2022 Winter Qualifying Exam in Algebra (AI-generated)

Proof.

Suppose and . Then and for some . Since is normal, . Therefore

Thus , so the product does not depend on the chosen representatives.

Proof.

Let . The action of on the left cosets of gives a homomorphism

Its kernel is normal in and is contained in . The map cannot be injective because

when . It is not trivial because the coset action is transitive on more than one coset. Hence is a nontrivial proper normal subgroup, so the given result implies . Thus . Since has index and is proper, necessarily .

Proof.

Let be the number of Sylow -subgroups. Then or .

If , the Sylow -subgroup is normal. Let be any Sylow -subgroup, of order . Then and

so .

If , the four Sylow -subgroups contain eight distinct nonidentity elements. Only three nonidentity elements remain. Every Sylow -subgroup has order , so its three nonidentity elements must be precisely those remaining elements. Thus the Sylow -subgroup is unique and normal. If is a Sylow -subgroup, then again and . Hence in this case as well.

Proof.

Put

Then and , so is an eigenvector with eigenvalue . The vector is a generalized eigenvector of rank .

If were diagonalizable, then so would . For a diagonalizable operator , one has

as is immediate from a diagonal representation. But here while , a contradiction. Thus is not diagonalizable.

Proof.

The quotient is an integral domain. It is also Artinian by the descending-chain hypothesis. Let . The chain

stabilizes, so for some . Hence for some , and

Since is a domain and , we get . Thus every nonzero element of is invertible, so is a field. Therefore is maximal.

Proof.

The units of are exactly and . Indeed, a unit of is a nonzero rational constant, and both it and its reciprocal lie in only when they are .

Every positive-degree divisor of in has degree and therefore has the form with . But for every integer ,

and both factors lie in and are nonunits. Thus no positive-degree divisor of is irreducible.

If were a finite product of irreducibles, the sum of their degrees would be , so one factor would have positive degree. That factor would be a positive-degree divisor of , contradicting the preceding paragraph. Hence no factorization into irreducibles exists.

Proof.

Over ,

The quadratic factor is irreducible because its discriminant is

which is not a square modulo . The Chinese remainder theorem gives

The multiplicative group of a finite field is cyclic, so the unit group is

Proof.

(a) The bijection sends to the automorphism

(b) The exponents are the nonzero quadratic residues modulo . Put

The subgroup of quadratic residues, of order , fixes . A nonresidue sends it to the sum over the five nonresidues. Since

and the classical quadratic Gauss-period calculation gives

the two periods are the roots of , whose discriminant is . Thus

up to the choice of square root. It is not rational, so its orbit has size and

Proof.

(a) In the infinite dihedral group

both and have order , while has infinite order.

(b) No. The image of the divisible group under a homomorphism is divisible, while the only divisible subgroup of is . More explicitly, if , then for every , so is divisible by every positive integer and must be zero. It follows that is the zero map.

(c) No. Taking determinants would give

which is impossible over .

(d) Take

Both successive extensions are quadratic and therefore Galois. The top extension over is not normal because it does not contain the conjugate of .