If the order of $\langle a\rangle$ is finite, then the distinct elements of $G$ are $e,a,a^2,a^3,...,a^{n-1}$

Question

Suppose $G=\langle a\rangle$. If the order of $\langle a\rangle$ is finite, then the distinct elements of $G$ are $e,a,a^2,a^3,...,a^{n-1}$.

I am trying to prove this statement. What I understand so far is:

Suppose $m\in \mathbb{Z}$. By the Division Algorithm, we have:

$$m=nq+r$$

where $q,r\in \mathbb{Z}$ and $0\leq r<n$. (Ok, I got this.)

Thus, $m\equiv r\mod n$. It follows that $a^m=a^r$. (I got this too.)

Since $G$ is a cyclic group, every element in $G$ is equal to some $a^i$ for $0\leq i<n$.

Suppose $a^i=a^j$. For $0\leq i<n$ and $0\leq j<n$. Without loss of generality, we suppose $i<j$. We have:

$$i\equiv j\pmod n$$ (Everything makes sense so far.)

However, given the range of $i$ and $j$, this is $\underline{impossible}$.

My questions are:

1. Why is $i\equiv j\mod n$ impossible given the range of $i$ and $j$?
2. The textbook says "In particular, $|a|=|\langle a\rangle|$." I don't really understand this. What does $|a|$ have to do with $|\langle a\rangle|$? I am aware that the order of $a$ is the smallest positive integer $n$ such that $a^n=e$.

Thank you!

Answer 1

• If $i \equiv j \pmod{n}$ then the distance from $i$ to $j$ is an integer multiple of $n$, so, if $i$ and $j$ are in $\{0,\dots,n-1\}$, this is only possible if $i=j$.
• Once you know that $|\langle a \rangle| = n$ implies $\langle a \rangle = \{e,a,\dots,a^{n-1}\}$, then the smallest positive integer $m$ such that $a^m = e$ is $n$ because $e,a,\dots,a^{n-1}$ are all distinct.