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?



Do actions of BS(1,n) on finite sets factor through abelian quotients?
No, these groups have many finite non-abelian quotients. Recall that the Baumslag-Solitar group $B(1,n)$ has a presentation
$$
B(1,n)=\langle t,a\,|\,tat^{-1}=a^n\rangle.
$$
A homomorphism $\Phi: B(1,n)\to S\_m$ amounts to a pair $(\tau,\alpha)$ of 
permutations, where $\Phi(a)=\alpha$ and $\Phi(t)=\tau$, such that $\tau$ conjugates $\alpha$ to $\alpha^n$.Observe that in the symmetric group $S\_m$, a cycle $\alpha$ of length $k$ is conjugate to its $n$th power $\alpha^n$ provided $n$ and $k$ are relatively prime, and $\alpha^n\ne \alpha$ for $k$ not dividing $n-1$, so that the conjugating permutation $\tau$ does not commute with $\alpha$. This yields examples of homomorphisms with non-abelian image.


For a specific example, take $n=2, k=3$ and let $\Phi: B(1,2)\to S\_3$ map the 
generators $t$ and $a$ of  

$$
B(1,2)=\langle t,a\,|\,tat^{-1}=a^2\rangle
$$
into permutations $\tau=\Phi(t)=(2,3)$ and $\alpha=\Phi(a)=(1,2,3)$ in the cycle notation. Since $\Phi(t)=(2,3)$ conjugates $\Phi(a)=(1,2,3)$ into its square $\Phi(a)^2=(1,3,2)$ and moreover, $(2,3)$ and $(1,2,3)$ generate $S\_3$, this defines a group homomorphism whose image is $S\_3$. 


