Definition of special $p$-groups

Question

I would like to know if there is a standard definition for finite special $p$-groups. I would have two questions related to this:

1. In some sources (for instance, Gorenstein's Finite groups), it is said that a $p$-group $P$ is special if $P$ is elementary abelian or $P$ is of class 2 and $P'=Z(P)=\Phi(P)$ is elementary abelian. In the latter case, it won't matter if we specifically say that $P$ is of class 2. However, in Berkovich & Janko's Groups of Prime Power Order (vol. 3), a special $p$-group would be a non-abelian $p$-group $P$ such that $P'=Z(P)=\Phi(P)$ is elementary abelian. So, my question would be if there is a specific reason for which the abelian case is dropped in the latter reference?

2. For the non-abelian case, would it be sufficient to say that a $p$-group $P$ is special if the condition $P'=Z(P)=\Phi(P)$ holds (without specifying that the 3 subgroups are elementary abelian)? If so, how can we prove that $P'=Z(P)=\Phi(P)$ implies that $P'$ (or $Z(G)$ or $\Phi(P)$) is elementary abelian? We know that $P/\Phi(P)=P/Z(P)=P/P'$ is elementary abelian by the properties of the Frattini subgroup.

Thank you for considering these questions!

Answer 1

For 2: if $P'=Z(P)=\Phi(P)$, then they are all elementary abelian.

Indeed, because $P'=Z(P)$, $P$ will be of class two (we are assuming non-abelian), and so the commutator bracket is bilinear.

Since $\Phi(P)$ contains all $p$th powers and equals the center, it follows that all $p$th powers are central.

Now, $P'$ is generated by elements of the form $[x,y]$, $x,y\in P$. Then we have $$[x,y]^p = [x^p,y] = e$$ since $x^p\in \Phi(P)=Z(P)$. Thus, $P'$ is abelian and generated by elements of exponent $p$, so it is elementary abelian, which is what we wanted to show.