Determining when a mapping is an isomporphism

[Bashyboy]

Homework Statement

Suppose that $\phi : \mathbb{Z}_n \rightarrow \mathbb{Z}_n$, where the rule is $\phi([a]_n) = [ka]_n$. Formulate and prove a conjecture that gives necessary and sufficient conditions on the positive integers $k$ and $n$ which would guarantee that $\phi$ is an isomorphism.

Homework Equations

The Attempt at a Solution

I have already shown that is function is a homomorphism. After having worked with a few examples, I found that if either $n$ or $k$ can divide the other number, then $\phi$ would not be an isomorphism. When they say that they want me to give necessary and sufficient conditions, does that mean they want to prove that

"$\phi$ is an isomorphism iff $k$ or $n$ do not divide each other?"

The only thing that troubles me is that there may be more restrictions upon $k$.

 

[pasmith]

Consider the case n = 10 and k = 15. Neither divides the other, and yet [a]_{10} \mapsto [15a]_{10} = [5a]_{10} is not invertible, because [5]_{10} does not have a multiplicative inverse.

Consider the case k = 1 and n \geq 1. Then clearly 1 divides n, but [a]_n \mapsto [a]_n is trivially invertible.

You need to think more carefully about the circumstances in which \phi is invertible. To start with, you might consider what \phi^{-1} can possibly be.

 

[Bashyboy]

because $[5]_{10}$ does not have a multiplicative inverse.

Actually, the group operation is not multiplication, but the addition operator $\oplus$, which is defined as

Edit: For someone reason, latex on physicsforums has been very problematic for me. I had the break the following up into three segments of latex code.

$[a]_n$ $\oplus$ $[a+b]_n$

 

[pasmith]

Actually, the group operation is not multiplication, but the addition operator $\oplus$, which is defined as

The reason that [a]_{10} \mapsto [5]_{10}[a]_{10} is not an invertible additive homomorphism is that [5]_{10} has no multiplicative inverse.

The operation which takes [a]_n to \phi([a]_n) is not addition of k mod n, but multiplication by k mod n, and that is the operation you need to invert.

 

[Bashyboy]

I don't quite understand. We have not even discussed multiplication with respect to the group $\mathbb{Z}_n$, only the addition operator $\oplus$. Why is $[a]_{10}$ being mapped to $[5]_{10} [a]_{10}$?

 

[pasmith]

I don't quite understand. We have not even discussed multiplication with respect to the group $\mathbb{Z}_n$, only the addition operator $\oplus$. Why is $[a]_{10}$ being mapped to $[5]_{10} [a]_{10}$?

This is the homomorphism you are dealing with:

Homework Statement

Suppose that $\phi : \mathbb{Z}_n \rightarrow \mathbb{Z}_n$, where the rule is $\phi([a]_n) = [ka]_n$.

Note that [ka]_n = [k]_n[a]_n, because (k + pn)(a + qn) = ka + n(kq + pa + pqn).

 

[Bashyboy]

$(k+pn)(a+qn)=ka+n(kq+pa+pqn)$

I don't understand what this demonstrates.

This is my problem: at this point in my algebra course, it does not even make sense to write $[k]_n [a]_n$, unless you wish to define this as $[k]_n [a]_n := [k]_n \oplus [a]_n$, which I do not believe is your import.

So, for all intents and purposes, I do not know what it means to talk about multiplication with respect to $\mathbb{Z}_n$, nor do I know about multiplicative inverses. My point is, I do not believe the problem requires these concepts, otherwise it would have introduced them.

 

[Bashyboy]

Okay, this problem is still troubling my mind. Could someone else possibly take a look at it? My professor said I was on the right track with both of these conjectures:

$\phi$ is an isomorphism iff $k$ is a prime number such that $k \nmid n$

and

$\phi$ is an isomorphism iff $gcd(k,n) =1$.

However, I ran into difficulties with both proofs.