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

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Theorem[edit]

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:

= # elements in
o(g) =

Proof[edit]

Since is cyclic, and has elements.

By diagram,

0. .
1. Let n = o(g), and m =
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.

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 .
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. -