# Theorem

Define Order of an Element g of Finite Group G:

o(g) = the least positive integer n such that gn = e

Define Order of a Cyclic Subgroup generated by g:

${\displaystyle o(\langle g\rangle )}$ = # elements in ${\displaystyle \langle g\rangle }$
o(g) = ${\displaystyle o(\langle g\rangle )}$

# Proof

Since ${\displaystyle \langle g\rangle }$ is cyclic, and has ${\displaystyle o(\langle g\rangle )}$ elements.

${\displaystyle g^{o(\langle g\rangle )}=e}$

By diagram,

0. ${\displaystyle g^{o(\langle g\rangle )}=e=g^{o(g)}}$.
1. Let n = o(g), and m = ${\displaystyle o(\langle g\rangle )}$
2. gn = gm
3. gn – m = e
4. Let n – m = sn + r where r, n, s are integers and 0 ≤ s < n.
5. gsn + r = e
6. ${\displaystyle [g^{n}]^{s}\ast g^{r}=e}$

By definition of n = o(g)

7. gr = e

As n is the least that makes gn = e and 0 ≤ r < n.

8. r = 0

Lemma: Let ${\displaystyle k,m\in \mathbb {Z} }$.
${\displaystyle g^{k}=g^{m}}$ if and only if ${\displaystyle k\equiv m(\,{\bmod {\ }}o(g))}$.
Proof: Let ${\displaystyle n=o(g)}$.
${\displaystyle g^{k}=g^{m}}$ if and only if ${\displaystyle g^{k-m}=e}$.
By Euclidean division: ${\displaystyle k-m=qn+r}$, some integers ${\displaystyle q,r}$ with ${\displaystyle 0\leq r.
We have ${\displaystyle g^{r}=g^{k-m-qn}=g^{k-m}{(g^{n})}^{-q}=g^{k-m}}$, hence ${\displaystyle g^{r}=e}$ if and only if ${\displaystyle g^{k}=g^{m}}$.
But ${\displaystyle g^{r}=e}$ if and only if ${\displaystyle r=0}$ (i.e. if and only if ${\displaystyle n\mid k-n}$),
since, by definition, ${\displaystyle n=o(g)}$ is the least positive integer satisfying ${\displaystyle g^{n}=e}$.
Hence the result. ${\displaystyle \square }$

By definition: ${\displaystyle \langle g\rangle =\{g^{r}:r\in \mathbb {Z} \}}$.
Therefore, ${\displaystyle g,g^{2},\ldots ,g^{n}}$ (where ${\displaystyle n=o(g)}$) all lie in ${\displaystyle \langle g\rangle }$ – furthermore, by lemma above, these are pairwise distinct.
Finally, any element of the form ${\displaystyle g^{r}}$, ${\displaystyle r\in \mathbb {Z} }$ equals one of ${\displaystyle g,g^{2},\ldots ,g^{n}}$ (again by lemma).
We conclude that ${\displaystyle g,g^{2},\ldots ,g^{n}}$ are precisely the elements of ${\displaystyle \langle g\rangle }$,
so ${\displaystyle o(\langle g\rangle )=n=o(g)}$, as required.
- Q.E.D. -