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

TheoremEdit

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) =  

ProofEdit

 
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. -