The group homomorphism from Z to Q* Let F be non trivial group homomorphism F: Z -> Q*. Want to prove that either Ker(F)={0} or Ker(F)= 2Z.
Okay here is what i did;
since i know that if there is homomorphism between two groups then there should be and isomorphism T such that, T: Z/Ker(F) -> Image(F).
so i took Z/{0}=Z implies Z is isomorphic to Image(F). and i took Z/2Z={2Z, 1+2Z} is isomorphic to Image(F). and i take Z/3Z to be isomorphic to Image(F). Then since all three are isomorphic to Image(F) then they should be isomorphic to each other i.e. Z/3Z should be isomorphic to Z/2Z. (hence we arrive to a contradiction!).
Is my method correct? If not then i need some directions. And once I'm done answering this question, can i deduce from it that (Q*, x) is not cyclic? Because to show that we need to show isomorphism of Q* with Z..
 A: Your method is not correct. It proves(?) for example that necessarily $\operatorname{Ker}F= \{0\}$.
Here is a proof:
Note that the only finite subgroups of $\mathbb{Q}^*$ are $\{1\},\{1,-1\}$. Also the only subgroups of $\mathbb{Z}$ are $n \mathbb{Z}, \mathbb{Z}$ and $\frac{\mathbb{Z}}{n \mathbb{Z}} $ has $n$ elements. (Why ?)
Now as you correctly said whatever $F$ is, it induce a group isomorphism $T:\frac{\mathbb{Z}}{\operatorname{Ker} F} \to \operatorname{Im}F$.
Since $\operatorname{Im}F$ is a subgroup of $\mathbb{Q}^*$, if $\operatorname{Im}F$ is finite we only have two possibilities for $\operatorname{Im}F$. But $F$ is non  trivial so $\operatorname{Im}F \neq \{1\}$. So we left with only one possibility   $\operatorname{Im}F = \{1,-1\}$. Therefore $\operatorname{Card}(\frac{\mathbb{Z}}{\operatorname{Ker} F})=2$.
Now if $\operatorname{Im}F$ is infinite $\operatorname{Card}(\frac{\mathbb{Z}}{\operatorname{Ker} F})= \infty$. 
This mean $\operatorname{Ker}F= 2 \mathbb{Z}$ or $\operatorname{ker}F = \mathbb{Z}$.
A: A homomorphism on $Z$ is completely determined by the image of 1. So if the homomorphism sends 1 to $q$ in $Q^*$, then it sends an arbitrary integer $n$ to $q^n$. The kernel consists of those $n$ for which $q^n=1$. If $q$ is not equal to plus or minus 1, then $q^n$ will never equal 1, except trivially at $n=0$. If $q=1$, then the homomorphism is trivial. If $q=-1$, then the kernel consists of those $n$ for which $(-1)^n=1$, i.e. the even integers.
