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let $G$ be a finite abelian group such that $\left | G \right | = p^n$ where $\left | G \right |$ is the cardinal of $G$, $n \in \mathbb{N}$ and $p$ is a prime number.

we note $\phi (G)$ the intersection of all maximal subgoups of $G$ (which is called Frattini subgroup). I want to prove that $\phi (G) = p.G$ where $p.G = \left \{ p.g \mid g \in G \right \}$. I have proved that if $H$ is a maximal subgroup of $G$ then $p.G \subset H$ :

if $g$ is not in $H$ then $H + \left \langle g \right \rangle = G $ but $H$ is maximal, so $G/H$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$ but $H + \left \langle g \right \rangle / H$ is also isomorphic to $\left \langle g \right \rangle/H\cap\left \langle g \right \rangle$ then $p.g \in H$. so $p.G \subset \phi (G)$

But I can't prove that $\phi (G) \subset p.G$ please if anybody can help.

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You want to prove that if $g$ is not in $p G$, then there exists a maximal subgroup of $G$ that does not contain $g$. The key is to think about what the quotient $\bar{G}=G/pG$ looks like. It is an elementary abelian $p$-group, i.e. you can think of it as an $\mathbb{F}_p$-vector space. If $g$ is not contained in $pG$, then it projects to a non-trivial element $\bar{g}$ in that vector space. Now, take any hyperplane (i.e. a codimension $1$ subspace) $\bar{H}$ in that vector space that does not contain $\bar{g}$, and take its preimage $H$ in $G$. Clearly, $H$ does not contain $g$, since this statement is already true for the respective images in a quotient of $G$. Moreover, it is clear that $\bar{H}$ is a maximal subgroup of $\bar{G}$, hence $H$ is a maximal subgroup of $G$.

More generally, if $G$ is a $p$-group, then the Frattini subgroup of $G$ is $\Phi(G)=[G,G]\cdot pG$, where $[G,G]$ is the derived subgroup. $\Phi(G)$ is the smallest subgroup of $G$ that is normal with elementary abelian quotient.

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