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Let $G$ be an abelian group, $S$ a subgroup. Let $H\cap S=0$ and $H$ is maximal among subgroups satisfying this relationship. Prove $G/(H+S)$ is torison.

This is not clear to me. If $S$ is a direct summand of $G$, then the question is trivial; otherwise in the finite generated case we would have $G=\prod \mathbb{Z}/p_{i}^{a_{i}}\mathbb{Z}$, and $S=\prod \mathbb{Z}/p_{i}^{b_{i}}\mathbb{Z},b_{i}\le a_{i}$. And $H$ cannot have any torsion elements. But I am not sure how to prove this in the general abelian group case.

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up vote 1 down vote accepted

If the class $\bar{\sigma}$ of some $\sigma\in G$ in the quotient $G/(H+S)$ has infinite order, then $(H+\mathbb Z \sigma)\cap S=0$.

Indeed, if $s\in (H+\mathbb Z \sigma)\cap S$, then $s=h+k\sigma$ with $h\in H, k\in \mathbb Z$. So $k \bar{\sigma}=0\in G/(H+S)$ and $k=0$. Thus $s=h$ and $s=0$.

As $H$ is maximal, we get $\sigma\in H$, hence $\bar{\sigma}=0$.

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