2001 Spring Qualifying Exam in Algebra (AI-generated)

Problem 1.

Let have order . Let act by left multiplication on .

(a) Explain how this action gives a homomorphism .

(b) If , show that .

(c) Assuming , explain how identifies with .

(d) Prove that . You may assume and that is simple.

Proof.

(a) The index of is

Thus has elements. After labeling them , left multiplication gives a permutation of these labels and hence a homomorphism

(b) If , then fixes every coset, in particular the coset . Thus , which means . Hence .

(c) If , then is injective and, because domain and codomain both have order , it is an automorphism. The subgroup is the stabilizer of its own coset in the coset action, so is a point stabilizer in . Every point stabilizer is isomorphic to .

(d) The subgroup is normal in . Now , so simplicity gives or . The latter is impossible because and

Thus . The quotient map restricts injectively to , so . A normal subgroup of order would be central, but for . Therefore .

Problem 2.

Let be primes and let .

(a) Show that is a semidirect product of a subgroup of order and a normal subgroup of order .

(b) If , show that the product is direct and is abelian.

(c) If , construct a nonabelian group of order .

Proof.

(a) Sylow's theorem gives

Since , this forces . Let be the normal Sylow -subgroup, and let be a Sylow -subgroup. Then and , so

(b) We have and

The action homomorphism must be trivial if . Hence .

(c) If , the cyclic group has an element of order . Let act on the additive group of by multiplication by . The semidirect product

has order and is nonabelian because the action is nontrivial.

Problem 3.

Let be a complex representation of a finite group on .

(a) Show that is conjugate to a unitary representation.

(b) Show that every eigenvalue of a unitary operator has absolute value .

(c) If is the character of , prove that for all , and that equality implies for a root of unity .

Proof.

(a) Begin with any Hermitian inner product and average it:

This is positive definite and -invariant. Choosing an orthonormal basis for it makes every unitary. Relative to the original basis, this is simultaneous conjugation.

(b) If and , then

so .

(c) Let be the eigenvalues of . Then

Equality in the triangle inequality occurs only when all have the same argument, so they are all equal to some . Since has finite order, is a root of unity. Hence .

Problem 4.

Let have order , let be its number of conjugacy classes, let be its number of irreducible complex representations, and let their degrees be .

(a) State the numerical relations among these numbers.

(b) Give the character table of .

(c) Use (a) to prove that every irreducible representation of an abelian group is one-dimensional.

Proof.

(a) The fundamental relations are

and

(b) On the conjugacy classes represented by , the character table is

(c) If is abelian, every element is its own conjugacy class, so . Hence . Since every and

each .

Problem 5.

Let have characteristic and put .

(a) Express in terms of and .

(b) Show that every extension of is separable.

(c) Show that is Galois and that all fields with elements are isomorphic.

(d) Find an automorphism of order and conclude that the Galois group is cyclic of order .

Proof.

(a) As an -dimensional vector space over ,

(b) Frobenius is injective and hence surjective on every finite field. Thus finite fields are perfect, so their algebraic extensions are separable.

(c) Every element of satisfies

This polynomial has derivative and its roots are precisely the elements of . Thus is its splitting field over and is Galois. In an algebraic closure, the roots of form the unique field with elements, proving uniqueness up to isomorphism.

(d) The Frobenius automorphism

fixes and has order exactly : its th power fixes at most elements for , while . Hence

Problem 6.

Let be Euler's totient function.

(a) Calculate for prime .

(b) If are rings, show that .

(c) If , derive a formula for .

Proof.

(a) Among the residue classes modulo , exactly are divisible by . Therefore

(b) A pair is invertible exactly when both components are invertible, in which case its inverse is . Thus

(c) The Chinese remainder theorem gives

Taking units and cardinalities, then using (a) and (b), yields

Problem 7.

Find the Galois group of the splitting field over of .

Proof.

Let and . The roots are , so the splitting field is

Eisenstein at gives , while . Their intersection has degree dividing both and , so it is . Hence

Every automorphism has the form

where and . Thus

where acts faithfully on by multiplication. This is the nonabelian Frobenius group of order .

Problem 8.

Prove that

is irreducible over .

Proof.

Reduce modulo :

The standard finite-field irreducibility criterion for a monic polynomial of prime degree says that is irreducible exactly when

and

Direct repeated-squaring computations give

and

Hence is irreducible in . Gauss's lemma then implies that is irreducible over .