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 g∈G 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= ( g⋅g⋯g ⏟ 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∈ℤ.
  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, ℤ

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

HW: Show that ℤ is an abelian group.

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

HW: Show that ℤ is a cyclic group.

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

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

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

Example. The subgroup 5ℤ of the integers ℤ consists of all multiples of 5. 5ℤ= …,-10, -5,0,5, 10,… The subgroup 15ℤ is contained in the subgroup 5ℤ. 5ℤ= …,-10, -5,0,5, 10,… ⊇ 15ℤ= …,-30, -15,0, 15,30,… . The sets 0+5ℤ= 5+5ℤ= …,-10, -5,0,5, 10,… =5ℤ, 1+5ℤ= -4+5ℤ= -9+5ℤ= …,-9, -4,1,6, 11,16,… , 2+5ℤ= 32+5ℤ= -23+5ℤ= …,-13, -8,-3,2, 7,12,17, 22,27,32, … , 3+5ℤ= -7+5ℤ =8+5ℤ = …,-7, -2,3, 8,13, … , 4+5ℤ= 404+5ℤ= -236+5ℤ= …,-6,-1, 4,9,14,… . are all cosets of the subgroup 5ℤ in the group ℤ. In fact ℤ/5ℤ= 0+5ℤ, 1+5ℤ, 2+5ℤ, 3+5ℤ, 4+5ℤ is the set of cosets of 5ℤ in ℤ. As a group ℤ/5ℤ≅ℤ5.

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

HW: show that ker ϕm=ℤ if m=0.

HW: Show that ϕm is injective if m≠0.

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 ℤ, Autℤ= ϕ1,ϕ-1 ≅ℤ2

HW: Show that the inner automorphisms of ℤ are Inℤ= ϕ1

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

The finite cyclic groups Cn≅μn≅ ℤ/nℤ, n≥1

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

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 ∈ℂ, with the operation of multiplication of complex numbers. In the complex plane the elements of μn all lie on the circle S1= z∈ℂ∣ z=1 .

    1 ξ5 ξ4 ξ3 ξ2 ξ

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

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

HW: Show that the group homomorphism ϕ:Cn→ℤ/nℤ given by ϕgi= i ‾ is an isomorphism.

  1. The subgropus of the cyclic group Cn are the subgroups generated by the elements gm, ⟨gm⟩, 0≤m≤n-1.
  2. Let 0≤m≤n-1 and let d=gcdm,n . Then ⟨gm⟩= ⟨gd⟩ and ⟨gd⟩ =n/d.
  3. Let 0≤m,k≤n-1. Then ⟨gn⟩⊆ ⟨gk⟩ if and only if gcdk,n divides gcdm,n.
  4. Let 0≤d≤n 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 𝒵12= ⟨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 ℂ× =ℂ∖ 0 with the operation of multiplication of comples numbers and let n be a positive integer. Every homomorphism from Cn→ ℂ× is of the from ϕk: Cn → ℂ× 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 Algèbres 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)

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