are done.

Theorem. If $G$ is a finite group of odd order, then $|G| \equiv c(G) \mod{8}$.

Proof, with character theory: $|G|$ is the sum of the squares of the dimensions of the complex irreducible representations of $G$. The number of these is $c(G)$. Since $|G|$ is odd, these dimensions are odd. Now reduce modulo $8$ and obtain the congruence.

Proof, without character theory: Instead of lifting $S_{3}$ to $Aut(F_{2})$, lift the dihedral group of order $8$, which is generated by the involutions $(x,y) \to (x^{-1},y)$ and $(x,y) \to (y,x)$. It suffices to check that none of the $5$ involutions of this group has a fixed point in the action on the non-commuting pairs of elements of $G$. This is easy to do, and it involves recalling that, since $|G|$ is odd, $t^{2}=1$ implies $t=1$ for $t \in G$.

This is all well and good, but, it is not the end of the story:

Theorem. The number of rows in the character table of $G$ which are entirely real-valued is the number of conjugacy classes $C$ of $G$ such that $x \in C$ iff $x^{-1} \in C$.

Corollary. If $|G|$ is odd, then the only entirely real-valued character of $G$ is the trivial character.

This leads to a strengthening of the character theory argument used above, so that it now proves that $|G| \equiv c(G) \mod{16}$.
In fact, $16$ is the highest power of $2$ that works here, since for any prime $p \equiv 3 \mod{8}$, we can let $G$ be a nonabelian group of order $p^{3}$. This gives $|G| - c(G) = p^{3} - (p^{2}+p-1) = (p^{2}-1)(p-1) \equiv 16 \mod{32}$.

The really tantalizing thing here is that $(p^{2}-1)(p-1)$ is the $p$-free part of the order of the general linear group $GL_{2}(\mathbb{Z} / (p))$.
I do not know whether $|G|-c(G)$ is always a multiple of $(p^{2}-1)(p-1)$ when $G$ is a $p$-group (though the theorems whose proofs I outlined above establish it when $p = 2$ or $p = 3$), and I do not know if some analogue of the lifting argument works, possibly with $Aut(F_{2})$ replaced by $Aut(B(2,p))$ or $Aut(B(2,p^{k}))$, where $B(r,n)$ denotes the rank $r$, exponent $n$ Burnside group.

Also see Bjorn Poonen's paper:
Congruences relating the order of the group to the number of conjugacy classes, American Mathematical Monthly, 105(1995), 440-442.