SECTION 4.2 Basics of Group Theory 227
o(1) = 1, o(2) = 3, o(3) = 6, o(4) = 3, o(5) = 6, o(6) = 2.Note that if the elementghas ordern, then{gk|k∈Z} = {e, g, g^2 ,···,gn−^1 }and all of the elements of{e, g, g^2 ,···,gn−^1 }are distinct. To see
this, note that when we divide any integerk byn we may produce a
quotientq and a remainderr, where 0 ≤ r ≤ n−1. In other words
we may expressk =qn+r, which implies thatgk =gqn+r =gqngr =
(gn)qgr=egr=gr.Therefore we already conclude that{gk|k∈Z} =
{e, g, g^2 ,···,gn−^1 }. Next, ife, g, g^2 ,···,gn−^1 aren’t all distinct, then
there must exist integersk < m, 0 ≤k < m≤n−1 such thatgk=gm.
But thene= gmg−k =gm−k. But clearly 0 < m−k ≤n−1 which
contradicts the definition of the order ofg. This proves our assertion.
Note that in general, o(g) = 1 precisely when g = e, the identity
element of G. Also, ifG is a finite group with n elements, and if G
has an elementgof ordern, thenGis cyclic andgis a generator ofG
(Exercise 4).
Exercises
- The two groups you computed in Exercise 1 of Subsection 4.2.4
both have order 4: one is cyclic and one is not. Which one is
cyclic? What are the generators of this group? - LetGbe a group and letgbe an element of finite ordern. Show
that ifgm=ethenmmust be a multiple ofn, i.e.,n|m. - Assume thatGis a group and thatg∈Gis an element of finite
ordern. Assume thatk is a positive integer which isrelatively
primeton(see page 60). Show that the elementgkalso has order
n.