For $[a]_n$ in $\Bbb Z_n$, what is the order of an invertible element? So far I know that if $$[a]_n[b]_n = [1]_n$$ for some congruence class $[b]_n$, then $[a]_n$ is an invertible element of $\Bbb Z_n$.
Also for the definition of an order: Let $\sigma$ be in $S_n$. The least positive integer $m$ such that $(\sigma)^m = (1)$ is called the order of $\sigma$.
So does this mean that the order is infinite?
I can't seem to put these two together.

By Euler's Theorem, we know that
$$a^{\varphi(n)} \equiv 1 \pmod{n}$$
whenever $a$ and $n$ are relatively prime, and where $\varphi$ is Euler's totient function. In the language of congruence classes, this means that
$$[a]_n [a^{\varphi(n) - 1}]_n = [1]_n$$
Now an element of $\mathbb{Z}_n$ is invertible if and only if $\gcd{(a, n)} = 1$ (prove this!), so the order of any invertible element can't be bigger than $\varphi(n)$. In particular, it's very finite (more generally, in any finite group, every element has finite order).
