Let $G$ be a non-abelian finite $p$-group. I wonder if $Z(G) \leq \Phi(G)$, where $\Phi(G)$ is the Frattini subgroup of $G$? I'd known that it is true for non-abelian groups of order $p^3$ but I doubt it is true in general or not.
I understand that it isn't true in general. As showed below it isn't true for some group of order $p^4$.
Let $H$ be the nonabelian group of order $p^3$ and exponent $p$. Let $C_p$ be a cyclic group of order $p$. Let $G=H\times C_p$. Then $Z(G)$ is a direct product of two copies of $C_p$, one contained in $H$, the other the second direct factor of $G$.
However, $H$ is maximal in $G$, and so $\Phi(G)\subseteq H$. Thus, $Z(G)$ is not contained in $\Phi(G)$.
More generally, given any nonabelian $p$-group $K$, letting $G=K\times C_p$ gives you a group with center $Z(G) = Z(K)\times C_p$, and with $K$ maximal and hence $\Phi(G)\subseteq K$, hence a counterexample to the assertion to $Z(G)\subseteq \Phi(G)$.