Proof

"$$\Rightarrow$$"

Let $$(G,\ast)$$ be a cyclic group generated by the element $$g$$. We have to find a surjective group homomorphism from the set $$\mathbb Z$$ of all integers to $$G$$. First, we define for all $$k\in\mathbb Z$$ a map: $f:\begin{cases} \mathbb Z&\mapsto G \\ k&\mapsto g^k \end{cases}$ For any two integers $$k,l\in\mathbb Z$$ we then have

$f(k+l)=g^{k+l}=g^k\ast g^l.$ It follows that $$f$$ is a group homomorphism. Since the sum $$k+l$$ can represent any integer $$j\in\mathbb Z$$ (for a given $$j$$, fix $$k$$ and set $$l:=j-k$$), $$g^j$$ can generate all (even infinitely many) elements of $$G$$. Therefore, $$f$$ is surjective.

"$$\Leftarrow$$"

Let $$f:\mathbb Z\mapsto G$$ be a surjective group homomorphism, i.e.

$f(k+l)=f(k)\ast f(l)\in G,$ for any integers $$k,l\in\mathbb Z$$ with $$im(f)=G$$. We have to show that $$G$$ is cyclic. In particular, we have to prove that there exist an integer $$j$$, for which $$G=\langle f(j) \rangle$$, i.e. for which each element of $$G$$ can be represented as some power of $$f(j)$$. We will show that $$j=1$$ will do the trick. First of all, note that because $$f$$ is a group homomorphism, we can generate all elements $$f(k)\in G$$ with $$k \ge 1$$ as "powers" of $$f(1)$$ as follows:

$\begin{array}{ccl} f(1)&=&f(1)^1\\ f(2)&=&f(1+1)=f(1)\ast f(1)=f(1)^2\\ \vdots\\ f(k)&=&f(\underbrace{1+\ldots+1}_{k\text{ times}})=\underbrace{f(1)\ast\ldots \ast f(1)}_{k\text{ times}}=f(1)^k.\\ \end{array}$

Following the properties of group homomorphism, also $$f(-1)$$ can be generated from $$f(1)$$ by the equation

$f(-1)=f(1)^{-1}.$

Therefore, we can also generate all elements $$f(k)\in G$$ with $$k \le -1$$ as "powers" of $$f(1)$$ as follows:

$\begin{array}{ccl} f(-1)&=&f(1)^{-1}\\ f(-2)&=&f(-1-1)=f(1)^{-1}\ast f(1)^{-1}=f(1)^{-2}\\ \vdots\\ f(-k)&=&f(\underbrace{-1-\ldots-1}_{k\text{ times}})=\underbrace{f(1)^{-1}\ast\ldots \ast f(1)^{-1}}_{k\text{ times}}=f(1)^{-k}.\\ \end{array}$

Finally, again following the properties of group homomorphism, also the identity $$e_G$$ of the group $$G$$ can be generated from $$f(1)$$ by the equation

$e_G=f(0)=f(-1)\ast f(1)=f(1)^{-1}\ast f(1)^1=f(1)^0.$

Because $$f$$ is surjective and we have explored all integers $$k\in\mathbb Z$$, we must have generated all elements of $$G$$ by only the element $$f(1)\in G$$. Thus, $$G=\langle f(1) \rangle$$.

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