Show that <a, b|a^5=b^2=e, ba=(a^2)b> is isomorphic to Z2.

Let the game begin...(Happy) :

\(\displaystyle bab^{-1}=a^2\Longrightarrow ba^2b^{-1}=\left(bab^{-1}\right)^2=a^4\) , but also \(\displaystyle ba^2b^{-1}=b\left(bab^{-1}\right)b^{-1}=b^2ab^{-2}=a\) , so we get \(\displaystyle a=a^4\Longrightarrow a^3=1\) , which together with the given data \(\displaystyle a^5=1\)

means that \(\displaystyle a=1\) ,and thus in fact \(\displaystyle \left<a,b\;/\;a^5=b^2=1\,,\,ba=a^2b\right>=\left<b\;/\;b^2=1\right>\cong\mathbb{Z}_2\)

Tonio