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: = # elements in o(g) = Proof thumb|200px| 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. -