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Here $F_2$ is the free group on two generators $x,y$. I'm interested in examples of finite groups $G$ such that $Aut(F_2)$ acts transitively on the set of surjections $F_2\rightarrow G$. (In particular I'm interested in $G$'s where automorphisms of $F_2$ of determinant 1 act transitive on surjections $F_2\rightarrow G$)

In this case the conjugacy class of the image of the commutator $xyx^{-1}y^{-1}$ is an invariant of any $Aut(F_2)$-orbit, so a necessary condition for transitivity of the action is that any two commutators of generating pairs of $G$ are conjugate in $G$

If $G$ is simple then one expects that "most" pairs of elements of $G$ will actually generate $G$. Thus, for alternating groups $A_n$ ($n > 5$), where every element is a commutator, one expects the action of $Aut(F_2)$ can't possibly be transitive.

For groups of the form $G = PSL_2(\mathbb{F}_q)$ it's proven in http://www2.math.ou.edu/~dmccullough/research/pdffiles/traces.pdf

that for $q > 11$ almost every element of $\mathbb{F}_q$ appears as the trace of a commutator of a generating pair, and so again $Aut(F_2)$ can't act transitively in this case.

Is this generally true for all finite nonabelian simple groups? More precisely..

  1. Does there exist a nonabelian simple group $G$ on which $Aut(F_2)$ acts transitively on surjections $F_2\rightarrow G$?

  2. What heuristics are there for describing the groups $G$ admitting such a transitive action (by $Aut(F_2)$? by automorphisms of determinant 1?)

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  • $\begingroup$ It is unclear to me whether the second part of your question refers to non-abelian finite simple groups or to (finite) $2$-generated groups. $\endgroup$ – Luc Guyot Jun 17 '16 at 22:25
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The answer to the first part of OP's question is "no", but possibly with a finite number of exceptions.

Indeed [Theorem 1.8, 4] established that the number of $T_2$-systems of a non-abelian finite simple group $G$ tends to infinity as the cardinality of $G$ tends to infinity. A $T_2$-system in the sense of B. H. Neumann and H. Neumann (T-system stands for transitivity system) is an orbit of $\operatorname{Aut}(F_2) \times \operatorname{Aut}(G)$ acting on the set $\operatorname{Epi}(F_2, G)$ of the epimorphisms $\pi: F_2 \twoheadrightarrow G$ in the following way: $\pi \cdot (\psi, \phi) = \phi^{-1} \circ \pi \circ \psi$. An orbit of $\operatorname{Aut}(F_2)$ is called a Nielsen equivalence class and clearly the number of such classes is bounded from below by the number of $T_2$-systems.

This result was already proved by M. Evans, R. M. Guralnick and I. Pak for the sequence $(PSL_2(\mathbb{F}_q))_q$ with $q$ the power of a prime, and was refined in [5], as noted in OP's question. I. Pak also proved in [Proposition 2.5.2, 1] the result for the sequence $(A_n)_n$ of the finite alternating groups and he conjectured with Guralnick the aforementioned general result in [2]. Estimates for the number of $T_2$-systems of these groups, and for the Suzuki groups as well, are collected in [Theorem 2.1, 3].

I don't know of any non-abelian finite simple group with exactly one $T_2$-system. In contrast, Wiegold's conjecture asserts that the number of $T_3$-systems should always be $1$ for such groups.

Regarding the second part of OP's question, I don't have any heuristics at hand. Stating the obvious, I would definitely try to know more about the classification of finite simple groups.

If the second part of the question concerns also two-generated finite groups which are not necessarily simple, then the following observation may help: the class of the finite groups $G$ for which $\operatorname{Aut}(F_2)$ acts transitively (the same holds if we require the determinant to be $1$) on $\operatorname{Epi}(F_2, G)$ is stable under quotient. This follows from Gaschutz'lemma [Lemma 2.1.5, 1] and imposes for instance that the order of the first invariant factor of $G_{ab} = G/[G, G]$ lies in $\{2, 3, 4, 6\}$ when $G_{ab}$ is not cyclic.


[1] "What do we know about the product replacement algorithm?", I. Pak, 2000.
[2] "On a question of B.H. Neumann", R. M. Guralnick and I. Pak, 2002.
[3] "Nielsen equivalence and stability graph of finitely generated groups", M. Evans, 2008.
[4] "Commutator maps, measure preservation, and T-system", S. Garion and A. Shalev, 2009.
[5] "Nielsen equivalence of generating pairs of $SL_2(q)$", D. Mc Cullough and M. Wanderley, 2013.

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  • $\begingroup$ Why is it that the first invariant factor of the abelianization is in that set? For example, doesn't Aut(F2) act transitively on surjections to any finite cyclic group (or at least those of prime order?). I'm assuming that by {1,2,3,4,6} you mean that the first invariant factor is a cyclic group of order in that set? $\endgroup$ – Will Chen Jun 18 '16 at 17:59
  • $\begingroup$ @oxeimon: Thanks, you're right. I have been slack. I edited my answer to address your comment. $\endgroup$ – Luc Guyot Jun 18 '16 at 19:07
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This perhaps should be a comment, but it is a bit too long. In

N. Dunfield, W. Thurston's, Finite covers of random 3-manifolds, Invent. Math. 166 (2006) 457-521

they study the set of surjections from the fundamental group of a genus $g$ surface to a fixed finite group. The automorphism group of a surface group is the mapping class group of a surface. They prove that for large $g$, the number of mapping class group orbits of surjections is independent of $g$ (and in fact give a precise formula for the number of such orbits). See the discussion around Corollary 6.17.

There is a strong analogy between the mapping class group and the automorphism group of a free group, so this discussion might be enlightening.

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  • $\begingroup$ As the rank of a closed surface group is two times the genus of the surface, we are left with the finite quotients of the fundamental group of a torus, i.e., the $2$-generated finite abelian groups. (For those groups, there is moreover no mystery about Nielsen equivalence, see e.g., Diaconnis-Graham, 1999.) $\endgroup$ – Luc Guyot Jun 17 '16 at 22:30
  • $\begingroup$ @luc guyot: I am aware that this says nothing interesting in the 2 generator case, which is why I said it was more of a comment than an answer. $\endgroup$ – Andy Putman Jun 18 '16 at 3:17

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