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April 18th, 2012, 10:14 PM   #1
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Let G be an abelian group of order . Define by , where . Prove if gcd(m,n) = 1 then is an isomorphism.
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April 22nd, 2012, 02:08 AM   #2
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Re: isomorphism

First, prove that is an homomorphism. Afterwards you need to prove is bijective thus on-to-one and onto (surjective).
To prove if is one-to-one (or injective) you have to show:

We know that thus you can write with (note that because of )
There's a theorem which says that if is a finite group then the order (in this case ) is a multiple of the order of every element that means and thus but because , has to be minimal and thus therefore and is on-to-one.

This is just an attempt so I'm not absolutely sure it's entirely correct, but maybe you can do something with my proof.
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July 2nd, 2012, 02:17 AM   #3
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Re: isomorphism

Note that:
(a) for each ;
(b) gcd(m,n)=1 is equivalent to the statement, there exist integers k and l such that km+ln=1;
(c) to show that is a bijection is to show its kernel is trivial and it is surjective.
Since for each it follows that:
1. and so is surjective;
2. If then and and so the kernel of is trivial ( is injective)
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