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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.

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They are both correct, $\frac{n}{gcd(n,d)}$ is just more general. –  user12205 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
    
Asaf, that seems rather trivial –  user12205 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
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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
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up vote 10 down vote accepted

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

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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
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@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
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