# A question in group theory.

• Nov 10th 2008, 10:22 AM
andreas
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?
• Nov 10th 2008, 11:01 AM
NonCommAlg
Quote:

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 $n=2k + 1.$ since G is finite, bijectivity of $f$ is equivalent to injectivity. so suppose $x^2=y^2.$ then: $x=x^{n+1}=x^{2k+2}=y^{2k+2}=y^{n+1}=y. \ \ \Box$
• Nov 10th 2008, 12:38 PM
andreas
Quote:

Originally Posted by NonCommAlg
so suppose $n=2k + 1.$ since G is finite, bijectivity of $f$ is equivalent to injectivity. so suppose $x^2=y^2.$ then: $x=x^{n+1}=x^{2k+2}=y^{2k+2}=y^{n+1}=y. \ \ \Box$

Thank you a lot!!!!