# Bounded group and direct summand

Let $G$ be a abelian group and $nG=0$ and $H$ is a cyclic subgroup of order $n$ then there exists a subgroup $K$ such that $H\oplus K=G$.

I try this:

Suppose that $\Lambda=\{L\leq G\,| \,L\cap H=0\}$ by Zorn's Lemma there is maximal element $M$ in $\Lambda$.

Now I want to show $M+H=G$.

Since $(M+H)/M\cong H$ so $(M+H)/M$ is a cyclic subgroup of order $n$ but I can not show that $M+H=G$.

Any suggestion?

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