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Let $F_2$ be the free group of rank 2. Let $K\le F_2$ be a characteristic subgroup, such that $G := F_2/K$ is finite.

Do there exist examples of such nonabelian $G$ such that the induced map $$Aut(F_2)\rightarrow Aut(G)$$ is surjective?

For example, if $C_n$ is the cyclic group of order $n$, then for every $n$, $G = C_n\times C_n$ is an abelian example of such a group.

EDIT: As Derek Holt pointed out in the comments, $C_n\times C_n$ is only an example in the cases $n = 1,2,3,4,6$.

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  • $\begingroup$ Would $K$ verbal work? The quotient would be the relatively free group of rank $2$ in the relevant variety of groups (your example being the case of $K$ the verbal subgroup generated by $xyx^{-1}y^{-1}$ and $x^n$). (You'd want a variety where the relatively free group is finite, of course...) $\endgroup$ – Arturo Magidin Sep 27 '16 at 18:16
  • $\begingroup$ @ArturoMagidin, iirc, there are no varieties of finite groups (and in particular no varieties in which the relatively free groups are finite). $\endgroup$ – HJRW Sep 27 '16 at 20:04
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    $\begingroup$ I think that $C_n \times C_n$ works only for small $n$ (probably $n=2,3,4,6$) because all automorphisms induced by elements of ${\rm Aut}(F)$ have determinant $\pm 1$. $\endgroup$ – Derek Holt Sep 27 '16 at 20:30
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    $\begingroup$ @HJRW: There are no varieties in which all groups are finite, but there certainly are varieties in which the finitely generated relatively free groups are finite (for example, the Burnside varieties of exponents $3$, $4$, and $6$ are locally finite, hence the finitely generated relatively free groups are finite); likewise, the relatively free groups of finite rank in $\mathfrak{N}_2\cap\mathfrak{B}_n$ are all finite (as are, trivially, the finitely generated relatively free groups in the varieties of abelian groups of exponent $n>1$). $\endgroup$ – Arturo Magidin Sep 27 '16 at 22:00
  • $\begingroup$ @ArturoMagidin, of course, thanks for the reminder. $\endgroup$ – HJRW Sep 28 '16 at 21:28
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I found one example of a verbal subgroup (as suggested by Arturo Magidin) that worked, but others that did not. Let $F/K$ be the largest quotient of $F=F_2$ that is a $2$-group and has exponent $2$ class $2$. In other words, $K=H^2[F,H]$, where $H=F^2[F,F]$.

Then $|F/K| = 2^5$ and has automorphism group of order $384$. (That is made of ${\rm GL}(2,3)$ of order $6$ acting on $F/H$, with a group of order $2^6$ that centralizes the layers $F/H$ and $H/K$.)

Now ${\rm Aut}(F)$ with $F = \langle a,b \rangle$ is generated by the three automorphisms (i) $a \mapsto b$, $b \mapsto a$; (ii) $a \mapsto ab$, $b \mapsto b$; and (iii) $a \mapsto a^{-1}$, $b \mapsto b$. It can be checked (I checked on a computer) that the induced automorphisms of $F/K$ generate the whole of ${\rm Aut}(F/K)$.

However, when I tried the same computation with exponent $3$ in place of exponent $2$, I found that $|{\rm Aut}(F/K)| = 34992 = |{\rm GL}(2,3)| \times 3^6$, but the subgroup induced by ${\rm Aut}(F)$ has order only $11664$.

I also tried the exponent $2$ class $3$ quotient of $F/L$ of $F$, where $L=K^2[F,K]$. Then $|F/L|=2^{10}$ and has automorphism group of order $6 \times 2^{16}$, but the subgroup induced by ${\rm Aut}(F)$ has order only $6 \times 2^{12}$.

This suggests that there might not be an abundance of examples.

Added later: Luc Guyot has pointed out that there is a quotient of $G/K$isomorphic to $Q_8$ which is also an example. I found also that the intersection of the kernels of the homomorphisms from $F$ to $S_3$, which has index $108$ in $F$, is a further example.

