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

I have the 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 $n$ 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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  • $\begingroup$ Identify $\Bbb Q/\Bbb Z$ with the torsion subgroup of $S^1=\Bbb C^{\times}$ and consider all elements of order $n$. $\endgroup$
    – lhf
    Mar 19, 2012 at 16:01
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    $\begingroup$ Abstract Duplicate (Covered in Mariano's answer): math.stackexchange.com/q/66145/21436 $\endgroup$
    – user21436
    Mar 19, 2012 at 16:05

3 Answers 3

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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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  • $\begingroup$ Dear Arturo! Let me ask you the following question: I know that every finitely generated subgroup of $\mathbb{Q}$ is cyclic. But how to show that the same is true for $\mathbb{Q}/\mathbb{Z}$. I was trying to prove it rigorously but no results. Could you show it, please? $\endgroup$
    – ZFR
    Nov 18, 2018 at 22:48
  • $\begingroup$ Consider the natural mapping $\varphi: \mathbb{Q}\to \mathbb{Q}/\mathbb{Z}$ and suppose that $\bar{G}\leq \mathbb{Q}/\mathbb{Z}$ with $\bar{G}=\langle \bar{g}_1,\dots, \bar{g}_n \rangle$. I was trying to consider its preimage in $\mathbb{Q}$ but it does not give something good. $\endgroup$
    – ZFR
    Nov 18, 2018 at 22:57
  • $\begingroup$ @ZFR: You are following up on a six year+ old post to ask someone whose profile says they no longer actively participate here so they can answer your personal query? Sigh. If $\overline{G}$ is generated by $\overline{g_1},\ldots,\overline{g_n}$, then the subgroup $G=\langle g_1,\ldots,g_n\rangle = \langle h\rangle$ is cyclic; and $\langle \overline{h}\rangle$ generates $\overline{G}$. $\endgroup$ Nov 18, 2018 at 23:24
  • $\begingroup$ @ZFR: PS and the preimage is generated by $g_1,\ldots,g_n$ and $1$, hence also finitely generated. I have no idea how it does not “give something good”. $\endgroup$ Nov 18, 2018 at 23:26
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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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  • $\begingroup$ Doesn't that Bezout formula only imply $\left<\dfrac{j a}b\right>\equiv \left<\dfrac{1}b\right>\:\ (mod\ \mathbb Z) $ ? $\endgroup$
    – vacant
    Mar 30 at 18:26
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    $\begingroup$ @vacant $\large\ \frac{1}b =j\left(\frac{a}b\right)+k\,\Rightarrow\, \bmod \Bbb Z\!:\,\langle\frac{1}b\rangle \subseteq \langle \frac{a}b\rangle.\,$ The reverse inclusion is clear. $\endgroup$ Mar 31 at 8:19
  • $\begingroup$ Ah, my mistake. I misinterpreted the angle brackets as denoting the class mod $\mathbb Z$ rather than denoting ideals. Thank you for the clarification. $\endgroup$
    – vacant
    Mar 31 at 17:25

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