(related to Proposition: Group Homomorphisms with Cyclic Groups)
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.
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, \(G=\langle f(1) \rangle\).
