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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?

Thanks in advance.

Student
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    They are both correct, $\frac{n}{gcd(n,d)}$ is just more general. –  Oct 30 '11 at 21:23
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    The variable $n$ is undefined. I assume it is the order of $G$? – Asaf Karagila Oct 30 '11 at 21:26
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    Asaf, that seems rather trivial –  Oct 30 '11 at 21:33
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    @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
  • He is probably assuming that $d$ is a divisor of $n$. In that case, everyone is right: $\operatorname{gcd}(n, d) = d$. – Dylan Moreland Oct 30 '11 at 21:47
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    @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
  • @Jeroen: See Zev's comment for why defining the variables is of utmost important. – Asaf Karagila Oct 30 '11 at 22:45

2 Answers2

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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 $$|H|\neq\frac{12}{5}$$

Zev Chonoles
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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 <$a^{\gcd(n,d)}$> is a subset of <$a^d$>, so from our results, we have: <$a^s$> = <$a^d$>, where $s = \gcd(n,d)$.

Your professor probably considered $d$ to be a divisor of $n$, meaning that they're perfectly correct, but aren't generalizing just yet.

danielbmancini
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