CYCLIC GROUPS

A group G is called a cyclic group if ∃ a ∈ G such that each element of G can be written as an integral power of a  i.e.,  if b ∈ G, then ∃ a ∈ G such that b = an for some integer n.
a is then called a generator of G.

Note: (i) If G is a cyclic group generated by a, we write it as G = <a>.
           (ii) If G is a group under addition, then G is called a cyclic group if each element of G is an integral multiple of a  i.e.,   if b ∈ G, then b = na for some integer n.
         (iii) A cyclic group is also called a Monic group.
         (iv) If G = <a> be a cyclic group of order n, then
                     G = {e, a, a²,..., a^n-1}    i.e.,  O(G) = O(a) = n.
         (v) If a is a generator of a cyclic group G, then a^-1 is also a generator of G, for, any 
x ∈ G, we have x = an , also x = (a^-1)^-n , where n, -n ∈ ℤ.

Example: Give an example of a group having all proper subgroups as cyclic but the group itself is not cyclic.
Sol. Consider the group
                 S₃ = {i, (12), (13), (23), (123), (132)}
the symmetric group on three numbers 1, 2 and 3. Then
       All the proper subgroups of S₃ are
                H₁ = {i, (12)} = <(12)>
                H₂ = {i, (13)} = <(13)>
                H₃ = {i, (23)} = <(23)>
                H₄ = {i, (123), (132)} = <(123)>
Which are cyclic, but the group S₃ is not cyclic as there does not exist an element  a ∈ G   s.t.  O(a) = 6.

Theorem: An infinite cyclic group has precisely two generators.
Proof. Let G = <a> be an infinite cyclic group.
Now, any element x ∈ G can be written as
                      x = a^m
also               x = (a^-1)^-m , m ∈ I.
∴   G has two generators a and a^-1.
Let b ∈ G be any other generator of G.
i.e., G = <b>.
Now a ∈ G    and    G = <b>
⇒ a = bⁿ    where   n ∈ I.
⇒ b = a^m   where   m ∈ I.
∴  a = bⁿ = (a^m)ⁿ = a^mn.
⇒ a^mn-1 = e                                                  [By cancellation law]
⇒ O(a) = mn - 1, which is finite.
But G is an infinite cyclic group.
∴ above result holds if mn - 1 = 0   i.e.,  mn = 1
⇒   m = (1/n)   but    m, n ∈ I.
⇒   m = n = ± 1
∴   b = a¹   or   a^-1.
Hence infinite cyclic group has precisely two generators.

Theorem: Every finite group of composite order pessesses proper subgroup.
Proof: Let G be a finite group of composite order mn.
i.e., O(G) = mn,     where   m > 1, n >1.

Case I. When G is cyclic.
    Let G = <a>     where O(a) = O(G) = mn
    ⇒ a^mn = e
    ⇒ (a^m)ⁿ = e.
    Let H be a cyclic group generated by a^m
   i.e.,   H = <a^m> = {a^m , a^2m , a^3m , ... , a^mn = e}
   ∴  O(H) = n.
   Also, since 2 ≤ n ≤ mn.
   Thus H is a proper subgroup of G.

Case II. When G is not cyclic.
    Since G is not cyclic.
    Therefore G has no element of order mn.
    Moreover order of each element of G is less than mn.
    ∴ ∃ an element a of G such that O(a) = k, 
    where 2 ≤ n ≤ mn.
    Let H = <a> be a cyclic group generated by a.
    Then   O(H) = O(a) = k < mn.
    Hence H is a proper subgroup of G.

Properties of Cosets

If H is a subgroup of a group G. then

Property I.       (i) Ha = H    iff   a ∈ H.
                            (ii) aH = H    iff   a ∈ H.

Proof: (i) We prove that Ha = H  iff a ∈ H.
Firstly, suppose that Ha = H                                                 ... (1)
Since H is a subgroup of G, so e ∈ H, where e is the identity element of H.
∴ ea ∈ Ha
⇒ a ∈ Ha
⇒ a ∈ H                                                                          (From (1))
∴ Ha = H      ⇒ a ∈ H.

Conversely: Suppose that a ∈ H.
We shall prove that  Ha = H.
Let x ∈ Ha be an arbitrary element.
∴ x = ha for some h ∈ H.
∴ h, a ∈ H
⇒ ha ∈ H, since H is a subgroup of G.
⇒ x ∈ H
∴ x ∈ Ha
⇒ x ∈ H
⇒ Ha ⊆ H.                                                                               ... (2)
Now let x ∈ H.
Since a also belongs to H and H is a subgroup.
∴ xa^-1 ∈ H
⇒ (xa^-1)a ∈ Ha
⇒ x(a^-1a) ∈ Ha
⇒ xe ∈ Ha
⇒ x ∈ Ha
∴ x ∈ H
⇒ x ∈ Ha
⇒ H ⊆ Ha.                                                                               ... (3)
From (2) and (3) we get Ha = H.

(ii)  Its proof is similar to (i).

Property II. (i)  Ha = Hb    iff  ab^-1 ∈ H
                         (ii) aH = bH    iff  a^-1b ∈ H

Proof: (i) We prove that Ha = Hb    iff ab^-1 ∈ H.
Firstly, let  Ha = Hb.
Since H is a subgroup of G, so e ∈ H.
∴ ea ∈ Ha          i.e., a ∈ Ha
⇒ a ∈ Hb,          since       Ha = Hb
⇒ a = hb             for some    h ∈ H
⇒ ab^-1 = (hb)b^-1 = h (bb)^-1 = he = h ∈ H
∴ ab^-1 ∈ H.

