TITLE: Do actions of BS(1,n) on finite sets factor through abelian quotients?
QUESTION [8 upvotes]: Suppose $BS(1,n)$ is the Baumslag-Solitar group and $S_m$ is the
symmetric group.  If $\Phi: BS(1,n) \to S_m$ is a homomorphism, must the
image of $\Phi$ be abelian?

REPLY [7 votes]: Another way to see the answer is no: the dihedral group $D_{2k}$ (sometimes written $D_k$) of order $2k$ is generated by $a$ ('rotation through $2\pi/k$') and $t$ (a reflection) and satisfies the relation $tat^{-1}=a^{-1}=a^{k-1}$. This is clearly a finite non-abelian group provided $k>2$, and taking $k=n+1$, we see that $D_{2k}$ is an image of $\mathrm{BS}(1,n)$ under the obvious map. 
Finally, since $D_{2k}$ acts on the vertices of a regular $k$-gon, $D_{2k}$ embeds in $S_m$ for any $m\geq k$. This gives a negative answer to the question provided $n\geq 2$ and $m\geq 3$.