Finite p-group with a cyclic frattini subgroup. I have a question about the following theorem that I found in some research.

Is it possible that $E$ is the identity?
I just found this elaborated proof that might help.


Yes, it is possible that $E$ is the identity.
Take $P$ non-abelian of order $p^3$. In this case $U = \Phi(P) \cap \Omega_1(Z(P)) = \Omega_1(Z(P)) = Z$ and so $E=1$. $Q=P$ has $\Phi(Q) = Z(Q)$ cyclic of order $p$. Furthermore such a $P$ is not a direct product of any proper non-identity subgroups, and so $E = 1$ is not only possible, it is necessary.
This is just @anon's comment with an explicit answer.
