# Order of Cyclic Subgroups

Let $G$ be a cyclic group with $n$ elements be generated by $a$. Then call the cyclic subgroup $\langle a^{d} \rangle = H$. My professor says that $|H| = \frac{n}{d}$ while the book says the $|H| = \frac{n}{gcd(n, d)}$. Are they the same or am I very confused?

-
They are both correct, $\frac{n}{gcd(n,d)}$ is just more general. – user12205 Oct 30 '11 at 21:23
The variable $n$ is undefined. I assume it is the order of $G$? – Asaf Karagila Oct 30 '11 at 21:26
Asaf, that seems rather trivial – user12205 Oct 30 '11 at 21:33
@Jeroen: At this level? Sure. However vaguely and ill-defined variables are like Baobab trees on the Asteroid B612. One must uproot them at their very infancy, or else they will take over the small planet. – Asaf Karagila Oct 30 '11 at 21:43
@Jon: If $n$ is the order of $G$, then both answers will be false unless $G$ is a cyclic group and $\langle a\rangle=G$ (in my answer, I assumed you meant $n=\text{ord}(a)$, which agrees in this case). For example, in the group $$G=(\mathbb{Z}/6\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z}),$$ for which $|G|=12$, the element $a=(1,1)$ has order 6, and $$a^5=(1,1)+(1,1)+(1,1)+(1,1)+(1,1)=(5,1),$$ hence $$H=\langle a^5\rangle=\langle(5,1)\rangle$$ has $|H|=6$, but $$|H|\neq\frac{12}{1}=\frac{12}{\gcd(5,12)}$$ and $$|H|\neq\frac{12}{5}$$ – Zev Chonoles Oct 30 '11 at 22:02

The book is correct; your professor is correct when $d$ divides (i.e. goes into) $n$.

Note that for any $g\in G$, $$|\langle g\rangle|=\text{ord}(g).$$ The general relation is that, if $\text{ord}(a)=n$, then $$\text{ord}(a^d)=\frac{n}{\gcd(d,n)}.$$ For example, consider the cyclic group $G=\mathbb{Z}/6\mathbb{Z}=\{0,1,2,3,4,5\}$ with operation $+\,$, and let $a=1$. It has order $n=6$. Let $d=5$; then $a^5$ means $$a+a+a+a+a=5$$ so $H=\langle a^5\rangle=\langle 5\rangle=G$, so $|H|=6=\frac{6}{\gcd(5,6)}$. In contrast, the statement that $|H|=\frac{6}{5}$ doesn't even make any sense.

If you intended $n$ to be the order of $G$, then both answers will be false unless $G$ is a cyclic group and $\langle a\rangle=G$ (in my answer above, I assumed you meant $n=\text{ord}(a)$, which agrees in this case, i.e. $\text{ord}(a)=n=|G|$ if and only if $G$ is cyclic and $G=\langle a\rangle$). For example, in the group $$G=(\mathbb{Z}/6\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z}),$$ for which $|G|=12$, the element $a=(1,1)$ has order 6, and $$a^5=(1,1)+(1,1)+(1,1)+(1,1)+(1,1)=(5,1),$$ hence $$H=\langle a^5\rangle=\langle(5,1)\rangle$$ has $|H|=6$, but $$|H|\neq\frac{12}{1}=\frac{12}{\gcd(5,12)}$$ and $$|H|\neq\frac{12}{5}$$

-
I would rather say that the professor is correct if $d$ divides $n$. We may not be getting the full report... – Arturo Magidin Oct 30 '11 at 21:48
@Arturo: That's a good phrasing, I've edited. – Zev Chonoles Oct 30 '11 at 21:50
@Zev: Thanks for another clear answer :) – Student Oct 30 '11 at 21:57
@Zev: Your first response was what I was looking for, and I apologize for not labeling my variables carefully. – Student Oct 30 '11 at 23:39

The book is correct- it is the statement of the Fundamental Theorem of Cyclic Groups. Its proof is rather simple:

Let $t$ belong to <$a^d$>, then $t$ = $a^{dq}$, where $q$ is an integer. Let $s = \gcd(n,d)$. Then $d = sp$, for some integer $p$. Then, $t = a^{spq},$ so $t= (a^s)^{pq}$, so $t$ belongs to <$a^s$> = <$a^\gcd(n,d)$>. Thus, <$a^d$> is a subset of <$a^{\gcd(n,d)}$>.

Let $r$ belong to <$a^{\gcd(n,d)}$>, then $r = a^{sm}$, for some integer $m$. Now, by Bezout's identity, there exist $x,y \in \mathbb{Z}$ such that $\gcd(n,d)= s = nx + dy$, so $r= a^{nx + dy}= a^{dy}$, since $a^{nx} =e$. But, $a^{dy}$ belongs to <$a^d$>, so, we are done, that is, we have proved that is a subset of <$a^d$>, so from our results, we have: = <$a^d$>, where $s = \gcd(n,d)$.

Your Professor probably considered $d$ to be a divisor of $n$, then he is perfectly correct, but is not generalizing just yet.

-