# 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:

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

# Proof

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

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

By diagram,

0. $g^{o(\langle g\rangle )}=e=g^{o(g)}$ .
1. Let n = o(g), and m = $o(\langle g\rangle )$
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. $[g^{n}]^{s}\ast g^{r}=e$

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

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