Show that it is torsion

Question

I am trying to solve the following exercise

Let $G$ be an abelian group, and let $S\subset G$ be a subgroup. If $H$ is maximal with $H\cap S=\{0\}$, prove that $G/(H+S)$ is torsion.

My attempt:

I assume that there is nonzero $x\in G$ with $nx\notin H+S$ for every $n\geq1$. Let $A=\{nx:n\geq1\}$, and $A'=\langle A,H\rangle$. Therefore, $H\subsetneq A'$ and $A'\cap S\neq\{0\}$. So, there is $y\in S$ and $y\in A'$. Therefore, I am looking for some relation between $y$ and $nx$, so that I can reach a contradiction.

Any help?

Answer 1

You are nearly there.

Let $x \in G$. If $x \notin H$, you have that $\langle H, x \rangle > H$, so that $\langle H, x \rangle \cap S \ne \{ 0 \}$. That is, there are an integer $n$, and $h \in H$ and $s \in S$ such that $0 \ne nx + h = s$, that is $n x = -h + s \in H + S$. Clearly $n \ne 0$, as $H \cap S = \{ 0 \}$.