$\newcommand{\ZZ}{{\mathbb{Z}}} \newcommand{\QQ}{{\mathbb{Q}}} \newcommand{\FF}{{\mathbb{F}}} \newcommand{\PP}{{\mathbb{P}}} \newcommand{\RR}{{\mathbb{R}}} \newcommand{\CC}{{\mathbb{C}}} \newcommand{\ra}{{\rightarrow}} \newcommand{\eps}{{\epsilon}}$
Show that the quotient group $\QQ^+/\ZZ^+$ cannot be decomposed into the direct sum of cyclic groups.
What I had was:
Suppose $\QQ^+/\ZZ^+$ decomposed into the direct sum of cyclic groups $\bigoplus H_i$. Patently $\QQ^+/\ZZ^+$ is not cyclic because if $r>0$ were the generator, then $r/2$, which is rational, would not be included. Thus we know if it were to decompose it must decompose into at least two proper nontrivial subgroups and any two groups must intersect trivially. Let $H_k$ be the cyclic subgroup in the decomposition that is generated by $\frac{a}{b}$ where $a,b \in \ZZ$ and $gcd(a,b)=1$. In fact, if $a\neq1$ then it must be contained in the direct sum $\bigoplus_{i\neq k} H_i$ which contains $\frac1b$ and thus it would contain $\frac{a}{b}$. Thus we know that all subgroups $H_i$ must be generated by an element of the form $\frac1b$. Now say a subgroup $H_k$ is generated by $\frac1b$, then it is contained in the direct sum $\bigoplus_{i\neq k} H_i$ because in $\bigoplus_{i\neq k} H_i$ must be $\frac1{b^2}$ since $\bigoplus H_i = \QQ^+/\ZZ^+$. Thus we can return to the original argument, for an arbitrary cyclic subgroup in the decomposition of $\QQ^+/\ZZ^+$, it is generated by some positive element $r$, and we know there is a smaller element $r/2 \in \QQ^+/\ZZ^+$ that this element will not generate. This smaller element thus is generated by the direct sum of all the other subgroups in the decomposition, and the sum of $r/2+r/2=r$ so that original cyclic subgroup cannot be in the direct sum decomposition. A contradiction! Thus $\QQ^+/\ZZ^+$ cannot be decomposed into the direct sum of cyclic groups.
However, I'm not sure this works. Please help!