Commutator width of soluble Baumslag Solitar groups 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).
 A: 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. 
A: 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 
$$b^\epsilon a^{km}b^{-\epsilon}b^\epsilon a^{-k}b^{-\epsilon} = b^{\epsilon - 1}a^k b^{1-\epsilon}b^\epsilon a^{-k}b^{-\epsilon} = b^{\epsilon - 1}a^k b a^{-k}b^{-\epsilon} = [b^{-1},b^\epsilon a^k].$$
A: 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.
