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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...." –  Arturo Magidin Apr 30 '12 at 19:40
    
@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... –  Steven-Owen Apr 30 '12 at 19:52
    
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!) –  Arturo Magidin Apr 30 '12 at 19:58
    
@ArturoMagidin Please see my new answer –  Steven-Owen Apr 30 '12 at 21:09
    
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. –  MJD May 1 '12 at 5:09

2 Answers 2

up vote 2 down vote accepted

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. –  Steven-Owen Apr 30 '12 at 20:14
    
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 lines –  Steven-Owen Apr 30 '12 at 20:21
    
@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) –  Arturo Magidin Apr 30 '12 at 20:21
    
@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 $\frac{1}{b}+\mathbb{Z}$. –  Arturo Magidin Apr 30 '12 at 20:23
    
@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. –  Arturo Magidin Apr 30 '12 at 20:25

$\newcommand{\ZZ}{{\mathbb{Z}}} \newcommand{\QQ}{{\mathbb{Q}}} \newcommand{\FF}{{\mathbb{F}}} \newcommand{\PP}{{\mathbb{P}}} \newcommand{\RR}{{\mathbb{R}}} \newcommand{\CC}{{\mathbb{C}}} \newcommand{\ra}{{\rightarrow}} \newcommand{\eps}{{\epsilon}}$ 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. –  Arturo Magidin Apr 30 '12 at 21:12
    
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. –  Arturo Magidin Apr 30 '12 at 21:20
    
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. –  Arturo Magidin Apr 30 '12 at 21:37
    
@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? –  Steven-Owen Apr 30 '12 at 21:44
    
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. –  Arturo Magidin Apr 30 '12 at 21:46

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