Can someone give an explicit example of a group with two generators $a$, $b$, such that $a^2 = b^3 = 1$ and $a b$ has infinite order, but which is not isomorphic to the free product of $\mathbb{Z}_2$ and $\mathbb{Z}_3$?
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asked Oct 12, 2010 at 17:30
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$\begingroup$ Isn't the free product of $C_2$ and $C_3$ a just-infinite group? $\endgroup$– Steve DCommented Oct 12, 2010 at 18:07
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4$\begingroup$ @Steve No. The group $\langle a, b | a^2 = b^3 = (ab)^7 \rangle$ is the fundamental group of a closed hyperbolic $2$-orbifold, and is infinite. $\endgroup$– Autumn KentCommented Oct 12, 2010 at 18:17
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2$\begingroup$ @Richard Are you missing at ``$=1$" in your example? Otherwise I don't think the example you give is a quotient of the group described in the question. $\endgroup$– Peter TingleyCommented Oct 12, 2010 at 18:27
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$\begingroup$ To perhaps make Mark's answer below a little more explicit: you can quotient your free product by almost any 'sufficiently complicated' element, and you will get another infinite hyperbolic group with the properties you want. 'Sufficiently complicated' means something like 'satisfying a suitable small-cancellation condition'. $\endgroup$– HJRWCommented Oct 12, 2010 at 18:29
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2$\begingroup$ @Peter Oops. Yeah, all those words should be set equal to 1. $\endgroup$– Autumn KentCommented Oct 12, 2010 at 18:33
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4 Answers
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It is straightforward to calculate that the commutator subgroup $G' = D$ of $G = \langle a,b \mid a^2, b^3 \rangle$ is a free group on the generators $x=bab^{-1}a$, $y=b^{-1}aba$, where $|G:D|=6$.
Now $(ab)^6$ is equal to the commutator $x^{-1}yxy^{-1}$, which lies in $D'$ but not in $D''$, so if we add any nontrivial element of $D''$ as an extra relator of $G$, then we will get an example with the required property.
answered Oct 12, 2010 at 21:16
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$\begingroup$ Wow, this is a really great example; it gives everything I was asking for. Thanks very much! $\endgroup$ Commented Oct 12, 2010 at 22:23
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The free product $\mathbb Z_2$ and $\mathbb Z_3$ (i.e. PSL(2, $\mathbb Z$)) is Gromov-hyperbolic (as every virtually free group) and non-virtually cyclic. Therefore by a result of Olshanskii, "SQ-universality of hyperbolic groups". (Russian) Mat. Sb. 186 (1995), no. 8, 119--132; translation in Sb. Math. 186 (1995), no. 8, 1199–1211, it is SQ-universal, that is every countable group embeds into a factor group of PSL(2, $\mathbb Z$). In "most" of these groups (by construction) $ab$ will have infinite order. Thus, in particular, there are uncountably many groups of the type you want.
Update 1: An explicit example would be this. Take $G=PSL(2,\mathbb Z)$, and any word $w(a,b)$ satisfying very small cancelation (that it no subword of length, say, $\frac{1}{10000}|w|$ occurs twice in $w$ (considered as a cyclic word). Then consider the group $G/\langle\langle w\rangle\rangle$. It is what you want. Geometrically, you just kill the large loop in the standard $K(\pi,1)$ for $PSL(2,\mathbb{Z})$ of course.
Another example, as far as I remember, is the R. Thompson group $V$ (it is generated by an element $a$ of order 2 and an element $b$ of order 3 such that $ab$ has infinite order (Mason?). It should be written in the Cannon-Floyd-Parry's survey on Thompson groups, but I do not have it with me.
Update 2: I cannot find the reference to the result about $V$. It is not in Cannon-Floyd-Parry. But here is a paper where it is proved that $SL(n,{\mathbb Z})$ is generated by an element of order 2 and an element of order 3, provided $n\ge 13$: Sanchini, Paolo; Tamburini, M. Chiara, Constructive $(2,3)$-generation: a permutational approach. Rend. Sem. Mat. Fis. Milano 64 (1994), 141–158 (1996).
