4

How does one see that the cyclic group $C_n$ of order $n$ has $\phi(d)$ elements of order $d$ for each divisor $d$ of $n$?

(where $\phi(d)$ is the Euler totient function)

Martin Sleziak
  • 53,687
Student
  • 2,755
  • 1
    I think a good hint is to start from the definitions. A group is cyclic if it can be generated from one element. Now playing with that element, you should be able to compute the order of any other element. Also, is that a homework? – David Kohler Mar 31 '11 at 20:30
  • 2
    @David I have never been in a mathematics department and hence will not get mathematics homeworks :) – Student Mar 31 '11 at 20:37
  • 1
    you don't have to be in a mathematics department to get mathematics homework :) – David Kohler Apr 02 '11 at 23:04

2 Answers2

2

Let $g$ be a generator of $C_n$. What is the order $g^a$?

$(g^a)^k = g^{ak} = 1$ if and only if $n|ak$. But $$\begin{align*} n|ak &\Longleftrightarrow n|ak\text{ and }a|ak\\ &\Longleftrightarrow \mathrm{lcm}(n,a)|ak\\ &\Longleftrightarrow a\left.\left(\frac{n}{\gcd(a,n)}\right) \right| ak\\ &\Longleftrightarrow \left.\frac{n}{\gcd(a,n)} \right|k \end{align*}$$ so the order of $g^a$ is exactly $\displaystyle \frac{n}{\gcd(a,n)}$.

So you are trying to count the number of integers $a$, $0\leq a \lt n$, such that $n = d\gcd(a,n)$.

Added. Alternatively, if you can show that a cyclic group of order $n$ has a unique subgroup of order $d$ when $d|n$, and no subgroups of order $d$ when $d$ does not divide $n$, then you turn the problem into finding how many generators the cyclic group of order $d$ has, which gives th result immediately.

Arturo Magidin
  • 398,050
1

Recall $g^{\large i}$ has order $\: n/(i,n)\ $ for a generator $\:g\:$ of $\ C_n\,$ (or, additively in isomorphic group $\Bbb Z_n\,$ $ $ recall $\,\ i\ \,$ has order $\,n/(i,n),\,$ i.e. $\,k\cdot i\equiv 0\pmod{\!n}\iff n\mid ki\iff n/(i,n)\mid k)$

Therefore $\ \ \displaystyle\ \ d\ =\ \frac{n}{(i,n)},\quad\ \ \ \ \ 0 \le i \le n$

$\quad\displaystyle\iff\quad\ (id,\ nd)\ =\ n,\ \ \ \ \ \, 0 \le i \le n$

$\quad\displaystyle\iff\quad \bigg(\frac{i\:d}{n},d\bigg)\ =\ 1,\ \ \ \ \: 0 \le i \le n,\ \ n\ |\ i\:d$

$\quad\displaystyle\iff\quad\ \ \ (\ j,\ d)\ \ =\ \ 1, \ \ \ \ \:0\le j \le d$

where we used the gcd distributive law above.

Bill Dubuque
  • 272,048
  • Sorry. I am a bit confused about your argument. Are you saying that there is a one-to-one correspondence between the $j's$ that solve $(j,d) =1$ and $i's$ that solve $n = d(i,n)$ ? (I guess that the set of $j's$ is typically very different from the set of $i's$) – Student Apr 06 '11 at 18:31
  • @Anirbit: Yes, that's what the above equivalences show. Note that $\rm\ j\ =\ i:d/n\ $ above. The equivalences prove that the $\rm:g^i:$ has order $\rm:d:$ iff $\rm:(j,d)= 1:,\ 0\le j\le d:.$ As is well-known, there are precisely $\rm:phi(d):$ such $\rm:j:.$ – Bill Dubuque Apr 06 '11 at 18:43