Thread: A question in group theory.

I am given a finite group G with order n. The following mapping is considered G to G: f(x)=x^2. I have to prove that this mapping is bijective only if only n is odd.


I easily proved that mapping is not bijective when n is even using Cauchy's theorem. How should I prove that it bijective when n is odd?

Originally Posted by andreas

I am given a finite group G with order n. The following mapping is considered G to G: f(x)=x^2. I have to prove that this mapping is bijective only if only n is odd.


I easily proved that mapping is not bijective when n is even using Cauchy's theorem. How should I prove that it bijective when n is odd?

so suppose since G is finite, bijectivity of is equivalent to injectivity. so suppose then:

Originally Posted by NonCommAlg

so suppose since G is finite, bijectivity of is equivalent to injectivity. so suppose then:

Thank you a lot!!!!