I was recently trying taking a look at the paper "Some two-generator one-relator non-hopfian groups" by Baumslag and Solitar where they introduce the groups now known as the Baumslag-Solitar groups given by the presentation
$$
G = \left< a,b \mid a^{-1}b^la = b^m \right>
$$
The paper is only 3-pages long and a bit skimpy on details (for me as someone without much group-theory know-how). I have a few questions about some of the claims in the paper.


In considering the case where there is a prime $p$ that divides $l$ but not $m$, the authors claim that the homomorphism given by $a \mapsto a, b \mapsto b^p$ has a nontrivial kernel, and namely, that the element
$$
[b^{l/p}, a]^p b^{l-m}
$$
which is in the kernel of the above map is actually nontrivial in $G$ (here the commutator convention is $[g\_1,g\_2] = g\_1^{-1}g\_2^{-1}g\_1g\_2$). With a little fussing, we can see that this map is surjective, and thus, if we see that that element is nontrivial, we have a non-hopfian group (which was a punchline/motivation in the above paper). My first question is, why is this element nontrivial?


My second question is with regards to a claim made on the first page. Taking $l = 2, m = 3$ we can see that $a$ and $b^4$ generate $G$. The claim is that for the surjection
\begin{align\*}
F\langle x,y\rangle &\to G \\
x &\mapsto a \\
y &\mapsto b^4
\end{align\*}
the kernel is finitely normally generated, but not by a single element. Why is this kernel finitely normally generated but not by a single element?


I bit of playing around yields some elements of the kernel, and I imagine that there is some folding-style way of seeing if a given element of a free group is in the normal closure of some finite list of other elements, but I don't know how to do that. If I did, I could probably show that the kernel is not normally generated by a single element.



Some questions on a paper of Baumslag and Solitar
I'll write out the answer to your first question (solving the word problem in one-relator groups) for how one might do this in practice. I'll focus on the case $\ell = 2, m=3, p=2$, but you'll hopefully be satisfied that the methods are sufficiently general!


Let $G = \langle a, b \mid a^{-1} b^2 a b^{-3} =1 \rangle$. Then we wish to check that the word
$$
W = [b, a]^2 b^{-1} = b^{-1} a^{-1} b a b^{-1} a^{-1} b a b^{-1}
$$
is non-trivial in $G$. To do this, we first notice that the letter $a$ has exponent sum zero in the defining relation $a^{-1} b^2 a b^{-3}$, which is usually denotet by $\sigma\_a(a^{-1} b^2 a b^{-3}) = 0$. Now any word $w$ with $\sigma\_a(w) =0$ can be rewritten, by an easy trick, as a product of conjugates $b\_i = a^i b a^{-i}$. For example,
$$
a^{-1} b^2 a b^{-3} = (a^{-1} b a)(a^{-1}b a)(b^{-3}) = b\_{-1}^2 b\_0^{-3}.
$$
The way this factorisation is found in practice is to look at the word $a^{-1} b^2 a b^{-3}$ and replace every occurrence of a letter $b$ by $b\_i$ when the exponent sum of the $a$'s preceding that occurrence equals $i$, and then removing all $a$'s.


Now it is not hard to check that $G$ is an HNN-extension of the (torus knot) group
$$
H = \langle b\_{-1}, b\_0 \mid b^2\_{-1} b\_0^{-3} = 1\rangle,
$$
where the stable letter is $a$, identifying the two (free) subgroups $\langle b\_{-1} \rangle$ and $\langle b\_0 \rangle$. Note that this new group is (always!) a one-relator group with a shorter defining relation than $G$.


Now the key insight is that a word equals $1$ in $G$ if and only if it can be written as a product of $b\_i$'s and this product equals $1$ in $H$. So to check if $W = 1$ in $G$, we note (again) that a word $v$ can be written as a product of $b\_i$'s if and only if $\sigma\_a(v) = 0$. As indeed $\sigma\_a(W) = 0$, we can rewrite $W$, using the same trick as above if necessary, as
$$
W = b^{-1} a^{-1} b a b^{-1} a^{-1} b a b^{-1} = (b^{-1})(a^{-1} b a)(b^{-1})(a^{-1}ba)(b^{-1}) = b\_0^{-1} b\_{-1} b\_0^{-1} b\_{-1} b\_0^{-1}.
$$


Now all we have to do is to solve the word problem in $H$ -- but this is easy (see e.g. [this answer](https://math.stackexchange.com/questions/828828/amalgamated-product) for the basics), as $H$ is just an amalgamated free product of two copies of $\mathbb{Z}$. In particular, we check quickly that the word $b\_0^{-1} b\_{-1} b\_0^{-1} b\_{-1} b\_0^{-1}$ is reduced, and so is non-trivial in $H$. Hence it is non-trivial in $G$, and we are done.




---


This idea is very reminiscent of Magnus' 1932 solution to the word problem for all one-relator groups. This did not use HNN-extensions, of course (predating the HNN paper) by nearly two decades!), but rather considered chains of normal subgroups -- the same idea is used however, i.e. to keep passing to a shorter defining relation and solving the word problem (and some more problems) there, then proceeding by induction. The same reduction, using $b\_i = a^i b a^{-i}$, is also there. An exposition of this is given in Magnus--Karrass-Solitar. Finally, sometimes one must deal with the case that there is no letter with exponent sum zero -- but this is not hard to deal with; one constructs a rewriting and then cranks the induction handle again, see e.g. [1]. For all Baumslag-Solitar groups, however, this is certainly not needed.


[1] *McCool, James; Schupp, Paul E.*, [**On one relator groups and HNN extensions**](http://dx.doi.org/10.1017/S1446788700014300), J. Aust. Math. Soc. 16, 249-256 (1973). [ZBL0288.20046](https://zbmath.org/?q=an:0288.20046).


