Abelian group w/ 1000 elements?

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Abelian group w/ 1000 elements?

My guess is...

Z mod 100 under addition.

I am new to algebra though...any thoughts?

Also, what about multiplication?
 
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Z mod 100 under addition is a group. But it has 100 elements, not 1000. Z mod 100 under multiplication isn't even a group. Why not? Do you just want an example of a group with 1000 elements?
 
Dick said:
Z mod 100 under addition is a group. But it has 100 elements, not 1000. Z mod 100 under multiplication isn't even a group. Why not? Do you just want an example of a group with 1000 elements?

Ahh, Z mod 1000 has a thousand elements and under addition is a group? How am I supposed to check a*(b*c)=(a*b)*c? Also, is it abelian?

Under mult., I don't know how to check either. I checked a small table and it worked...

I feel like any group of order n is isomorphic to Z mod n under addition (and/or mult) but I cannot find this in my book. I remember my prof saying it though.
 
Z mod 1000 is associative because (Z,+) is associative. And it's commutative, since (Z,+) is commutative. Under multiplication 0*a=0 for all a and 10*100=0. Thinks like that are going to pose problems for the inverse property of a group. You can't find the statement that any group of order is isomorphic to Z mod n in your book because it's not true. It isn't even true for abelian groups. It's only true if n is prime.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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