Let $G$ a abelian group. I recall that $G^p=\{g^p:g\in G\}$ ($p$ is primes) and $\Phi(G)$ is Frattini supgroup of $G$ i.e. intersection of maximal subgroup of $G$.
I want proof that $\Phi(G)=\displaystyle\bigcap\limits_{p\in\mathbb{P}}G^p$.
$(\supseteq)$ Let $M<\hspace{-2mm}\cdot$ $G$ (i.e. maximal subgroup of $G$) then $\lvert G:M\rvert=p$ for some $p\in\mathbb{P}$, then $G^p\le M$. Therefore $\displaystyle\bigcap\limits_{p\in\mathbb{P}}G^p\le\Phi(G)$.
I ask help for $(\subseteq$). Thanks