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$\mathbb{Q}/\mathbb{Z}$ has a unique subgroup of order $n$ for any positive integer $n$?

I have factor group $\Bbb Q/\Bbb Z$, where $\Bbb Q$ is group of rational numbers and $\Bbb Z$ is group of integers (operation in both of them is addition). It's necessary to prove that for every natural number $\Bbb n$ there exists one and only one subgroup of $\Bbb Q/\Bbb Z$ with order equal to $n$.

I've proved that such group exists (cyclic group suits this condition), but have almost no idea about how to prove its uniqueness.

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marked as duplicate by Kannappan Sampath, Arturo Magidin, anon, lhf, Brandon Carter Mar 19 '12 at 17:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Identify $\Bbb Q/\Bbb Z$ with the torsion subgroup of $S^1=\Bbb C^{\times}$ and consider all elements of order $n$. –  lhf Mar 19 '12 at 16:01
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Abstract Duplicate (Covered in Mariano's answer): math.stackexchange.com/q/66145/21436 –  user21436 Mar 19 '12 at 16:05

3 Answers 3

up vote 5 down vote accepted

Remember that every finitely generated subgroup of $\mathbb{Q}$ is cyclic; therefore, the quotient $\mathbb{Q}/\mathbb{Z}$ has the same property. Thus, a subgroup of $\mathbb{Q}/\mathbb{Z}$ of order $n$ must be cyclic, generated by some element $\frac{a}{b}+\mathbb{Z}$; we may assume without loss of generality that $\gcd(a,b)=1$, and $0\leq a\lt b$.

Since it is of order $n$, that means that $\frac{na}{b}+\mathbb{Z} = 0+\mathbb{Z}$, which in turn means that $\frac{na}{b}\in\mathbb{Z}$. Since $\gcd(a,b)=1$, then $b|n$. But since the order is exactly $n$, for any $k$, $1\leq k\lt n$, we have $\frac{ka}{b}\notin \mathbb{Z}$, so $b$ does not divide $k$.

That is: $b|n$, but $b$ does not divide any positive number smaller than $n$. That means that $b=n$.

So a subgroup of order $n$ is generated by an element of the form $\frac{a}{n}+\mathbb{Z}$, with $1\leq a\lt n$, $\gcd(a,n)=1$.

Since $\gcd(a,n)=1$, there exist $x,y\in\mathbb{Z}$ such that $ax+ny=1$. Therefore, $\frac{1}{n} = \frac{ax+ny}{n} = \frac{ax}{n}+y$. Thus, $\frac{1}{n}+\mathbb{Z} \in \langle \frac{a}{n}+\mathbb{Z}\rangle$; since $\frac{a}{n}+\mathbb{Z}\in\langle\frac{1}{n}+\mathbb{Z}\rangle$, the two subgroups are equal. So the only subgroup of order $n$ is $\langle \frac{1}{n}+\mathbb{Z}\rangle$.

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Hint: A generator will be of the form m/n, where m is an integer relatively prime to n. Show that all such generators actually generate the same group.

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Hint $\rm\ \ (a,b) = 1\: \Rightarrow\ \left<\dfrac{a}b\right>\equiv \left<\dfrac{1}b\right>\:\ (mod\ \mathbb Z)\: $ by $\rm\:b^{-1}\:\! (j\:a + k\: b = 1)\ $ by Bezout.

Remark $\ $ This Bezout reduction of fractions $\rm\:\! (mod\ 1)\:\! $ often proves handy, e.g. see here.

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