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Order of Cyclic Subgroups

If I have a group G where $\vert G \vert = n$ and I want to calculate the number of the elements in G of order n can I then deduce the following statement?

If G is a cyclic group then the number is $\phi (n)$ (Euler's totient function)

else the number is 0


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marked as duplicate by lhf, Grigory M, Austin Mohr, t.b., BenjaLim Sep 12 '12 at 4:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

How can we know if you can deduce it? I think that you are rather trying to ask if that statement is true and, in that case, how to prove it... – Mariano Suárez-Alvarez Apr 9 '12 at 22:56
This is very, very far from true. Did you try any examples? – Qiaochu Yuan Apr 9 '12 at 22:57
How could the number be independent of $k$? – joriki Apr 9 '12 at 22:57
If the group is not cyclic, do you think there is no element of ANY order? ;) – N. S. Apr 9 '12 at 23:06
I have edited the question. – bemyguest Apr 9 '12 at 23:08
up vote 2 down vote accepted

The statement is true; that is:

Let $G$ be a finite group of order $n$. Then either $G$ is cyclic and has exactly $\phi(n)$ elements of order $n$, or else $G$ is not cyclic and has no elements of order $n$.

Proof. $\langle g\rangle$ is a subgroup of $G$ of order $|g|$ for each $g\in G$. By definition, $G$ is cyclic if and only if there exists $g\in G$ such that $\langle g\rangle = G$. So for $G$ of order $n$ we have: $$\begin{align*} G\text{ is cyclic} &\iff \text{there exists }g\in G\text{ such that }\langle g\rangle = G\\ &\iff\text{there exists }g\in G\text{ such that }|\langle g\rangle|=|G|\\ &\iff \text{there exists }g\in G\text{ such that }|\langle g\rangle = n\\ &\iff\text{there exists }g\in G\text{ such that }|g|=n. \end{align*}$$ Thus, if $G$ is not cyclic, then there are no elments of order $n$. If $G$ is cyclic, then there is at least one element of order $n$, and every element is of the form $g^k$ for some $k\in\mathbb{Z}$, which we may assume satisfies $0\leq k\lt n$ (prove it!). Since $|g^k| = n/\gcd(n,k)$ (prove it!), then the number of elements of order $n$ is the number of integers $k$ such that $0\leq k\lt n$ and $n/\gcd(n,k)=n$, which is the number of integers $k$ such that $0\leq k\lt n$ and $\gcd(n,k) = 1$.

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