# Show that $\mathbb{Q}^+/\mathbb{Z}^+$ cannot be decomposed into the direct sum of cyclic groups.

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Show that the quotient group $\QQ^+/\ZZ^+$ cannot be decomposed into the direct sum of cyclic groups.

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.

#### Solutions Collecting From Web of "Show that $\mathbb{Q}^+/\mathbb{Z}^+$ cannot be decomposed into the direct sum of cyclic groups."

1. You are conflating elements of $\mathbb{Q}$ with elements of $\mathbb{Q}/\mathbb{Z}$. The latter are congruence classes modulo $\mathbb{Z}$. Better to keep them straight.

2. You assert that if $\mathbb{Q}=\oplus H_k$, and $\frac{a}{b}+\mathbb{Z}$ is a generator for one of the direct summands, then there is a direct summand that contains $\frac{1}{b}+\mathbb{Z}$; you have no warrant for that assertion; why can’t $\frac{1}{b}+\mathbb{Z}$ have more than one nontrivial coordinate in the direct sum? Better: show that $\frac{1}{b}+\mathbb{Z}\in\langle \frac{a}{b}+\mathbb{Z}\rangle$, by using the fact that $\gcd(a,b)=1$. That will prove that you can always select a generator of the form $\frac{1}{b}+\mathbb{Z}$, which seems to be what you are trying to do in the first place.

3. Your argument is murky and way too complex. It is simpler to show that any two nontrivial subgroups of $\mathbb{Q}/\mathbb{Z}$ must intersect, and so any direct sum decomposition $A\oplus B$ of $\mathbb{Q}/\mathbb{Z}$ must have $A$ or $B$ trivial. Then use your argument to show that the group $\mathbb{Q}/\mathbb{Z}$ is not cyclic in order to finish the proof.
Actually, this suggestion works for $\mathbb{Q}$ but no necesarily for $\mathbb{Q}/\mathbb{Z}$; for instance, the Prufer $p$-groups are subgroups of $\mathbb{Q}/\mathbb{Z}$ but they intersect trivially for distinct primes. Sorry about that.

Instead, use a similar argument to what you had before: if $\mathbb{Q}/\mathbb{Z} = \bigoplus H_k$, select $b_k\in\mathbb{Z}$, greater than $1$ without loss of generality (since $b_k=1$ means the subgroup generated by $\frac{1}{b_k}+\mathbb{Z}$ is trivial and can be omitted), such that $\frac{1}{b_k}+\mathbb{Z}$ generates $H_k$. Then select a particular direct summand, say $H_1$; consider the element $\frac{1}{b_1^2}+\mathbb{Z}$ of $\mathbb{Q}/\mathbb{Z}$. Since it is an element of $\mathbb{Q}/\mathbb{Z}=\oplus H_k$, then we can express it as:
$$\frac{1}{b_1^2}+\mathbb{Z} = \sum_k\left(\frac{a_k}{b_k}+\mathbb{Z}\right)$$
for some $a_k\in\mathbb{Z}$, with $b_k|a_k$ for almost all $k$ (since almost all components must be trivial). Then adding it to itself $b_1$ times we get that
$$\frac{1}{b_1}+\mathbb{Z}=b_1\left(\frac{1}{b_1^2}+\mathbb{Z}\right) = b_1\sum_k\left(\frac{a_k}{b_k}+\mathbb{Z}\right) = \sum_k\left(\frac{a_kb_1}b_k+\mathbb{Z}\right).$$
Therefore, equating components, we get that $b_k|a_kb_1$ for all $k\neq 1$, and $b_1|a_1b_1-1$. But the latter implies $b_1|1$, which is a bit of a problem.

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Incorporating what I have learned from Arturo, here is where I am now.

Suppose $\QQ^+/\ZZ^+$ decomposed into the direct sum of cyclic groups $\bigoplus H_i$. First we will show that $\QQ^+/\ZZ^+$ itself is not cyclic.

Lemma: A cyclic subgroup of $\QQ^+/\ZZ^+$ by definition has at least one generator, patently it will be of the form $\langle \frac{a}b\rangle$, and we can have that $\gcd(a,b)=1$. By Bezout’s identity, we know there are integer solutions in $x,y$ to the equation $ax+by=1$ if $\gcd(a,b)=1$. Thus take one of the pairs of solutions $(x_0,y_0)$: thus $\frac{1-by_0}{a}=x_0$. So, we have that $\langle \frac{a}b\rangle \times \frac{1-by_0}{a}= \langle \frac{a}b \times \frac{1-by_0}{a}\times \frac{1-by_0}{a}\rangle = \langle \frac1b \times 1-by_0 \times x_0\rangle = \langle \frac1b \rangle$. Thus for any cyclic subgroup of $\QQ^+/\ZZ^+$ there will exist an integer $b$ such that $\langle \frac1b\rangle$ generates it.

To show that $\QQ^+/\ZZ^+$ itself is not cyclic, we suppose it is. By the lemma there will exist some integer $b$ such that $\langle \frac1b\rangle$ generates it. However the only elements which this thing can generate with the addition operation are $\langle\frac{c}b\rangle$ where $c\in\{0, 1,\ldots,b-1\}$. Thus the equivalence class $\langle\frac1{2b}\rangle$ is not contained in the group generated by $\langle\frac1b\rangle$, which is a contradiction.

Now I know that if I had a direct sum decomposition into cyclic groups, there must be at least two proper non-trivial non-intersecting cyclic subgroup which generate all of $\QQ/\ZZ$. Please help!