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. 
 A: It is much better to build the isomorphism in the other direction, as
$$
\varphi([i]) = g^{i},
$$
if $[i]$ is the class of $i$ in $\mathbb{Z}_{6}$. You have to prove it is well-defined, but this is immediate.
Even better, start with the homomorphism
$$
\mathbb{Z} \to G, \qquad i \mapsto g^{i},
$$
and use the first isomorphism theorem.
A: Yes the homomorphism $g\mapsto 1 \mod 6$ is the correct homomorphism. Check the condition: $f(g^ig^j) = f(g^{i+j}) = i+j \mod 6 = (i \mod 6) + (j \mod 6) = f(g^i)+f(g^j)$.
A: When you write $f(g^{m})=mf(g)$ you write this in $\mathbb{Z}/6\mathbb{Z}$ so indeed $6=0$. Your construction completely defines $f$ and as explained in the answer above, it is not difficult to check that $f$ so defined is an isomorphism.
I don't know if you've heard about quotient morphisms. I prefer myself to define the isomorphism between $G$ and $\mathbb{Z}/6\mathbb{Z}$ that way. The application
\begin{align*}
\phi: & \mathbb{Z}\rightarrow G \\
& m \mapsto g^{m}
\end{align*}
is surjective with kernel $6\mathbb{Z}$. So the quotient mophism is bijective and maps $\mathbb{Z}/6\mathbb{Z}$ into $G$.
