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$\newcommand{\ZZ}{{\mathbb{Z}}} \newcommand{\QQ}{{\mathbb{Q}}} \newcommand{\FF}{{\mathbb{F}}} \newcommand{\PP}{{\mathbb{P}}} \newcommand{\RR}{{\mathbb{R}}} \newcommand{\CC}{{\mathbb{C}}} \newcommand{\ra}{{\rightarrow}} \newcommand{\eps}{{\epsilon}}$

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!

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    "Let $H_k$ be **the** cyclic subgroup in the decomposition that is generated by $\frac{a}{b}$..." is bad phrasing. How do you know there is such a subgroup? Presumably you meant "Let $H_k$ be **a** cyclic subgroup in the decomposition, and let $\frac{a}{b}$ be a generator, where...."2012-04-30
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    @ArturoMagidin You're right about the language. However, I am still unsure about doing what you say. I can't really say that there is an element $a/b$ that is the generator for some group. But then it seems like I cannot make an argument, since how can I talk about generators...2012-04-30
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    You are assuming that your group is a direct sum of cyclic groups. Then you let $H_k$ be one of the direct summands. Since $H_k$ is cyclic, it has generators, so you can pick an $\frac{a}{b}\in\mathbb{Q}$ with the property that $\frac{a}{b}+\mathbb{Z}$ is a generator of $H_k$. That accomplishes what I think you were trying to do with what you wrote (but what you wrote is technically nonsensical, because you seem to *first* fix the element $\frac{a}{b}$ and then look for "the" direct summand that is generatedby it; such a direct summand might not exist, even if the group *were* a direct sum!)2012-04-30
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    @ArturoMagidin Please see my new answer2012-04-30
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    You say: "if $r>0$ were the generator, then $r/2$, which is rational, would not be included." But 1/3 is in the subgroup generated by 2/3.2012-05-01
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    @MarkDominus yes that was a mistake. I showed below though that I can always find a generator for a cyclic subgroup of the form $\frac1b$ and this subgroup does not contain $\frac1{2b}$.2012-05-01

3 Answers 3

3

Some comments:

