# Gcd(m, n)=1 implies Zmn isomph to Zm x Zn

## Homework Statement

show that if m, n are relatively prime, that is, greatest common divisor of m and n is 1, then $\mathbb{Z} _mn \approx \mathbb{Z} _m \times \mathbb{Z} _n$

## Homework Equations

I need to show that $\theta$ is operation preserving, and I need to show that it is one to one and onto.

## The Attempt at a Solution

For theta, $\theta ([a]_{mn} + _{mn}) = \theta ([a+b]_{mn})=([a+b]_m,[a+b]_n)=$
$([a]_m+_m,[a]_n+_n)=([a]_m,[a]_n)+([b)_m,_n)= \theta ([a]_{mn}) + \theta (_{mn})$
Did I assume anything I shouldn't have there?
I'm going to consult my notes about proving 1-1. going to try the kernel thing.
As for onto, how do I show that?
I'm concerned that I haven't used the fact that m, n are relatively prime.

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I like Serena
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Hi Arcana!

You assumed your definition of theta, which I have to deduce from your work now.
Beyond that your proof of the preservation of the operation is correct.

So what's left is to proof that theta is 1-1, for which you need that m and n are relatively prime.

The definition of theta was given in a previous problem. I'm just tired of typing latex. Guess I need more practice so I'm not so slow at it.
going to consult notes about kernel thingy now for 1-1, but what about onto?

I like Serena
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How many elements does each set have?
Are they the same?

I'm not sure what your kernel thingy is, but if you can say each element is mapped onto a different element, you can deduce "onto" by the fact that each set has the same number of elements.

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difficulty:
theta is 1-1 iff ker(theta)= {e_Z_mn}
$$\theta ([a]_{mn})=(_m,_n) \Rightarrow$$
$$m \mid a \wedge n \mid a \Rightarrow$$
$$a=km \wedge a=jn$$

I know that being relatively prime is the reason a must =mn but I'm not sure why, so I don't know what to say next.

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You have n|a.

So from a=km, you know that n|(km).
Since n does not have any common factors with m, it must be that n|k.

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This means that k=i n.
And in turn that:
a=km=(in)m=i(mn)

What is i? 1? And why does k=in? Sorry I'm being dense here.

I like Serena
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What is i? 1? And why does k=in? Sorry I'm being dense here.
Since n|k, there must be a number i such that k=in.

I like Serena
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Not dense, I know how hard it is to follow other people's reasoning in algebra.
The trick is to set up your own reasoning.