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Viewing $\mathbb{Z}$ and $\mathbb{Q}$ as additive groups, I have an idea to show that $\mathbb{Q}/\mathbb{Z}$ has a unique subgroup of order $n$ for any positive integer $n$. You can take $a/n+\mathbb{Z}$ where $(a,n)=1$, and this element has order $n$.

Why would such an element exist in any subgroup $H$ of order $n$? If not, you could reduce every representative, and then every element would have order less than $n$, but where is the contradiction?

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4 Answers 4

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Find all elements of order a divisor of $n$ in $\mathbb Q/\mathbb Z$ (that is, all elements $g$ such that $ng=0$). There are exactly $n$ of them, and there is in fact at least one of them whose order is exactly $n$.

Show that those elements are in fact the elements of a subgroup.

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First $\mathbb{Q}\cong\prod_{p\in\Pi}\mathbb{Z}_{p^{\infty}}$, where $\Pi$ is the set of all prime numbers. Then for each $p$ prime $\mathbb{Z}_{p^{\infty}}$ has a unique subgroup of order $p^n$ for each $n\in\mathbb{N}$. At last apply the Fundamental Theorem of Arithmetic.

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Let $x$ be a real number in the open interval $(0,1)$. If $nx$ is an integer $k$ for some positive integer $n$, we have that $x = k/n$. You see that $k=1,\ldots,n-1$. I'll leave the rest to you.

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4  
Hm, $x$ an integer in the open interval $(0,1)$. There aren't too many of those. –  JSchlather Sep 20 '11 at 19:11
    
whoooooops. typo. –  Gooz Sep 20 '11 at 23:01

We can approach the problem using elementary number theory. Look first at the subgroup $K$ of $\mathbb{Q}/\mathbb{Z}$ generated by (the equivalence class of) $q/r$, where $q$ and $r$ are relatively prime.

Since $q$ and $r$ are relatively prime, there exist integers $x$ and $y$ such that $qx+ry=1$. Divide both sides by $r$. We find that $$x\frac{q}{r}+y=\frac{1}{r}.$$ Since $y$ is an integer, it follows that $\frac{1}{r}$ is congruent, modulo $1$, to $x\frac{q}{r}$. It follows that (the equivalence class of) $1/r$ is in $K$, and therefore generates $K$.


Now let $H$ be a subgroup of $\mathbb{Q}/\mathbb{Z}$ of order $n$. Let $h$ be an element of $H$. If $h$ generates $H$, we are finished. Otherwise, $h$ generates a proper subgroup of $H$. By the result above, we can assume that $h$ is (the equivalence class of) some $1/r_1$, and that there is some $1/b$ in $H$ such that $b$ does not divide $r_1$. Let $d=\gcd(r_1,b)$.

There are integers $x$ and $y$ such that $r_1x+by=d$. Divide through by $r_1b$. We find that $$x\frac{1}{b}+y\frac{1}{r_1}=\frac{d}{r_1b}.$$ It follows that (the equivalence class of) $d/(r_1b)$ is in $H$. But $r_1b/d$ is the least common multiple of $r_1$ and $b$. Call this least common multiple $r_2$. Then since $r_1$ and $b$ both divide $r_2$, the subgroup of $H$ generated by (the equivalence class of) $1/r_2$ contains both $1/r_1$ and $1/b$.

If $1/r_2$ generates all of $H$, we are finished. Otherwise, there is a $1/b$ in $H$ such that $b$ does not divide $r_2$. Continue.

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