It is known that if there is a non-abelian group of order $pq$, then it must be the case that $q\mid p-1$ and this group is isomorphic to $\langle a,b:a^p=b^q=1,ab=ba^u\rangle $ wherein $u$ is of order $q$ mod $p$ in $\mathbb{Z}_{p-1}^*$.
The book I am reading states that
this group is unique upto isomorphism, that it is independent of the choice of $u$, and
every element is distinctly written as $a^xb^y$ where $0\leq a< p$ and $0\leq b<q$.
But if $p=3$, $q=7$ and $u=2$, $ba^2b$ can be expressed as $baab=baba^2=b^2a^4$ but cannot be expressed in the form $a^ib^j$. I just want to confirm that $2$ is the false claim. Also essentially for $1$, I need to show that it doesn't matter if $u=2$ or $u=4$. I would appreciate if I can get the idea. Thank you!!