$BS(2,3)=\langle a,b \mid ba^2b^{-1}=a^3 \rangle$. Let $H$ be a finite group and $\alpha:BS(2,3)\to H$ a homomorphism. Let $g=[bab^{-1},a]$.

Prove $\alpha(g)=1$

My Attempt

$$\begin{align} \alpha(g)&=\alpha([bab^{-1},a])\\ &=\alpha((bab^{-1})a(bab^{-1})^{-1}a^{-1})\\ &=\alpha(bab^{-1}aba^{-1}b^{-1}a^{-1})\\ &=\alpha(b)\alpha(a)\alpha(b)^{-1}\alpha(a)\alpha(b)\alpha(a)^{-1}\alpha(b)^{-1}\alpha(a)^{-1} \end{align}$$

Let $H=\{1,h_1,h_2,...,h_k\}$. Since a homomorphism is defined by the generators, define, $$\alpha(a)=k_i\equiv \phi$$ $$\alpha(b)=k_j\equiv\psi$$ If either $\phi=1$ or $\psi=1$ then the answer is trivial. Similarly if $\phi=\psi$ then the answer is also trivial. Thus we can assume $\phi \neq \psi$.

From here I don't know how to proceed, I attempted to use the relation $ba^2b^{-1}=a^3$, but I seem to be running in circles. I believe I need to use the finiteness of $H$.

  • $\begingroup$ The First Isomorphism Theorem might be of use. $\endgroup$
    – Shaun
    Apr 14 at 16:12

1 Answer 1


$ba^2b^{-1}=a^3$. Let $n$ be the order of $\alpha(a)$, $n\ne0$. If $n=2k$, then conjugate $\alpha(a)^{2k}=1$ by $\alpha(b)$: $\alpha(a)^{3k}=1$, so $2k$ divides $3k$, impossible. So $n=2k+1$. Conjugating $\alpha(a)^{2k+1}=1$ by $\alpha(b)$, $\alpha(a)^{3k}\cdot \alpha(a^b)=1$, so $\alpha(a^b)$ commutes with $\alpha(a)$, and you are done.

  • 2
    $\begingroup$ Because it took me a half-second: $\alpha(a)^{3k}\cdot\alpha(a^b)=1$ implies that $\alpha(a^b)=\alpha(a)^{-3k}$ — in other words, that $\alpha(a^b)$ is a power of $\alpha(a)$ — and that's why they commute. $\endgroup$ Apr 14 at 16:36
  • $\begingroup$ What does $\alpha(a^b)$ mean? b is not a number $\endgroup$ Apr 14 at 16:37
  • $\begingroup$ @MinecraftPlayer69 $a^b$ is common group-theoretic notation for the conjugate $bab^{-1}$. $\endgroup$ Apr 14 at 16:38
  • 1
    $\begingroup$ $a^b=bab^{-1}$. $\endgroup$
    – markvs
    Apr 14 at 16:38

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