Abstract Algebra/Group Theory/Subgroup/Cyclic Subgroup/Order of a Cyclic Subgroup

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 = g^m

3. g^{n – m} = e

4. Let n – m = sn + r where r, n, s are integers and 0 ≤ s < n.

5. g^{sn + r} = e

6. $[g^{n}]^{s}\ast g^{r}=e$

By definition of n = o(g)

7. g^r = 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 $0\leq r<n$.
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. -