Cyclic Group Order of Elements In "Mathematical Methods for Physics and Engineering" by Riley, Hobson, and Bence, in the chapter on group theory, the order of the element $X$ in the finite group $G$ is the number $m$ such that $X^m=I$; thus, the order of the identity is always $1$. A cyclic group is then defined as a group generated from a single element $X$, i.e.,
$$G=\{I,X,X^2,X^3,\dots,X^{g-1}\},$$
where $g$ is the order of the group. They then write that

It is clear that cyclic groups are always Abelian and that each element, apart
from the identity, has order g, the order of the group itself.

I don't see why the second statement is true; for example, when $g=4$, then the element $X^2$ should have order $2$. Am I wrong?
Update: I reached out to the first author and he said that this error was corrected in the $18$th reprint of the $3$rd edition (I was looking at the $2$nd edition).
 A: You are right.

The main problem is in the given definition of order. In fact, the order $|a|\in\Bbb N$ of an element $a$ in a group is the smallest positive integer such that $a^{|a|}=e$.

Consider the cyclic group $\Bbb Z_6$ of six elements. Let $X$ be a generator. Then $X^2$ has order three and $X^3$ has order two.
A: The definition for order is a bit weird in the book. Generally you would say that the order of $X \in G$ is the least integer for which $X^m = I$ (if such an integer exists). I believe what they meant to articulate is that if $G$ is cyclic of order $n$, then raising any element of the group to $n$ will give you the identity, which is not the same as saying an element has that order, if you're using the standard definition.
To see why it's the case, let $G = \langle X \rangle$ with $|G| = n$, then for any $Y \in G$ note that $Y = X^k$ for some $1 \leq k \leq n$ by definition of being a cyclic group, then
$$
Y^n = (X^k)^n = X^{kn} = (X^n)^k = 1^k = 1.
$$
This is what I imagine they mean.
Edit: I base this off the sentence

the order of the element $X$ in the finite group $G$ is the number $m$ such that $X^m =I$

in your answer. It doesn't make sense to say the number $m$ such that $X^m =I$ since it certainly isn't unique, that's why the definition of order usually specifies minimal.
