1996 Spring Qualifying Exam in Algebra (AI-generated)

Proof.

Let have order . The quotient has order , so by Cauchy's theorem it contains an element of order . Since is finite, lifting this element and taking a suitable power gives an element of order .

Conjugation by defines a homomorphism

The source has order , while , so the homomorphism is trivial. Thus commutes with every element of . If generates , then has order .

Proof.

Choose . The cyclic subgroup cannot be proper, because every proper subgroup lies in . Hence is cyclic.

It cannot be infinite cyclic: if under an identification , then one can choose a proper subgroup not contained in . Thus is finite cyclic. The hypothesis says it has a unique maximal proper subgroup. A finite cyclic group has one maximal subgroup for each distinct prime divisor of its order, so its order has only one prime divisor. Therefore

for some prime , and is its unique maximal subgroup of order .

Proof.

Conjugation on the set of Sylow -subgroups gives a nontrivial homomorphism

Its kernel is normal, so simplicity makes the action faithful. Hence divides . But

because among only is divisible by . Therefore does not divide , and consequently cannot divide .

Proof.

Over , has minimal polynomial , while and have , and has characteristic polynomial . Thus only and could possibly be similar. But conjugation by a unimodular integer matrix preserves the ideal generated by all entries of . The entries of generate , while those of generate . Hence and are not similar. Thus no two distinct matrices in the list are similar over .

Over , . Both and are nonzero nilpotent matrices of rank whose square is zero, so both have the single Jordan form and are similar. Therefore the two similarity classes are

Proof.

Since , the polynomial splits into two distinct linear factors over . Hence

Its unit group has order

Proof.

Every commutator has trace zero. For a trace-zero matrix, Cayley-Hamilton gives

Thus every square of a commutator is scalar and lies in .

Conversely, every scalar matrix occurs. If , put and take

Then and .

If , put and take

Then

and its square is . Since is precisely the scalar matrices, the two sets coincide.

Proof.

Choose with . Put

The map , , is injective because implies

and it is surjective by definition. Moreover, , so is its inverse and is an isomorphism.

For any ,

where the first term lies in and the second in . Also because is injective. Hence .

Proof.

Writing transforms the equation into

so its splitting field is that of . The polynomial is irreducible over because is not a rational cube. Its discriminant is

which is not a square in . An irreducible cubic has Galois group when its discriminant is a square and otherwise. Therefore the Galois group is

Proof.

The field has characteristic . Hence

in . Its only root is , already in , so its splitting field is and

Proof.

Let be the root set. If , then is a finite nonempty multiplicatively closed subset of the field's multiplicative group. For any , two positive powers agree, and cancellation shows that and belong to . Thus is a finite subgroup of the multiplicative group of a field, hence cyclic. If , then is exactly the set of roots of , so the monic squarefree polynomial with root set is

If and contains a nonzero element, then is similarly a finite cyclic group, say of order . Therefore

There is one exceptional case omitted from the printed statement: if , then , which is not of either displayed form for a positive natural number . Thus the claimed conclusion is correct provided the root set contains a nonzero element; without that hypothesis, is a counterexample.