This is a continuation of this problem. I have part of a solution, but am unsure it is correct. Allow me to write the problem again:
Determine the integers $z$ for which $f : \mathbb{Z}/48 \mathbb{Z} \rightarrow Z_{36}$ with $f_z(\overline{1})= x^z$, where $x$ is the generator of $Z_{36}$, extends to a well-defined homomorphism.
First of all, it seems that proving this problem requires I prove that
$f_z$ is a well-defined homomorphism iff $f_z(\overline{b}) = x^{bz}$ and $z$ is such that $36$ divides $48z$.
Being the overly cautious person that I am, I searched the internet for a solution. I found this. This is somewhat embarrassing, but how does $36 | 48z$ imply $3|z$? I can only get $3|4z$.
My second question pertains to proving the above conjecture. I was able to show that $f_z(\overline{b}) = x^{bz}$ and $z$ is such that $36$ divides $48z$ implies $f_z$ is a well-defined homomorphism easily enough. However, I am having trouble with the second direction. Here is my attempt:
Suppose $f_z$ is a well-defined homomorphism with $f_z(\overline{1}) = x^z$. Let $\overline{b} \in \mathbb{Z}/46\mathbb{Z}$ be arbitrary. Then by the homomorphism property, we have
$f_z(\overline{b}) = f_z(\overline{1} + \dots + \overline{1}) = f_z(\overline{1}) \dots f_z(\overline{1}) = x^z \dots x^z = x^{bz}$.
Now, we want to show $z$ satisfies $36|48z$. Suppose the contrary, and let $\overline{b} = \overline{b'}$, which implies $b' = b +48k$, for some $k \in \mathbb{Z}$. Because $f_z$ is well-defined, we have
$f_z(\overline{b}) = f_z(\overline{b'})$
$x^{bz} = x^{b'z}$
$x^{bz} = x^{(b+48k)z}$
$x^{bz} = x^{bz} (x^{48z})^k$
$e = (x^{48z})^k$,
where we have assumed $x^{48z} \neq e$.
I feel that there is some contradiction lurking in this equations, but I cannot see it.