No group of following property. Is this true?

Question

Let $p$ be a prime greater than 3 and $G$ be group of order $p^5$. Is it true that there is no group $G$ of order $p^5$ such that the order of frattini subgroup is $p^3$ and the order of center is $p^2$?

If the answer is yes, how to prove it.

Answer 1

The class $3$ quotient of the Burnside group $B(2,p)$ has these properties. It has the presentation

$\langle a,b,c,d,e \mid a^p=b^p=c^p=d^p=e^p=1, [b,a]=c, [c,a]=d, [c,b]=e, d,e\ {\rm central}\ \rangle$

The Frattini subgroup is $\langle c,d,e \rangle$ and the centre is $\langle d,e\rangle$.

Answer 2

The Frattini subgroup is always characteristic, so if it's Abelian, then it must be central, which is a contradiction of the orders. So the Frattini subgroup is not Abelian: Apply Lemma 4 from http://msp.org/pjm/1960/10-1/pjm-v10-n1-p12-s.pdf