0

Let $\mathbb{Q}/\mathbb{Z}$ be the quotient group of the additive group of rational numbers. Find the order of the element $\frac{2}{3}+\mathbb{Z}$ in $\mathbb{Q}/\mathbb{Z}$.

I tried it by using facts that any quotient $G/H$ of $G$ has induced operation from $G$. So I can do $$\frac{2}{3}+\mathbb{Z} + \frac{2}{3}+\mathbb{Z} = \frac{4}{3}+\mathbb{Z} + \frac{2}{3}+\mathbb{Z} = 2+\mathbb{Z} =\mathbb{Z},$$ and $\mathbb{Z}$ being identity we get order three. But this way, to manually start computing elements, if correct is very unreliable in case of more difficult problem. So is there a generalized approach for quotient groups?

Michael Albanese
  • 99,526
Foggy
  • 1,259
  • 1
    If $p$ and $q$ are coprime, what can we say about the order of $\frac{p}{q} + \mathbb{Z}$? – Jonas Gomes Jan 24 '15 at 14:44
  • Perhaps order is q. – Foggy Jan 24 '15 at 14:46
  • 1
    Yes. Try to prove it. – Jonas Gomes Jan 24 '15 at 14:48
  • Is there generalized result for any quotient group? – Foggy Jan 24 '15 at 14:49
  • The denominator is the order. This is discussed in some of my posts on order and denominator ideals, e.g. this post. – Bill Dubuque Jan 24 '15 at 15:07
  • It is incorrect to write $$\frac{2}{3}+\mathbb{Z} + \frac{2}{3}+\mathbb{Z} = \frac{4}{3}+\mathbb{Z} + \frac{2}{3}+\mathbb{Z} = 2+\mathbb{Z} =\mathbb{Z},$$ as the expressions are not equal. Instead, you should write something like

    \begin{align} \frac{2}{3}+\mathbb{Z} + \frac{2}{3}+\mathbb{Z} &= \frac{4}{3}+\mathbb{Z}\ \frac{4}{3} + \mathbb{Z} + \frac{2}{3}+\mathbb{Z} &= 2+\mathbb{Z} = \mathbb{Z}. \end{align}

     – Michael Albanese Feb 01 '15 at 05:05

2 Answers2

2

Your calculation is correct. The order of $\frac{2}{3} + \mathbb{Z}$ in $\mathbb{Q}/\mathbb{Z}$ is three. You could simplify it by noting that $\left(\frac{2}{3} + \mathbb{Z}\right)^k = \frac{2}{3}k + \mathbb{Z}$.

As alluded to in the comments, if $p$ and $q$ are coprime and $q \neq 0$, then $\frac{p}{q} + \mathbb{Z}$ has order $q$ in $\mathbb{Q}/\mathbb{Z}$; this follows from an analogous calculation.

One way to attack the problem of finding the order of an element of $G/H$ is to use the following result:

If $g \in G$ has finite order $n$, $H \unlhd G$, and $g + H \in G/H$ has order $m$, then $m \mid n$.

So if you know $n$, you can determine the order of $g + H$ by computing $(g + H)^k$ where $k$ ranges over the factors of $n$, starting with the smallest nontrivial factor $k > 1$.

Note, if $g \in G$ has infinite order, $g + H$ may or may not have infinite order.

Michael Albanese
  • 99,526
0

You should think of $\Bbb Q/\Bbb Z$ as a subgroup of $\Bbb R/\Bbb Z$. Let’s look at the latter: from the homomorphism $t\mapsto\exp(2\pi it)$ from $\Bbb R$ to the unit circle-subgroup of $\Bbb C^*$, which you see is onto with kernel $\Bbb Z$, you get the result that $\Bbb R/\Bbb Z$ is isomorphic to the circle-group of complex numbers of absolute value $1$. What’s the subgroup $\Bbb Q/\Bbb Z$? It corresponds to the rotations that are rational multiples of $2\pi$, in other words, rotations that are described as rational number of degrees. With this understanding, $2/3 +\Bbb Z$ becomes the complex number at unit distance from the origin with argument $240^\circ$, namely $(-1-i\sqrt3\,)/2$. To get to the identity $1$, you have to cube it: your quantity $2/3+\Bbb Z$ is of order three.

Lubin
  • 62,818