# For any finite group $H$ and homomorphism $\alpha:BS(2,3)\to H$, prove $\alpha([bab^{-1},a])=1$

$$BS(2,3)=\langle a,b \mid ba^2b^{-1}=a^3 \rangle$$. 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 $$H$$.

• The First Isomorphism Theorem might be of use. Apr 14 at 16:12

$$ba^2b^{-1}=a^3$$. 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.
• 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. Apr 14 at 16:36
• What does $\alpha(a^b)$ mean? b is not a number Apr 14 at 16:37
• @MinecraftPlayer69 $a^b$ is common group-theoretic notation for the conjugate $bab^{-1}$. Apr 14 at 16:38
• $a^b=bab^{-1}$. Apr 14 at 16:38