3
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ The First Isomorphism Theorem might be of use. $\endgroup$
    – Shaun
    Apr 14 at 16:12

1 Answer 1

4
$\begingroup$

$ba^2b^{-1}=a^3$. 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.

$\endgroup$
4
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.