Non-abelian group of order $p^3$ without semidirect products

Question

I am trying to read a proof that there are at most two non-abelian groups of order $p^3$ if $p$ is an odd prime. The proof presents it as two cases: the first, where every non-identity element has order $p$, is fairly straight forward and results in the group generated by $a, b, c$, each of order $p$, with $ac=ca, bc=cb,$ and $aba^{-1}b^{-1} = c$.

The part that I am confused about is the second case, where there is an element of order $p^2$, called $a$. The author asserts that $G$ is then generated by $a$ and some $b$, where $b$ has order $p$. Why is this necessarily true?

Answer 1

Choose $b\notin <a>$. Can $ab \in <a>$?

Answer 2

I found the answer in another post The initial problem allows use of semidirect product terminology, but the linked proof does not require it.