Help with using gcd() and Integers mod n

[daon]

How would I go about proving: For a,b in Z mod n, If gcd(a,n)=1 and gcd(b,n)=1 then gcd(ab,n) = 1.

I'm not sure if this is correct because I never used the fact that a and b are in Z mod n, just that they are in Z, but I gave it a shot:

[informal]
gcd(a,n) = a(x) + n(y), some x,y in Z mod n
gcd(b,n) = b(x') + n(y'), some x',y' in Z mod n

gcd(a,n)*gcd(b,n) = 1*1 = 1 = (ax + ny)(bx' + ny')
1 = abxx' + + axny' + nybx' + nnyy' = ab(xx') + n(axy'+ybx'+nyy')

So now I have: 1 = ab(c) + n(k), for the above c,k

Does this justify me saying that gcd(ab,n) = ab(c) + n(k) = 1?

-Daon

 

[pka]

I am not sure what is meant by mod n here.
Are you dealing with a finite field, in which case n is prime?
Or, are you just talking about multiplication modulo n?
If a is relatively prime to n and b is relatively prime to n the clearly their product must be relatively prime to n in the integers. But I still do not see where the mod n comes into play. Maybe you can post the entire question.

 

[daon]

I'm not sure where it comes in, either. I am dealing with an equivilance class of size n.

The entire question was:

\(\displaystyle \L \text{For a,b } \in \mathbb{Z}_n \text{, if gcd(a,n)=1 and gcd(b,n)=1 then gcd(ab,n)=1.}\)

We had just finished going over \(\displaystyle \mathbb{U}_n\) when this was assigned. I'm not sure if this has anything to do with it.

 

[daon]

If I have: 1 = ab(c) + n(k), is that enough to say gcd(ab,n) = 1?