  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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    So is it more like: Take some nontrivial proper subgroup $A$, call its generator $\frac{a}b$, where $\gcd(a,b)$. I know that $\frac1b \in A$ since for some integer $n$, there is an integer solution for $x$ in $a x = nb+1$, i.e. $\frac{nb+1}{a}$ will eventually be an integer? ....I'm not really sure how to connect this with my previous argument.2012-04-30
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    Is it once I show that $\frac{1}{b}$ is in $A$, I say $\frac{1}{b}$ must also be a generator since it clearly generates what was assumed to be another generator $\frac{a}b$. Now that $\frac1b$ is a generator though, I am still not sure where to go grom there. How do I know something like $\frac1{b^m}$ won't be in $A$. You seem to be saying to focus on just two nontrivial subgroups and show that they must intersect, but surely this is not the case. I feel like I want to say something along the lines2012-04-30
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    @jake: You seem to be trying to argue that if $\frac{1}{b}+\mathbb{Z}$ is a generator of one of the cyclic summands, then $\frac{1}{2b}$ cannot be represented by the direct sum; your first step is to argue that generators can *always* be taken to be of the form $\frac{1}{b}+\mathbb{Z}$, as opposed to the more general $\frac{a}{b}+\mathbb{Z}$. Your argument is flawed, because it asserts that $\frac{1}{b}$ has to be *in* one of the cyclic factors, which is not true a priori. I'm suggesting instead that you use the fact that there exist $x,y\in\mathbb{Z}$ with $ax+by=1$ (cont)2012-04-30
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    @jake: so that $\frac{1}{b}+\mathbb{Z}=\frac{ax+by}{b}+\mathbb{Z} = (\frac{ax}{b}+y)+\mathbb{Z} = \frac{ax}{b}+\mathbb{Z} = x(\frac{a}{b}+\mathbb{Z})\in\langle \frac{a}{b}+\mathbb{Z}\rangle\subseteq \langle \frac{1}{b}+\mathbb{Z}\rangle$, in order to show that the subgroup generated by $\frac{a}{b}+\mathbb{Z}$ is the same subgroup as the one generated by $\frac{1}{b}+\mathbb{Z}$.2012-04-30
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    @jake: You cannot say "take some nontrivial proper subgroup, call its generator $\frac{a}{b}$". **First:** because arbitrary proper subgroups need not be cyclic, so talking about a generator for "some nontrivial proper subgroup" is nonsense. **Second:** because you should not say "its generator", since there are generally *lots* of generators; **Third:** because the elements of your group are not rationals, they are **classes** of rationals.2012-04-30
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    @jake: You make a bad argument in order to show that $\frac{1}{b}+\mathbb{Z}$ is in $\langle \frac{a}{b}+\mathbb{Z}\rangle$. Your argument is bad, because it assumes that $\frac{1}{b}+\mathbb{Z}$ must be an element of a cyclic factor; *you have no warrant for that assertion*. As to my suggestion that you focus on "just two nontrivial subgroups", that's a suggestion for how to prove the result without going down into the analysis you are making, which is currently murky, confusing, and flawed.2012-04-30
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    Like I said, I agree that my initial argument for that was shoddy, but I see now that it is easy to remedy this and put myself in a position to assert that: you can take some cyclic subgroup (we are assuming for sake of contradiction that it decomposes into CYCLIC subgroups) in the direct sum decomp, it will have "a" generator of the form $\frac{a}b$, which can be simplified so that $\gcd(a,b)=1$. By Bezout's identity I can solve $ax+by=1$ with integer solutions for $x,y$ and thus this group contains $\frac1b$ which obviously generates the previous generator so we can say $\frac1b$ generates.2012-04-30
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    @Jake: You are still confusing elements of $\mathbb{Q}$ with elements of $\mathbb{Q}/\mathbb{Z}$. "Generates the previous generator" is bad writing; you mean $\frac{a}{b}+\mathbb{Z}$ and $\frac{1}{b}+\mathbb{Z}$ generate the same subgroup. That argument, which can be made *outside* (independent) of the assumption that your group is decomposed as a direct sum of cyclic groups is just: "If $C$ is a cyclic subgroup of $\mathbb{Q}/\mathbb{Z}$, then there exists an integer $b$ such that $\frac{1}{b}+\mathbb{Z}$ generates $C$". Phrase it that way instead of embedding it in a contradiction proof.2012-04-30
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    So this doesn't seem to work...2012-04-30
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    @jake: What is "this"?2012-04-30
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    I understand up until you say "Then $\frac{1}{b_1^2}+\mathbb{Z} = \sum\left(\frac{a_i}{b_i}+\mathbb{Z}\right)$ for some $a_i$, with $b_i|a_i$ for almost all $i$." What is this $b_1$? How did what you did before get you to that point? Are you just saying that $\frac1{b_k^2}$ isn't in $H_k$ so it must be able to be written using the other summands in the direct sum decomp? I think you changed your indices somewhere along the way, and it's confusing me.2012-04-30
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    @jake: $b_1$ is a particular generator, say that of a "first" subgroup. Since $\frac{1}{b_1^2}+\mathbb{Z}$ is an element of $\mathbb{Q}/\mathbb{Z}$, it must be expressible as an element of the direct sum. An element of the direct sum is a sum of multiples of the generators of the direct factors, with almost all multiples equal to $0$; that's what $b_i|a_i$ gives (that the summand is zero). I'm saying that $\frac{1}{b_1^2}$ must be expressible as an element of the direct sum, and I'm using the expression to derive a contradiction.2012-04-30
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    How do you get $b_1$ divides $a_1b_1-1$ it seems like it just divides $a_1b_1$...what it seems like you're saying is that since we are assuming that that this is a direct sum decomposition we must have that $H_1 \cap \bigoplus_{k\neq1} H_k$ is trivial and thus we must have that $b_k|a_kb_1$ for all $k\neq1$...but then you go on to say $b_1|a_1b_1-1$ where is that?2012-05-01
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    @jake: Because two direct sum expressions are equal if and only if they are identical coordinate by coordinate. So the $1$th coordinate of the left hand side, which is $\frac{1}{b_1}+\mathbb{Z}$, must equal the $1$th coordinate of the right hand side, which is $\frac{a_1b_1}{b_1}+\mathbb{Z}$. Therefore, $\frac{1}{b_1}+\mathbb{Z} = \frac{a_1b_1}{b_1}+\mathbb{Z}$, so $\frac{a_1b_1}{b_1}-\frac{1}{b_1}=\frac{a_1b_1-1}{b_1}\in\mathbb{Z}$, that is, $b_1$ divides $a_1b_1-1$. Did you write it out and try it yourself?2012-05-01
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    It just seems like you could have said $b_1|a_1b_1$ and thus $\frac1{b_1}+\ZZ=\frac{a_1b_1}{b_1}+\ZZ=\ZZ$ a contradiction...obviously, it's the same, but it just seems nicer.2012-05-01
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$\newcommand{\ZZ}{{\mathbb{Z}}} \newcommand{\QQ}{{\mathbb{Q}}} \newcommand{\FF}{{\mathbb{F}}} \newcommand{\PP}{{\mathbb{P}}} \newcommand{\RR}{{\mathbb{R}}} \newcommand{\CC}{{\mathbb{C}}} \newcommand{\ra}{{\rightarrow}} \newcommand{\eps}{{\epsilon}}$ 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!

