Question

Theorem 4.27. Suppose G is a finite cyclic group of order n. Then G is isomorphic to Rn if n ≥ 3, S2 if n = 2, and the trivial group if n = 1.

Most of the previous results have involved finite cyclic groups. What about infinite cyclic groups?

Answer 1

Let G be an infinite cyclic group generated by a ,then we have

G = 〈a〉 = {a^k : k∈Z } ,Z is group of integers

Let us define the mapping : m

m :Z→G by m(k)=a^k   

we will now show that m is isomorphism (Z is isomorphic to G)

m(x+y) = a^x+y =a^xa^y =m(x)m(y) m is a homomorphism

As G is cyclic, every element of G is a power of a for some a∈G such that G=〈a〉. ∀x∈G ∃k∈Z st x=a^k .By the definition of m: m(k)= a^k  = x . m is surjective

∀m,n∈Z : m ≠ n ⟹ a^m ≠ aⁿ . m is injective .

m is a homomorphism  G≅(Z,+) .

Yes ,the result is true for cyclic groups of infinite orders .