Let be a primitive th root of unity in . Let be a primitive root modulo , so the field automorphism defined by generates the Galois group. Fix dividing , let , and let where . Thus is the fixed field of , the unique element of the Galois group of order . We have and .

A few days ago, while reading Burnside’s proof (in Section 252 of his wonderful book, *Theory of groups of finite order*) that a permutation group of prime degree is either 2-transitive or a Frobenius group contained in an affine general linear group , I needed the identity

where . One rather sneaky proof is to look at the intended conclusion of Burnside's argument, namely that is Frobenius of order , and work back. I've put in details of the construction of the character table below, but taking this as read, the proof is essentially one line.

#### Character tables of affine Frobenius groups

Let . Setting , we have . In the action of on the cosets of , the non-identity elements of act as -cycles, and all elements of have a unique fixed point. Therefore is a Frobenius group. The conjugacy class of is , and the non-identity conjugacy classes have representatives

and are of sizes respectively.

There are linear characters of inflated from . Let be defined by . Let . Then

and more generally, for all . It follows that are distinct characters of . By Frobenius reciprocity

The sum of the squares of the degrees of the characters constructed so far is , so we have found all irreducible characters.

#### Conclusion

Let . From

we get

for . If , the left-hand side is and we get instead .

#### Two related well known tricks

In Wielandt’s well-known proof of Sylow’s Theorem, one needs to know that if is coprime to then mod . This follows immediately from Lucas’ Theorem, or one can follow Wielandt and use a short direct argument, but it is in the spirit of his proof to fix a set of size and let have disjoint cycles. Then clearly has fixed points, in its action on the set of subsets of of size and all other orbits have size divisible by . So modulo .

Another nice argument, applying linear algebra to algebraic number theory, shows that if and are algebraic numbers then is also algebraic. Indeed, is an eigenvalue of where and are irreducible integer matrices having and as eigenvalues. (For example, take the companion matrices of the minimal polynomials.) This trick also shows that is an eigenvalue of an integer matrix, say, and so is an eigenvalue of . Therefore is algebraic, and multiplying through by (using the result just proved), we get that is algebraic.