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The number of elements of order $n$ in a finite cyclic group of order $N$ is $0$ unless $n|N$, in which case it is $N/n$.

Is "the number of elements of order $n$" referring to the number of elements of the subgroup that is of order $n$?

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No, it refers to the number of elements $x$ such that $x^n$ is the identity element, but $x^m$ is not for any $m<n$. –  Alex Becker Aug 18 '12 at 2:49
    
You move just a bit too quick for me! I write my last answer to your other question, and you enhance the question while I'm writing it. You write me a comment (thank you) asking me about the second half, so I type up a response, and then you edit it out of the question into another! But I caught up! –  mixedmath Aug 18 '12 at 2:58
    
The answer by mixedmath addresses the details, but in short the statement in the question is completely wrong (where did you get this from?). It should read: The number of elements of order $n$ in a finite cyclic group of order $N$ is $0$ unless $n \mid N$, in which case it is $\phi(n)$ (Euler phi function). –  Marc van Leeuwen Aug 18 '12 at 4:05
    
@mixedmath: I know it is perfectly "legal" for someone to ask 4 questions their first day here, but surely something can be done to at least discourage people from doing as you describe in your comment? I mean, is anybody learning anything from this? –  user641 Aug 18 '12 at 6:15

1 Answer 1

You say that if $n\mid N$, then there are $N/n$ elements of order $n$ in the cyclic group of order $N$. And this is just wrong. For example, let's consider the cyclic group of order $5$. From above (I mean your previous question, we know there are $\varphi(5) = 4$ elements of order $5$, and the last element is of order $1$ (it's the identity). (And this is correct). Let's test your second statement against this.

If we ask ourself how many elements of order $5$ there are, as $5 \mid 5$, we think that there are $5/5 = 1$... How many of order $1$? There are $5/1 = 5$. Oh dear, this is terrible. In fact, we've now given the orders of $6$ different elements of our $5$-element group.

It is true that if $n\mid N$, then there are elements of order $n$, and if $n \not | \;\, N$, then there are none. Briefly, this is seen easiest by the fact that if $N = kn$, and $g$ generates $G$, then $g^k$ will be or order $n$. If $n \not | \;\,N$, then Lagrange's theorem prevents there from being any elements of order $n$.

So we are left with a question: if $n \mid N$, how many elements of order $n$ are there? Well, if $G$ is cyclic, there is a unique subgroup of order $n$, and this will have $\varphi(n)$ generators (which are elements of order $n$). This reasoning can be formalized, so that there are $\phi(n)$ elements of order $n$ if $n \mid N$.

To answer your exact question, we say an element $g$ of $G$ is of order $n$ if $g^n = 1$, the identity, but $g^m \neq 1$ for any $m < n$. The number of elements of order $n$ is not the same as the number of elements in the subgroup of order $n$ (which is... $n$).

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You proved the statement, but I don't think you answered the question. –  William Aug 18 '12 at 3:05
    
@William: Ah, perhaps you're right. I'll add a line. Is that better? –  mixedmath Aug 18 '12 at 3:06

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