Equivalence of definitions of cyclic group

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This article gives a proof/explanation of the equivalence of multiple definitions for the term cyclic group
View a complete list of pages giving proofs of equivalence of definitions

The definitions that we have to prove as equivalent

Definition in terms of modular arithmetic

A group is said to be cyclic (sometimes, monogenic or monogenous) if it is either isomorphic to the group of integers or to the group of integers modulo n for some positive integer n.

Definition in terms of generating sets

A group is termed cyclic (sometimes, monogenic or monogenous) if it has a generating set of size 1.

Definition as a quotient

A group is termed cyclic if it is a quotient of the group \mathbb{Z}, in other words, there exists a surjective homomorphism from \mathbb{Z} to the group.

Proof of equivalence

From modular arithmetic to generating sets

This is direct: \mathbb{Z} is generated by the element 1 \in \mathbb{Z}, and \mathbb{Z}/n\mathbb{Z} is generated by the element 1.

From generating sets to modular arithmetic

Given: A group G with a generating set \{ g \}

To prove: G is isomorphic either to \mathbb{Z} (the group of integers) or to \mathbb{Z}/n\mathbb{Z} (the group of integers modulo n)

Proof: We consider two cases.

Case 1: g has finite order. Thus, there exists a minimal positive integer n such that g^n is the identity element. Consider now the map \varphi: \mathbb{Z}/n\mathbb{Z} \to G that sends a to the element g^a. We want to prove that \varphi is an isomorphism.

We first show that \varphi(a + b) = \varphi(a)\varphi(b). For this, observe that if a and b add up to less than n as integers, then g^{a+b} = g^ag^b by definition. If the sum of a and b as integers is at least n, then \varphi(a + b) = g^{a+b-n} =g^ag^bg^{-n} = g^ag^b (since g^{-n} is the identity element).

Similarly, \varphi(0) = g^0 = e by definition, and \varphi(-a) = \varphi(a)^{-1}, again because g^n = e.

Surjectivity: Since g generates G, every element of G can be written as a power of g, say g^m for some integer m. Writing m = nq + r where q,r are integers and r \in \{ 0,1,2,\dots,n-1 \}, we get that g^m = g^r = \varphi(r). Thus, \varphi is surjective.

Injectivity: Finally, if \varphi(a) = \varphi(b) with a < b both in \{ 0,1,2,\dots,n-1 \}, then g^{b - a} = e, contradicting the assumption that g has order n.

Thus, \varphi is an isomorphism of groups.

Case 2: g does not have finite order. In that case, consider the map \varphi:\mathbb{Z} \to G that sends n to g^n.

Clearly, by definition, \varphi(a + b) = \varphi(a)\varphi(b), \varphi(-a) = \varphi(a)^{-1}, and \varphi(0) = e.

Surjectivity: Since g generates G, every element of G can be written as g^n for some integer n.

Injectivity: If \varphi(a) = \varphi(b) for a < b, then g^{b - a} = e, contradicting the assumption that g does not have finite order.