prove :a group is divisible if and only if it has no maximal subgroups.
Let $G$ be a group. We denote by $\Phi(G)$ the intersection of all maximal subgroups of the group $G .$ Clearly $\Phi(G)$ is a subgroup of $G$ (called the Frattini subgroup) and $\Phi(G)=G$ if and only if $G$ has no maximal subgroups. We shall use the following lemma: For a group $G$ we have $\Phi(G)=\bigcap_{p \in \mathbf{P}} p G$ It is clear then that $G$ is divisible if and only if $p G=G$ for all $p \in \mathbf{P},$ or equivalently, if $\Phi(G)=\bigcap_{p \in \mathbf{P}} p G=G,$ which implies $G$ has no $p \in \mathbf{f}$ maximal subgroups.($\mathbf{P}$ is the set of all prime numbers)
and we know If a group has no maximal subgroups, then all elements are non-generators.
I want proof conversely. If a group $G$ has no maximal subgroups then $G$ is divisible
