I recently came up with the following statement: Let $P$ be a finite $p$-group, then $Z(P) \cap \Phi(P) = 1$ if and only if $\Phi(P) = 1$.
For the proof I used induction, the idea being the following: Suppose $Z(P) \cap \Phi(P) = 1$. Let $N$ be a minimal normal subgroup of $P$ contained in $Z(P)$. As $N \cap \Phi(P) = 1$ there exists some maximal subgroup $M$ of $P$ such that $P = NM$ and $N \cap M = 1$. As $M$ is a maximal subgroup, it is normal and of index $p$. Hence $N$ is of order $p$ and $P \cong N \times M$. Now use the fact that $Z(P) = Z(N) \times Z(M)$ and $\Phi(P) = \Phi(N) \times \Phi(M)$ to conclude that $Z(M) \cap \Phi(M) = 1$. After finitely many steps we see that $P$ is elementary abelian.
Can you think of a different proof? Maybe one that does not rely on induction?
The Frattini subgroup of $G$ is a normal subgroup, so we may use the following fact: for any $p$-group $G$ and a nontrivial normal subgroup $N\subset G$, we have $N\cap Z(G)\neq \{1\}$.
To see this, $N$ admits a $G$ action, namely conjugation. Thus $$ |N|=p^k=\sum_{O_k}|O_k|, $$ where $O_k$ are the distinct orbits of the action. Now by the orbit stabilizer theorem, each $|O_k|$ is a power of $p$. Reducing mod $p$ and using the fact that the identity element has a trivial orbit gives the fact that there must be other central elements in $N$.