Is Group G Abelian if f(a)=a^-1 is a Homomorphism?

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Homework Statement


Prove that Group G is abelian iff the function f:G to G given by f(a)=a^-1 is a homomorphism

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The Attempt at a Solution



Group G must be communitative for it to be abelian I have no idea where to start to begin to prove this I know that homomorphic means that complete the operation first on a in the group gets mapped to an element that the operation is completed 2nd. Help... I really need a jumping point.
 
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You can start by writing out some relevant definitions and basic facts along with the facts you're given. Often this is also where you finish. In this case, remember that the condition for f to be a homomorphism is f(ab)=f(a)f(b). Now use the definition of f.
 
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ldelong said:
Group G must be communitative for it to be abelian


When I see things like that I worry. Commutative and abelian are synonyms.
 
Actually there's not much to prove. It is, as StatusX already said, a matter of writing out the definition of homeomorphism and plugging in commutativity of the group and the definition of f at the right points.

So just start writing something down, then tell us where you get stuck.
 
I got it thanks
 
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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