# Well-defined Homomorphism

1. Sep 20, 2016

### Bashyboy

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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...

2. Sep 20, 2016

### Bashyboy

Note, $x$ is the generator of $Z_{36}$.

3. Sep 20, 2016

### pasmith

You haven't specified the codomain of $f_k$. Is it $\mathbb{Z} / 36 \mathbb{Z}$ as your attempt suggests?

4. Sep 20, 2016

### Bashyboy

Yes, I am sorry. It is actually $Z_{36}$, which is of course isomorphic to it.

5. Sep 20, 2016

### pasmith

You don't need to prove two directions separately; you can do both at once.

$f_k$ is well-defined if and only if $f_k(\overline b) = f_k(\overline c)$ whenever $b$ and $c$ are in the same equivalence class. Thus you need $x^{bk} = x^{(b+ 48q)k} = x^{bk}x^{48qk}$ for each integer $q$.

Your hypothesis is that well-definedness of $f_k$ is governed by whether 36 divides 48k. So set $48k = 36p + r$ for integers $p \in \mathbb{Z}$ and $r \in \{0, 1, \dots, 35\}$. For which values of $r$ can you satisfy the above condition for all $q$?

6. Sep 20, 2016

### Bashyboy

What is $q$? Should it be $p$? The only $r$ that would satisfy the equation is $r=0$, right?

Last edited: Sep 20, 2016
7. Sep 20, 2016

### Bashyboy

Sorry I misread what you wrote. I was able to work problem and it agrees with what you suggested. THANKS!