Let G be a Finite group written mlutiplicatively. Prove that if the order of G is odd, then every has a square root. Conclude the proof with there exists exactly one with
I think i need to show that squaring is an injective funciton from and use the pigeonhole principle...
Well, if by "using the pigeonhole principle" you mean that a function from a finite set to itself is injective iff it is onto then yes: you need that.
Now define . Since the order of G is odd the function is injective (because if then, as g has odd order, and then we get x has even order, which is impossible in a group of odd order) we're done.