In this post, I will state the following proposition about finite field and give its proof:
Proposition The number of cubes in the finite field with elements is if , and if .
Proof Let be the group . So, . If , then there is no subgroup of order 3 by the Lagrange’s theorem. So there is no element such that other than , because otherwise, the powers of , of which there are distict elements in total, forms a cyclic group. This means that for , the kernel is . Since the kernel of is trivial, is surjective. Now we have shown that if does not divide , then the number of cubes in the group is .
For the case is divisible by , we will use the fact
is a cyclic group.
Since divides , we can write where is a positive integer. So , to which we have the following claim.
Claim. In a cyclic group of order , any subgroup , formed by all elements such that , where , is of order . In addition, the set is exactly
Proof Of Claim We consider the isomoprhism between and . It suffices to show that , the set of such that , is equal to If , then . So it is established that . Conversely, since , from which it follows that for some hence
So by the 1st isomorphism theorem and Lagrange’s theorem.
Perceivably, this fact generalises to the th power.