Here is my problem:
Let $H=\langle a,b| a=bab, b=aba\rangle $ and $\frac{H}{A}\cong\ Q_8$ wherein $A\leq Z(H)\cap H'$. Show that $H\cong Q_8$.
Working on the elements, I could see that $H\cong Q_8=\langle a,b| a^4=1, a^2=b^2,ab=ab^3\rangle$. In fact, the $G$'s relations can be yielded by $H$'s.
Yet; I don't see the additional points, above problem wanted to give me? Thanks.