Is there a simple proof that $BS(m,n)$ is residually finite when $n | m$ or $m | n$?

Question

In their orginal paper, SOME TWO-GENERATOR ONE-RELATOR NON-HOPFIAN GROUPS , G. Baumslag and D. Solitar mention that, if $n$ or $m$ divides the other, then $BS(m,n)$ is residually finite, but they do not give a proof of this fact.

Are you aware of a simple proof of this fact?

Answer 1

The claim in the original paper of Baumslag and Solitar is incorrect. The correct result, due to Meskin, Nonresidually finite one-relator groups, Trans. Amer. Math. Soc. 164 (1972), 105-114 (open access), is:

Theorem. $BS(m, n)$ is residually finite if and only if either $|m|=1$ or $|n|=1$ or $|m|=|n|$.

In fact, Meskin proves something slightly stronger, where the relator $b^{-1}a^mba^n$ is replaced with $u^{-1}v^muv^n$, where $u$ and $v$ are non-commuting words. Meskin's paper is only 10 pages, so quite accessible (although the proof in Moldavanskii's survey is shorter).

Answer 2

Yes, there is a simple proof given in the article On the residual properties of Baumslag–Solitar groups by David Moldavanskii.