# Thread: A question in group theory.

1. ## 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?

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

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