Cyclic Groups

Cyclic Groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 6 December 2010

Definition. A cyclic group is a group G that contains an element gG such that the group generated by g is G, g=G .

The following facts follow from the definition
  1. If G is cyclic with generator g then all elements of g are of the form gk= ( ggg k   factors or g-k= g-1 g-1 g-1 k   factors for some nonnegative integer k.
  2. If G is cyclic with generator g and G is finite and G=n then G= 1, g, g2, , gn-1 .
  3. If G is cyclic then G is abelian since gigj= gi+j for all i,j.
  4. If G is cyclic then all subgroups of G are normal since G is abelian.

HW: Let G be a group of order p where p is prime. Show that G is cyclic.

The integers,

Definition. The group of integers is the set = , -2, -1, 0, 1, 2, with the operation of addition.

HW: Show that is an abelian group.

HW: Show that both the element 1 and the element -1 generate .

HW: Show that is a cyclic group.

HW: Show that every element of is in a conjugacy class by itself.

  1. Let H be a subset of the integers . Then H is a subgroup of if and only if H=m for some nonnegative integer m.
  2. Let m and n be positive integers. Then mn if and only if n divides m.
  3. Let n be a positive integer. Then the quotient group /n/n.

HW: Show that every subgroup of is a normal subgroup of .

Example. The subgroup 5 of the integers consists of all multiples of 5. 5= ,-10, -5,0,5, 10, The subgroup 15 is contained in the subgroup 5. 5= ,-10, -5,0,5, 10, 15= ,-30, -15,0, 15,30, . The sets 0+5= 5+5= ,-10, -5,0,5, 10, =5, 1+5= -4+5= -9+5= ,-9, -4,1,6, 11,16, , 2+5= 32+5= -23+5= ,-13, -8,-3,2, 7,12,17, 22,27,32, , 3+5= -7+5 =8+5 = ,-7, -2,3, 8,13, , 4+5= 404+5= -236+5= ,-6,-1, 4,9,14, . are all cosets of the subgroup 5 in the group . In fact /5= 0+5, 1+5, 2+5, 3+5, 4+5 is the set of cosets of 5 in . As a group /55.

Every homomorphism from to is of the form ϕm: n mn, for some positive integer m.

HW: show that kerϕm= if m=0.

HW: Show that ϕm is injective if m0.

HW: Show that ϕm is bijective if and only m=1 or m=-1.

HW: Show that ϕ1 is the identity mapping.

HW: Show that the automorphism group of , Aut= ϕ1,ϕ-1 2

HW: Show that the inner automorphisms of are In= ϕ1

The group of integers is isomorphic to the free group on one generator.

The finite cyclic groups Cnμn /n, n1

Definition. The cyclic group of order n, Cn is the set 1, g, g2, , gn-1 with the operation given by gigj= gi+jmodn.

There are other representations of the group Cn which are useful.

  1. Let μn be the group given by μn= 1, ξ, ξ2, , ξn-1 where ξ=e 2πin , with the operation of multiplication of complex numbers. In the complex plane the elements of μn all lie on the circle S1= z z=1 .

    1 ξ5 ξ4 ξ3 ξ2 ξ

  2. Let /n be the group given by /n= 0 , 1 ,, n-1 with operation given by i + j = i+jmodn . This operation is called addition modulo n.

HW: Show that the group homomorphism φ:Cnμn given by φgi=ξi is an isomorphism.

HW: Show that the group homomorphism ϕ:Cn/n given by ϕgi= i is an isomorphism.

  1. The subgropus of the cyclic group Cn are the subgroups generated by the elements gm, gm, 0mn-1.
  2. Let 0mn-1 and let d=gcdm,n . Then gm= gd and gd =n/d.
  3. Let 0m,kn-1. Then gn gk if and only if gcdk,n divides gcdm,n.
  4. Let 0dn and suppose that d divides n. Then Cn/ gd Cd/n .

Example: The subgroup lattice of the group C12 is given by

Orders Inclusions 12 6 4 3 2 1 C12= g g2= 1, g2, g4, g6, g8, g10 g3= 1, g3, g6, g9 g4= 1, g4, g8 g6= 1, g6 1

The set of cosets of C12/g3 = H,gH,g2H where H= 1, g3, g6, g9 , gH= g, g4, g7, g10 , and g2H= g2, g5, g8, g11 .

Let × = 0 with the operation of multiplication of comples numbers and let n be a positive integer. Every homomorphism from Cn × is of the from ϕk: Cn × g ξk , where ξ= e 2πin .

  1. The cyclic group Cn= 1, g, g2, , gn-1 of order n is generated by the element g and g satisfies gn=1.
  2. The cyclic group Cn has a presentation by generators and relations of the form Cn= x xn=1 .

Let S be a circular necklace with n equally spaced beads b0, b1, , bn-1, numbered counterclockwise around S.

  1. There is an action of the cyclic group Cn on the necklace S such that g acts by rotating S counterclockwise by an angle of 2π/n.
  2. This action has one orbit, Cnb0= b0, b1, , bn-1 and the stabiliser of each bead is the group 1.

b0 b5 b4 b3 b2 b1 g

References

[Bou] N. Bourbaki, Groupes et Alg├Ębres de Lie, Masson, Paris, 1990.

[GW1] F. Goodman and H. Wenzl, The Temperly-Lieb algebra at roots of unity, Pacific J. Math. 161 (1993), 307-334. MR1242201 (95c:16020)

page history