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:

$o(⟨g⟩)$ = # elements in $⟨g⟩$

o(g) = $o(⟨g⟩)$

Proof

Since $⟨g⟩$ is cyclic, and has $o(⟨g⟩)$ elements.

$g^{o(⟨g⟩)}=e$

By diagram,

0. $g^{o(⟨g⟩)}=e=g^{o(g)}$.

1. Let n = o(g), and m = $o(⟨g⟩)$

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 $k\equiv m(modo(g))$.
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 $0\le r<n$.
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. $◻$

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