Let $G$ be a finite cyclic group. Prove that if $n$ is a divisor of $|G|$ then there exists an element of order $n$ in $G$.

Question

Let $G$ be a finite cyclic group. Prove that if $n$ is a divisor of $|G|$ then there exists an element of order $n$ in $G$. How many such elements are there?

My attempt:

Let $|G| = m$.

Since $n$ is a divisor of $|G| = m$ there exists some integer $k$ such that $\frac{m}{n} = k$. We want to show that there's an element $h \in G$ such that $h^n = e$.

Let $g \in G$ be the generator of this cyclic group.

$$g^{\frac{m}{n}} = h \Rightarrow g^m = h^n = e$$

For any $s < n$ we have $h^s = g^{s \frac{m}{n}} \neq e$ and therefore the order of the element $h$ is $n$.

My questions:

1. Is that reasoning correct?
2. How do I figure out how many such elements are there?

Answer 1

As an answer regarding the existence is already given, I will only answer your second question on how many such numbers exist. A number m in $\mathbb Z _n$ is of order $d$ if $gcd(n,m) = \frac n d$, for we want $m * k$ for a minimal $k$ to be 0 modulo n. From here we know that the number of such elements is $\varphi (\frac {n}{d})$, (where $\varphi(n)$ is the number of positive integers up to n that are relatively prime to n) as we want elements of the form $k * d$, where gcd(k,$\frac n d $) = 1.

Answer 2

Question: "Let G be a finite cyclic group. Prove that if n is a divisor of |G| then there exists an element of order n in G."

Answer: Let $G:=\{e,g,g^2,..,g^{m-1} \}$ with $g^m=e$ hence $\#G=m$ and assume $m=kn$. Let $u:=g^k$. It follows $u^n=g^{nk}=g^{m}=e$ hence

$$H(n):=\{e,u,u^2,..,u^{n-1}\} \subseteq G$$

is a subgroup with $n$ elements. Hence for any integer $n$ dividing $\# G$ there is a subgroup $H(n) \subseteq G$ with $n$ elements.

Comment: "You didn't show that the order of u is actually n. – aschepler 2 hours ago"

Note: If $u^i=e$ for $i \leq n-1$ it follows $u^i=g^{ik}=e$ and $ik< m$, hence $g$ is not a generator of $G$. Hence $n$ is the smallest integer with $u^n=e$.