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

< Abstract Algebra | Group Theory | Subgroup

# TheoremEdit

Define Order of an Element *g* of Finite Group G:

- o(
*g*) = the least positive integer n such that*g*^{n}=*e*

Define Order of a Cyclic Subgroup generated by *g*:

- = # elements in

- o(
*g*) =

- o(

# ProofEdit

By diagram,

- 0. .

- 1. Let n = o(g), and m =

- 2. g
^{n}= 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.

By definition of n = o(g)

- 7. g
^{r}=*e*

As n is the least that makes g^{n} = *e* and 0 ≤ r < n.

- 8.
*r*= 0

**Lemma:** Let .

if and only if .
**Proof:** Let .

if and only if .

By Euclidean division: , some integers with .

We have , hence if and only if .

But if and only if (i.e. if and only if ),

since, by definition, is the *least* positive integer satisfying .

Hence the result.

By definition: .

Therefore, (where ) all lie in - furthermore, by lemma above, these are pairwise distinct.

Finally, any element of the form , equals one of (again by lemma).

We conclude that are *precisely* the elements of ,

so , as required.

- Q.E.D. -