Suppose that $G$ is a finite cyclic group. Let $m=|G|$. Assume that $m\ge3$. Let $S=\{a\in G:|a|=m\}$. Prove that the cardinality of $S$ is even.

Question

Yeah, NO IDEA WHERE TO GO FROM HERE. $|S|$ has to be equal to $2k$ with $k$ being a positive integer, but someone please offer a hint on how I can get started on this.

Answer 1

Hint: $S$ is the set of generators of $\mathbb{Z}_n$, and the number of generators is exactly $\varphi(n)$, where $\varphi$ is the Euler phi function.

Answer 2

Theorem : If $G$ is a cyclic group of order $m$ and $d$ is a divisor of $m$, then the number of elements in $G$ that has order $d$ is $\phi (d)$.

$\phi (d)$ = the length of $U(d)$, where $U(d)$ = the set of positive integers that are relative prime to $d$. example : $U(8) = {1,3,5,7}$, $\phi (8) = 4$.

Hint : $S$ is the set of all the elements in $G$ that has order $m$, whats the Cardinality?

Answer 3

If $a\in S$ then $a^{-1}\in S$. Since $m\ge 3$, we can't have $a=a^{-1}$. Hence the elements in $S$ appear in pairs and so $S$ has an even number of elements.

This arguments works for groups in general: there is an even number elements of order $m\ge3$. No need to assume that the group is cyclic or that $m=|G|$.