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Do the soluble Baumslag-Solitar groups have finite commutator width? A soluble Baumslag-Solitar group is given by a presentation of the from

$$\mathrm{BS}(1,m) = \langle a,b \mid \mbox{ } a^{-1}ba = b^m\rangle.$$

We also know that these groups are boundedly generated, in particular all elements can be written as $g=a^z b^t$, with $z,t$ integers.

If $G$ is a non-soluble BS group then it maps onto a free product and these have by a result of Rhemtulla infinite commutator width (edit (YCor): it has infinite commutator width, see Misha's answer).

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    $\begingroup$ Two remarks (I assume $m\ge 2$): 1) you cannot write $aba^{-1}$ (which, in intuition, is $b^{1/m}$) as $a^zb^t$ with $z,t$ integers. So I'm not sure $BS(1,m)$ is boundedly generated. $\endgroup$ – YCor Jan 7 '14 at 19:39
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    $\begingroup$ 2) $BS(m,m+1)$ never maps onto a nontrivial free product. Indeed being 2-generated, since the rank (minimal number of generators) is additive under free products, it would mean a free product of 2 nontrivial cyclic groups, say of order $u,v$. But the abelianization of $BS(m,m+1)$ being cyclic, hence $u,v$ are coprime. Hence the free product is a residually finite, large group. But the largest residually finite quotient of $BS(m,m+1)$ is solvable, whence a contradiction. Still, it is very plausible that an argument based on the HNN decomposition works to discard bounded generation. $\endgroup$ – YCor Jan 7 '14 at 19:41
  • $\begingroup$ It has to be a bit more subtle to be boundedly generated. More details are for example here: arxiv.org/pdf/1307.4861v1.pdf Thanks for pointing out that BS(m,m+1) does not map onto the free product. By the paper mentioned below it has infinite commutator width. $\endgroup$ – Elisabeth Fink Jan 7 '14 at 21:09
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    $\begingroup$ The solvable groups $BS(1,n)$ are linear. The commutator subgroup consists of the nilpotent matrices. Computing a generic commutator, I think it ought to be easy to show that every nipotent matrix in the group is a commutator, but I haven't worked it out. For the representatio, see: en.wikipedia.org/wiki/Baumslag-solitar_group $\endgroup$ – Ian Agol Jan 8 '14 at 22:11
  • $\begingroup$ The bounded generation assertion is not true (except in $BS(1,\pm 1)$). This can be checked by showing that the "orbit" of the product of two cyclic subgroups on the Bass-Serre tree is not cobounded. This does not affect the question anyway. $\endgroup$ – YCor Dec 2 '19 at 7:33
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Any solvable group BS(1,m) is metabelian. A.H. Rhemtulla proved in (Commutators of certain finitely generated soluble groups. Canad. J. Math., 21 (1969), 1160-1164) that every finitely generated soluble of class $\leq 3$ group has a finite commutator width. Moreover, P.W. Stroud (Topics in the theory of verbal subgroups.PhD Thesis, Univ. of Cambridge, 1966) proved that every verbal subgroup $w(G)$ of a finitely generated abelian-by-nilpotent group $G$ has a finite width. See for instance: D. Segal. Words: notes on verbal width in groups. London Math.Soc. Lect.Notes Ser. 361, Cambridge Univer. Press., 2009. I can add that the commutator width of the free metabelian group of rank $r \geq 2$ has the commutator width equal to $r.$ Hence every f.g. metabelian group has the commutator width $\leq r.$

Let $M$ be the free metabelian group with base $x, y$. Then the derived subgroup $M'$ is generated as a module over $Z[M/M']$ by $u = [x,y].$ Every element of $Z[M/M']$ can be written as $k + \alpha (1-x) + \beta (1-y).$ Then every element of $M'$ can be written as $u^{k}[u^{\alpha },x][u^{\beta}, y]= [x,y^k] [u^{\gamma}, x][u^{\delta}, y].$ It gives bound 3. After some improving we get 2.

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Scl for Baumslag-Solitar groups was analyzed, to some extent, in this paper by Clay, Forester and Louwsma. In particular, all nonsolvable B-S groups contain elements with nonzero scl, see formula (1) on page 2.

However, scl formula (1) on page 2 of the paper does not apply in the solvable case (say, $m=1$) as it gives a negative value; solvable case was explicitly excluded by the authors. I suggest, you read the paper in detail to find out what is going on. You can also use the results from the second part of the paper (on well-aligned elements) to determine if scl is nonzero in the solvable case.

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  • $\begingroup$ Thanks for reminding me of the paper. The problem is just that scl=0 does not imply finite commutator width. $\endgroup$ – Elisabeth Fink Jan 7 '14 at 21:17
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    $\begingroup$ @ElisabethFink: I was using the fact that nonzero scl implies infinite commutator width. $\endgroup$ – Misha Jan 7 '14 at 21:21
  • $\begingroup$ Yes, that's certainly helpful. I couldn't find a criterium in the paper which forced any of the soluble BS groups to have positive scl. $\endgroup$ – Elisabeth Fink Jan 8 '14 at 20:42
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Although this question is a few years old, and finite commutator width has been established by Vitaly, it seems worth adding that $BS(1,m)$ actually has commutator width 1. Indeed, any element of $BS(1,m)$ may be written as $b^\epsilon a^y b^{-\epsilon}b^z$, with $\epsilon, y,z$ integers (and $\epsilon$ nonnegative). Since the relator $b^{-1}aba^{-m}$ has $b$-exponent 0 and $a$-exponent $1-m$, any word on $\{a,b\}$ which represents an element of the derived subgroup must have $b$-exponent 0 and $a$-exponent a multiple of $m-1$. Therefore, each element of the derived subgroup may be written as $b^\epsilon a^{k(m-1)} b^{-\epsilon} $, which is equal to $$b^\epsilon a^{km}b^{-\epsilon}b^\epsilon a^{-k}b^{-\epsilon} = b^{\epsilon - 1}a^k b^{1-\epsilon}b^\epsilon a^{-k}b^{-\epsilon} = b^{\epsilon - 1}a^k b a^{-k}b^{-\epsilon} = [b^{-1},b^\epsilon a^k].$$

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  • $\begingroup$ More generally every abelian-by-cyclic group has commutator width $\le 1$ (namely $=1$ if it's not abelian and $=0$ if abelian). Indeed, write it as $G=A\rtimes_T\mathbf{Z}$ where the positive generator acts as the automorphism $T$ of $A$. Then the derived subgroup of $G$ equals the image of $T-1\in\mathrm{End}(A)$, which consists of commutators in $G$. $\endgroup$ – YCor Dec 2 '19 at 5:07
  • $\begingroup$ This is a nice generalization. I can see that $G'=(T-1)(A)$ by commutator identities and an induction argument, but it feels like I'm missing something; is there a more intuitive way to see this? $\endgroup$ – Richard Mandel Dec 3 '19 at 2:57
  • $\begingroup$ Clearly $(T-1)A$ consists of commutators, hence is contained in $G'$. Also it's a normal subgroup of $G$ (as it's $T^\pm$-invariant) and the action of $T$ on $A/(T-1)A$ is clearly trivial, so the quotient is abelian, so $G'\subset (T-1)A$. $\endgroup$ – YCor Dec 3 '19 at 7:52
  • $\begingroup$ Thanks YCor! Can’t believe I missed that. $\endgroup$ – Richard Mandel Dec 3 '19 at 13:19

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