Let G be a group of odd order Prove that the mapping x x2 f

Let G be a group of odd order. Prove that the mapping x --> x2 from G to itself is one-to-one.

Solution

since the mapping is from G to itself, the order is the same, so we\'re showing that the kernel of this mapping is 0.

suppose x belongs to the kernel of the mapping, we want to show that x must be the identity, e. this implies that the mapping is injective.

Hence, 1 to 1 since we are mapping from G to G. then, x^2 =e. order of the element divides the order of the group, but |x| =2 => does not divide the order of the group, which is odd.