- #1

Bashyboy

- 1,421

- 5

## Homework Statement

Determine the integers ##k## for which ##f_k : \mathbb{Z}/ 48 \mathbb{Z}## with ##f_k (\overline{1}) = x^k## extends to a well-defined homomorphism.

## Homework Equations

## The Attempt at a Solution

My claim is that ##f_k## extends to a well-defined homomorphism iff ##f(\overline{b}) = x^{bk}## for every ##\overline{b} \in \mathbb{Z}/48\mathbb{Z}## and ##k## is such that ##36 divides ##48k# (which is equivalent to ##3## dividing ##k##). I was able to prove that if ##f_k(\overline{b})=x^{kb}## and ##k## is such that ##36## divides ##48k##, then ##f## is a well-defined hommorphism. However, I am having difficulty with the other direction.

Suppose that ##f## is a well-defined homomorphism. Then

##

f_k(\overline{b}) = f_k(\overline{1} + \dots + \overline{1}) = f_k(\overline{1}) \dots f_k(\overline{1}) = x^k \dots x^k = x^{kz}##

Now we want to show ##k## is such that ##36## divides ##48k##. Suppose the contrary, and suppose ##\overline{b} = \overline{b'}##, which implies ##b' = b + 48m## Then by the well-defined property,

##f_k(\overline{b}) = f_k(\overline{b'})##

##x^{bk} = x^{b'k}##

##x^{bk} = x^{(b+48m)k}##

##x^{bk} = x^{bk} (x^{48k})^m##

##e = (x^{48k})^m##

This is where I get stuck...