Update 3: The paper cited in Update 2 follows this paper: Tamburini, M. Chiara; Wilson, John S.; Gavioli, Norberto On the $(2,3)$-generation of some classical groups. I. J. Algebra 168 (1994), no. 1, 353–370. The result there is quite general (and nice), the generating matrices are explicitly given. To check that $ab$ has infinite order, one just needs to find the characteristic polynomial of $ab$ and show that some roots are not roots of unity. That should be straightforward (using any CAS).
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$\begingroup$ That's very interesting Mark (and thank you), but an explicit example would be very nice. Explicit geometric examples would be most welcome! $\endgroup$ Commented Oct 12, 2010 at 18:29
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$\begingroup$ Does one need the full force of Olshanskii's result? $\endgroup$ Commented Oct 12, 2010 at 18:29
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$\begingroup$ @Andreas: No, one can use the fact that PSL(2,Z) is "large" (that is it has finite index free subgroup). Then by an old result of Pride (?) it is SQ-universal. $\endgroup$– user6976Commented Oct 12, 2010 at 18:32
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$\begingroup$ These look like serious contenders in your update, Mark -- thanks very much! I didn't know about Thompson's group $V$, and I would have to do some reading up here to convince myself. But I'm on the verge of accepting this as an answer. $\endgroup$ Commented Oct 12, 2010 at 19:35
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1$\begingroup$ Sorry to have to eat my words, Mark. I appreciate the time you put into this, and I wish I could accept your answer as well. But Derek Holt's solution was a bit easier for me to follow. $\endgroup$ Commented Oct 12, 2010 at 22:26
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I don't know an explicit example off hand, but I would recommend looking at the generalized triangle groups
$\langle a,b \ | \ a^2 = b^3 = 1 = w^k \rangle$
where $w$ is a word in $a$ and $b$. Baumslag, Morgan, and Shalen given conditions on when this virtually surjects $\mathbb{Z}$ or a free group of rank two. I would suspect that it wouldn't be too tough to find an explicit example where $ab$ has infinite order.
See
Baumslag, Morgan, Shalen, "Generalized triangle groups" Math. Proc. Camb. Phil. Soc. (1987) 102, page 25
and
Fine, Rosenberger, "A note on generalized triangle groups" ABHANDLUNGEN AUS DEM MATHEMATISCHEN SEMINAR DER UNIVERSITÄT HAMBURG Volume 56, Number 1, 233-244
answered Oct 12, 2010 at 18:23
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$\begingroup$ Thanks, Richard; that seems very reasonable. I'll see whether I can get hold of that reference. $\endgroup$ Commented Oct 12, 2010 at 18:31
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I am hazarding a guess, I believe this should do the job
A = { {1, x, 0}, {0, -1, 0}, {0, y, 1} }
It really does not matter what x and y are, they can be chosen arbitrarily and can even be two formal symbols
B = { {0, 0, -$i$}, {$i$, 0, 0}, {0, 1, 0} }
Then A.A= B.B.B = Id
the order of $A.B$ would be infinite when $x$ and $y$ are suitably chosen, for example one can choose $x$ and $y$ so that the coefficient of the matrices $(A.B)^n$ unbounded ?
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4$\begingroup$ These two matrices may generate a group isomorphic to $PSL(2,\mathbb Z)$ (at least for some $x,y,\iota$? $\endgroup$– user6976Commented Oct 12, 2010 at 19:16
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$\begingroup$ I am still wondering how to prove that the order of A.B is infinite in a neat way. There must be a way to describe the action in a simple geometric way and conclude . $\endgroup$– VagabondCommented Oct 12, 2010 at 19:26
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$\begingroup$ Well, I surely believe that you get the desired relations if $x$ and $y$ are algebraically independent, but Mark's objection would still need to be addressed. $\endgroup$ Commented Oct 12, 2010 at 19:37
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$\begingroup$ True. Now, How does one find a relation ? I think I have ended up asking the same question as you originally asked !! $\endgroup$– VagabondCommented Oct 12, 2010 at 20:00
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2$\begingroup$ @Vagabond: In fact they probably always generate $PSL(2,\mathbb Z)$ if $x,y,\iota$ are sufficiently independent. But see updates 2 and 3 in my answer: there are two matrices of sizes 13 with integer coeff. satisfying $a^2=b^3=1$ that generate the whole $SL(13,\mathbb Z)$. $\endgroup$– user6976Commented Oct 13, 2010 at 7:58