Infinite homomorphic image of the integers A question asks if there is an infinite homomorphic image of the integers, then it is isomorphic to the integers. I thought about using the isomorphism theorem by stating that 
$\alpha: Z$ $\to$ $\alpha(Z)$ is a homomorphism 
then using the fact that 
$Z$/ker($\alpha$) $\cong$ $\alpha(Z)$. 
I'm not quite sure where to go from here. 
Can we state that $ker(\alpha) = \{1\}$? 
 A: You can use the first isomorphism theorem but remember the kernel is $\{0\}$. This can be shown as follows. Consider $\alpha(0)$ and suppose $n$ is some other element of the kernel. Then $0=\alpha(0)=\alpha(n+(-n))=\alpha(n)+\alpha(-n)=\alpha(-n)$. So every multiple of $n$ is also in the kernel as $\alpha(nk)=\alpha(n)+...+\alpha(n)=k\alpha(n)=k0=0$ (and similarly for $-n$).
But $\Bbb{Z}/n\Bbb{Z}$ is finite. Since $N=n\mathbb{Z}$ is normal in $G=\mathbb{Z}$ and $K=\ker\alpha$ is normal in $\mathbb{Z}$ containing $n\mathbb{Z}$, we have $N\subseteq K\subset G$ and by the third isomorphism theorem $(G/N)/(K/N)\cong G/K$. Since $G/N$ is finite and $K/N$ a subgroup of $G/N$, their quotient is also finite.
A: If one looks closely then we could easily see that even Z is a cyclic group with generators <1> and <-1>, Thus any group which has to be isomorphic to Z must also be cyclic 
In this case: Thus we modify our search as non cyclic infinite groups are ruled out .Thus now we have to prove isomorphism between a group Z to G whose genarator is a. This can be done easily even without using the isomorphic theorums by defining a function
f(K)=a^k and this function can be proven well defined, one one and onto thus proving the homomorphism.
