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.

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