Problem: Let $_{\mathbb{Z}}M$ be an abelian group. Prove that: If $_{\mathbb{Z}}M$ is divisible, then $\operatorname{Rad} M$$=M$.
My ideas: we know that \begin{align} \operatorname{Rad}M&=\bigcap\{K\leq M\mid K \text{ is maximal in }M \} \\[4px] &=\sum\{L\leq M\mid L\text{ is superfluous in }M \}. \end{align} Since the maximal submodule must be proper submodules, why $\operatorname{Rad} M=M$?
Suppose $N$ is a maximal submodule of $M$.
Then $N$ is a proper submodule of $M$, hence $M \setminus N \ne \large{Ø}$.
Let $x \in M \setminus N.$
Since $N$ is maximal, $M/N$ is cyclic of order $p$, for some prime $p.$
It follows that $pM \subseteq N.$
Since $M$ is divisible, there exists $y \in M$ such that $py = x.$
But then $py \in pM \implies py \in N \implies x \in N,\,$ contradiction.
Therefore $M$ has no maximal submodules.
Thus, $\operatorname{Rad}M$ is the empty intersection (of submodules of $M$).
It follows that $\operatorname{Rad}M = M,\,$ as was to be shown.
