1997 Fall Qualifying Exam in Algebra (AI-generated)

Problem 1.

Let be finite and suppose that for any subgroups , either or . Prove that is cyclic of order for some prime .

Proof.

If two distinct primes divided , Cauchy's theorem would give subgroups of orders and , neither containing the other. Thus .

Choose an element of maximal order. For any , the cyclic subgroups and are comparable. Maximality of the order rules out , so . Hence every element lies in , and is cyclic.

Problem 2.

Prove that no group of order , , is simple.

Proof.

The number of Sylow -subgroups satisfies

so or . If it is , the Sylow subgroup is normal.

If , conjugation on the six Sylow subgroups gives a nontrivial homomorphism . Were simple, its kernel would be trivial, so would divide . But with does not divide , which contains only one factor of . This is impossible.

Problem 3.

Let stabilize the set of vectors . Prove that is solvable.

Proof.

The stabilizer consists exactly of the invertible upper triangular matrices

Its normal subgroup

is abelian. The quotient by is isomorphic to the abelian diagonal group . Hence and , so the derived series reaches the identity after at most two steps.

Problem 4.

Over , find the monic gcd of

and find such that .

Proof.

The Euclidean algorithm gives the monic gcd

Indeed, direct expansion verifies the Bézout identity

Thus one may take

Problem 5.

Determine the unit group of subject to .

Proof.

The ring is . Since

and the quadratic factor is irreducible over , the Chinese remainder theorem gives

Therefore its group of units is

Problem 6.

Let , , and let be generated by

Prove that is finite and find its order.

Proof.

Use these generators as columns of

Its determinant is

Thus has full rank and the quotient is finite. For a full-rank sublattice over the Gaussian integers, the underlying abelian-group index is the Gaussian norm of the determinant. Hence

Problem 7.

Let be the splitting field of over . Determine and the number of its subfields.

Proof.

Because is odd, the roots are distinct. The splitting field is , where is the multiplicative order of modulo . We have

and no smaller positive exponent gives . Thus and

The subfields of are precisely for . Since has six positive divisors, has six subfields. Of these, the three corresponding to contain the base field .

Problem 8.

Find the degree over of the splitting field of .

Proof.

Let and . The splitting field is

Eisenstein at gives , while . Both fields contain

and the unique quadratic subfield of is . Hence their intersection is exactly and has degree . Therefore

Problem 9.

Let be separable and irreducible of degree . Determine the Galois group of its splitting field when that group has order .

Proof.

The Galois group acts faithfully and transitively on the four roots, so it is a transitive subgroup of of order . Such a subgroup is a Sylow -subgroup of . All Sylow -subgroups are conjugate, and a standard one is the symmetry group of a square, isomorphic to the dihedral group . Hence

In its action on the roots, it is conjugate in to the usual action of on the four vertices of a square.

Problem 10.

For

show that its splitting field over is obtained by adjoining one root, and find its Galois group.

Proof.

The rational-root test shows that has no rational root, so it is irreducible. A direct computation shows

If is a root, then is also a root. Iterating this transformation produces all three roots. Hence all roots lie in , so this degree- field is already the splitting field.

Since the splitting field has degree , its Galois group has order and therefore

The generator cyclically permutes the three roots by .