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  • $\begingroup$ Doesn't $F/K$ have exponent $4$ (though the images of the two generators have exponent $2$), as the subvariety of $\mathfrak{N}_2$ ought to correspond to the identities $x^4=[y,z]^2=[x,y,z] = 1$ ? (since a group of exponent 2 is necessarily abelian, after all). $\endgroup$ – Arturo Magidin Sep 27 '16 at 20:02
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    $\begingroup$ Yes it has exponent $4$. The exponent-$p$-class (I should probably have hyphenated exponent-$2$) of a $p$-group is the length of its so-called $p$-central series, which is computed by the well-known $p$-quotient algorithm, and descends in elementary abelian layers. I think you also want the identity $[x^2,y]=1$. $\endgroup$ – Derek Holt Sep 27 '16 at 20:28
  • $\begingroup$ Oh, okay; that explains it. The identity $[x^2,y]=1$ follows from $[x,y,z]=1$ and $[y,z]^2=1$, since $[x^2,y] = [x,y]^x[x,y]=[x,y][x,y,x][x,y]$. $\endgroup$ – Arturo Magidin Sep 27 '16 at 22:07
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The quaternion group $Q_8 = \langle x, y \,\vert \, xyx^{-1}y = yxy^{-1}x = 1\rangle$ has $24$ generating pairs, all Nielsen equivalent, and $24$ automorphisms. It looks like a good match, certainly the one with the smallest cardinal.

There is a reason why constructions of the type $G = F_2/K$ with $K$ a verbal subgroup of $F_2$ are not so abundant. For such a finite group $G$, the automorphism group acts transitively on the set of generating pairs. Because of the surjectivity assumption $Aut(F_2) \twoheadrightarrow Aut(G)$, this implies that $G$ possesses only one Nielsen equivalence class of generating pairs. So does its abelianization $G/[G, G] \simeq C_n \times C_n$ by a famous lemma of Gaschütz. As already noted by Derek Holt, this holds only if $n \in \{1, 2, 3, 4, 6\}$. Thus, examples of this kind suffer a significant restriction on their abelianization.

The following article of G. Rosenberger seems to be a reference for this kind of problems: "Automorphismen und Erzeugende für Gruppen mit einer definierenden Relation", 1972. (It may address only infinite groups though). This article is quoted in "Combinatorial Group Theory" of R. C. Lyndon and P. E. Schupp in Section I.4 and Section II.2; the key word is quasifree presentation. In more recent texts, some authors speak about tame automorphisms, others about induced automorphisms.

Afterthought: Looking at this post, it dawn on me that we have further elementary examples at hand. The Burnside group $B(2, 3)$ has $27$ elements and is isomorphic to the group $\text{Heisenberg}_2(R)$ of matrices of the form$$\begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$$for $x,y,z\in R = \mathbf{Z}/3\mathbf{Z}$. It is known that $\text{Heisenberg}_2(\mathbf{Z})$ is isomorphic to the two-generated 2-step free nilpotent group and that this group has a unique Nielsen class of generating pairs [1, Theorem 1.7]. It is easy to check that $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ has one only Nielsen equivalence class of generating pairs for $n \in \{2, 3, 4, 6\}$. The group $\text{Heisenberg}_2(\mathbf{Z}/2\mathbf{Z})$ is the dihedral group of order $8$, which is not a quotient of $F_2$ by a characteristic subgroup. But $B(2, 3) = \text{Heisenberg}_2(\mathbf{Z}/3\mathbf{Z})$, is certainly another example. For $n \in \{4, 6\}$, I didn't check whether $\text{Heisenberg}_2(\mathbf{Z}/n\mathbf{Z})$ is a quotient of $F_2$ by a characteristic subgroup.


[1] "Andrews–Curtis and Nielsen equivalence relations on some infinite groups", A. Myropolsky, 2016.

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  • $\begingroup$ But is $\langle xyx^{-1}y, yxy^{-1}x\rangle^{F_2}$ characteristic in $F_2$? $\endgroup$ – Arturo Magidin Sep 27 '16 at 22:02
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    $\begingroup$ Yes, because the above description of generating pairs implies that the kernel of any surjection $F_2 \twoheadrightarrow Q_8$ is the same. $\endgroup$ – Luc Guyot Sep 27 '16 at 22:06
  • $\begingroup$ I think it is a similar example to Derek Holt, or at worst a small quotient of Derek's example, since $Q_8$ lies in the corresponding subvariety and is a quotient of the relatively free group of rank $2$. $\endgroup$ – Arturo Magidin Sep 27 '16 at 22:13
  • $\begingroup$ Yes, it's a quotient of my example. $\endgroup$ – Derek Holt Sep 27 '16 at 22:22
  • $\begingroup$ Proofs that the kernel of the map to $Q_8$ is characteristic were given here: mathoverflow.net/questions/109679/… . $\endgroup$ – HJRW Sep 28 '16 at 4:38

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