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    $\LaTeX$ comment: `<` and `>` are operators, not delimiters. The delimiters are `\langle` and `\rangle`. Note the spacing and the formatting.2012-04-30
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    My suggestion about subgroups doesn't work in $\mathbb{Q}/\mathbb{Z}$ (it works in $\mathbb{Q}$). Sorry about that. The rest has been cleaned up considerably.2012-04-30
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    Note that the examples you give don't give nontrivial non-intersecting subgroups. $\langle 0+\mathbb{Z}\rangle$ is trivial, and $\langle\frac{1}{3}+\mathbb{Z}\rangle = \langle\frac{2}{3}+\mathbb{Z}\rangle$, so they intersect.2012-04-30
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    @ArturoMagidin To your last comment, I am not using $\langle$ and $\rangle$ to denote the subgroup generated by these elements I using it to denote equivalence classes as I said earlier. Look at those two sets, those are the groups. The left is the subgroup isomorphic to $C_2$ and the other is isomorphic to $C_3$, these groups (using bars to denote equivalence classes instead) are $\{\bar 0, \bar\frac12\}$ and $\{\bar 0, \bar\frac13, \bar\frac23\}$. Is that clear?2012-04-30
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    That's bad notation again. $\frac{1}{2}+\mathbb{Z}$ is *already* an equivalence class: it is a coset, hence an element of the quotient; you would want to use $\langle\frac{1}{2}\rangle$ instead.2012-04-30
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    $\mathbb{Q}/\mathbb{Z}$ is divisible, and in particular has no finite index subgroups. Direct sums of cyclic groups, on the other hand...2012-05-01
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    @SteveD Direct sums of cyclic groups on the other hand what?..Can they not be divisible? What if the sum is infinite...2012-05-01
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    Can you always find a finite index subgroup?2012-05-01
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    @SteveD No you cannot, the case where the group is divisible is the counterexample. Although I appreciate the help, could you be marginally less vague please.2012-05-01
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    @jake: No, no... can you always find a proper subgroup of finite index in a direct sum of cyclic groups (remember "on the other hand...")? If you can always find a proper subgroup of finite index in a direct sum of cyclic groups, but never in a divisible group, then since $\mathbb{Q}/\mathbb{Z}$ is divisible...2012-05-01
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  1. The group $\mathbb{Q}$ is divisible.

  2. A homomorphic image of a divisible group is divisible.

  3. The group $\mathbb{Q}/\mathbb{Z}$ is divisible (by 1 and 2).

  4. A direct summand of a divisible group is divisible (by 2).

  5. No nonzero cyclic group is divisible, because nonzero divisible groups are infinite and $\mathbb{Z}$ is not divisible.

Therefore no nonzero cyclic group is a direct summand of $\mathbb{Q}/\mathbb{Z}$.