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.