Conversely: Let ab^-1 ∈ H
We shall prove that Ha =Hb.
Since ab^-1 ∈ H, so ab^-1 = h for some h ∈ H
⇒ (ab^-1)b = hb
⇒ a(b^-1b) = hb
⇒ ae = hb
⇒ a = hb
∴ Ha = H(hb)
           = (Hh)b
           = Hb,     since     h ∈ H,   so Hh =H.

(ii) Its proof is similar to that of (i).

Property III. Any two right (or left) cosets are either disjoint or identical.

Proof: Let H be a subgroup of G.
Let Ha and Hb be two right cosets of H of G, so that a, b ∈ G
We shall prove that either Ha =Hb        or          Ha ∩ Hb = ∅
If  Ha ∩ Hb = ∅, then we have nothing to prove.
So, let Ha ∩ Hb ≠ ∅.
In this case we shall prove that Ha =Hb.
Since Ha ∩ Hb ≠ ∅,       so ∃ at least one x ∈ Ha ∩ Hb
∴ x ∈ Ha         and               x ∈ Hb
⇒ x = h₁a        for some     h₁ ∈ H        and
    x = h₂a        for some     h₂ ∈ H
∴            h₁a = h₂b
⇒ h₁^-1 (h₁a) = h₁^-1 (h₂b)
⇒ (h₁^-1 h₁)a = (h₁^-1 h₂b)b
⇒ ea = h₃b      where h₃ = h₁^-1 h₂ ∈ H.
⇒ a = h₃b
⇒ Ha = H(h₃b)
           = (Hh₃)b
           = Hb     since    h₃ ∈ H,    so   Hh₃ = H
∴ Ha = Hb.
∴ If      Ha ∩ Hb ≠ ∅, then  Ha = Hb.
So, either  Ha ∩ Hb = ∅  or Ha = Hb.

Property IV. The group G is equal to the union of all right cosets of H in G.

Proof: Let e, a, b, c, ... be all the elements of G.
∴ He = H, Ha, Hb, Hc, ... are all the right cosets of H in G.
We shall prove that G = H ∪ Ha ∪ Hb ∪ Hc ∪ ... 
Let x ∈ G be any element.
∴ Hx is a right coset of H in G.
Since H is a subgroup of G, so e ∈ G, where e is the identity element of G.
∴ ex ∈ Hx                     i.e.,   x ∈ Hx
⇒ x ∈ H ∪ Ha ∪ Hb ∪ Hc ∪ ...  ∪ Hx ∪ ... 
∴ G ⊆ H ∪ Ha ∪ Hb ∪ Hc ∪ ...                                           ... (1)

Conversely: Let Ha be any right coset of H in G, where a ∈ G.
let x ∈ Ha.
∴ x = ha for some h ∈ H.
Since h ∈ H
∴ h ∈ G
Also a ∈ G.
⇒ ha ∈ G.
⇒ x ∈ G
∴ x ∈ Ha         ⇒ x ∈ G
⇒ Ha ⊆ G
∴ ∪ Ha ⊆ G   ∀ a ∈ G
⇒ H ∪ Ha ∪ Hb ∪ Hc ∪ ... ⊆ G                                          ... (2)
From (1) and (2), we get
G = H ∪ Ha ∪ Hb ∪ Hc ∪ ...  

Property V. There is one to one correspondence between any two right cosets of H in G.

Proof: Let Ha, Hb be two right cosets of H in G, where a, b ∈ G.
Define a map f : Ha → Hb     by
                       f (ha) = hb,    ∀ ha ∈ Ha.
f is one-one. Let x, y ∈ Ha such that f (x) = f (y)
since     x, y ∈ Ha
∴      x = h₁a           and y = h₂b    for some   h_1, h₂ ∈ H.
∴  f (x) = f (y)
⇒ f (h₁a) = f (h₂b)
⇒ h₁a = h₂b
⇒ h₁ = h₂                         by right cancellation law in the group G.
⇒ h₁a = h₂a
⇒ x = y
⇒ f is one-one.
f is onto. Let y ∈ Hb
∴      y = hb     for some   h ∈ H.
Take x = ha.
Since h ∈ H,  so ha ∈ Ha
⇒ x ∈ Ha,       where    x = ha ∈ Ha
∴ f (x) = f (ha) = hb = y
∴ f is onto.
∴ f : Ha → Hb is one-one and onto.
∴ Ha, Hb are in one-one correspondence.

Property VI. There is one-one correspondence between the set of left cosets of H in G and the set of right cosets of H in G.

Proof: Let L and M be respectively the set of left cosets and right c  osets of H and G.
∴ L = {aH : a ∈ G} and M = {Ha : a ∈ G}.
Define a map  f : L → M by
                        f (aH) = Ha^-1 ,    ∀ a ∈ G.
If a ∈ G, then a^-1 ∈ G and hence Ha^-1 ∈ M.
∴ f is a map from L to M.
We now prove that f is well defined.
Let a, b ∈ G such that aH = bH.
                                ⇔ a^-1b ∈ H.
                                ⇔ Ha^-1b = H.
                                ⇔ (Ha^-1b)b^-1  = Hb^-1
                                ⇔ H(a^-1b)b^-1  = Hb^-1
                                ⇔ Ha^-1(bb^-1) = Hb^-1
                                ⇔ Ha^-1e = Hb^-1
                                ⇔ Ha^-1 = Hb^-1
                                           ⇔  f (aH) = f (bH)
∴ f is well-defined.
The reverse steps shows that f is one-one.
We finally prove that f is onto.
Let Ha ∈ M be arbitrarily.
∴ a ∈ G.
⇒ a^-1H ∈ L. such that       f (a^-1H) = H(a^-1)^-1 =Ha.
∴ f is onto.
∴ The mapping f : L → M is in one-one and onto.
⇒ The set of left cosets of H in G and the set of right cosets of H in G are in one-one correspondence.