Building a proper homomorphism between groups. Suppose I have a cyclic group $G$ of order $6$. I want to show that it is isomorphic to $\Bbb {Z}_6$. So $G=\{e,g^2,g^3,g^4,g^5\}=\langle g\rangle$. Can I build a homomorphism $f:G \to \Bbb{Z}_6$ that way?
$f(x)=f(g^m)=mf(g)=m$ where $f(g)=1$ and $f(e)=0$. It is problematic because I might get that $f(g^6)=f(e)=6=0$... 6 $6$ is somehow $0$, but is it constructive and plausible?? Would appreciate your reply. 

Yes the homomorphism $g\mapsto 1 \mod 6$ is the correct homomorphism. Check the condition: $f(g^ig^j) = f(g^{i+j}) = i+j \mod 6 = (i \mod 6) + (j \mod 6) = f(g^i)+f(g^j)$.
