Equivalence of definitions of cyclic group

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.

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.