On the Frattini subgroup of a finite group Let $G=P \ltimes Q$ be a finite non-nilpotent group, where $P \in {\rm Syl}_p(G)$ and $Q \in {\rm Syl}_q(G)$ such that $P(=N_G(P))$ and $Q$ are cyclic such that $|Q|=q^2$. Then can we prove that $\Phi(P)=Z(G)$.($|P|=p^n$ with $n \geq 2$).
${\bf My Try}$. Since all Sylow subgroup of $G$ are cyclic, $G$ is supersolvable. Clearly, $Q=G^{\prime}$ and so $G^{\prime} \cap Z(G)=1$.
Thus $Z(G)<P$. So $Z(G)=1$ or $Z(G) \leq \Phi(P)$.
 A: I would like to first note that you can actually prove from the other assumptions that $p|q-1$ and $P\Phi(Q)$ is non-abelian.
A slightly more clear argument that $Z(G)\le \Phi(P)$ is that $Z(G)\le N_G(P)=P$, $P\ne Z(G)$ since $N_G(P)\ne G$ and $\Phi(P)$ is the unique maximal subgroup of $P$ so we must have $Z(G)\le \Phi(P)$.
Massive Correction (Also found by C Monsour)
Remarkably my post has been upvoted twice and marked correct despite being completely wrong. I made a mistake in the calculations.
In fact, you can't show that $\Phi(P)=Z(G)$, because it is not always the case.
Consider for example $C_4\ltimes C_{25}=\langle x,y|x^4=y^{25}=1,y^x=y^7\rangle$. This is a correct definition because $y^7$ generates $C_{25}$ and since $y^{x^2}=y^{49}=y^{-1}$, we have $y^{x^4}=y$ and one can check that $x^y=y^{17}x$, so $C_{25}\cap N_G(C_4)=1$ so $N_G(C_4)=C_4$. However $x^2\notin Z(G)$ so $Z(G)=1$.
A: I have a correct counterexample now.  Consider $Q$ cyclic of order 25 and $P$ cyclic of order 4, acting faithfully on $Q$ (which it can do because $Aut(Q)$ is cyclic of order 20).  To fix ideas, let the generator of $P$ act by the automorphism of raising to the 7th power. Then $\Phi(P)$, of order 2, acts non-trivially by conjugation on $Q$ (the non-identity element inverts the elements of $Q$), so it can't be in the center of $G$.  It remains to show that $N(P)=P$ in this case.  If not, then since $P$ is not normal, clearly $|N(P)=20|$ and $|Q\cap N(P)|=5$.  Also, since $P$ has no automorphism of order 5, N(P) is abelian.  But then $P$ commutes with the 5-element subgroup of $Q$, a contradiction, since raising to the 7th power does not fix any non-identity elements of $Q$.
