2023 Spring Qualifying Exam in Algebra

Solution.

Let be a group of order . By Sylow's theorem, the number of Sylow -subgroups satisfies

Thus . Hence the Sylow -subgroup is normal in .

Let be a Sylow -subgroup. Then

and since is normal, is a semidirect product

Such semidirect products are determined by homomorphisms

Now

There are exactly two possible images of a homomorphism : the trivial subgroup and the unique subgroup of order .

If the action is trivial, then

If the action is nontrivial, then the generator of acts on by inversion, giving

the dihedral group of order .

These two groups are not isomorphic: is abelian, while is nonabelian. Therefore, up to isomorphism, there are exactly two groups of order .

Solution.

Since both and are abelian groups, any homomorphism from to either group factors through the abelianization of .

For , the commutator subgroup of is , so

For part (a), homomorphisms correspond to homomorphisms

The image of a generator of must be an element of whose square is . The only possibilities are

Thus there are exactly homomorphisms: the trivial homomorphism and the sign homomorphism followed by the inclusion .

For part (b), homomorphisms correspond to homomorphisms

But has no nontrivial element of order dividing . Therefore only the trivial homomorphism exists.

Hence the answers are:

Solution.

Let be a subgroup of index . Consider the action of on the left cosets by left multiplication. This gives a homomorphism

Because the action on is transitive, the image is a transitive subgroup of .

The transitive subgroups of are and . If

then

would be an index subgroup of , since has index in . This contradicts the assumption that has no index subgroup.

Therefore

The group acts transitively on the three cosets, and this action is regular. Hence the stabilizer in of any coset is trivial.

Now is exactly the stabilizer in of the coset . Since the stabilizer in the image is trivial, we get

Therefore is the kernel of a group homomorphism, so is normal in .

Solution.

They are not isomorphic.

In the ring

the class of is nonzero and nilpotent, since

Thus has a nonzero nilpotent element.

Now consider

Over ,

The factors and are relatively prime in , and is irreducible over because its discriminant is

By the Chinese remainder theorem,

The ring has no nonzero nilpotent elements, because both and are fields.

Being reduced is preserved under ring isomorphism. Since has a nonzero nilpotent element and does not, the two rings are not isomorphic.

Solution.

Since is a PID, the ideal generated by and is principal. Thus there exists such that

Then divides both and .

Because is prime, it is irreducible. Therefore any divisor of is either a unit or an associate of . Hence is either a unit or an associate of .

Suppose were not a unit. Then would be an associate of . Since , this would imply

Because is prime, hence irreducible, this would force and to be associates, contradicting the hypothesis.

Therefore must be a unit. Hence

So , which means there exist such that

Solution.

Since

we have

Let

By rank-nullity,

Since , we get

Thus

so

This bound is achievable. For example, take a nilpotent Jordan matrix with three Jordan blocks of size and one Jordan block of size :

Then and each contributes rank , so

Therefore the largest possible rank is

Solution.

No. The -module is not finitely generated.

Suppose, for contradiction, that were generated as a -module by finitely many rational numbers

Write

where and . Let

Then every integer linear combination of the has denominator dividing . More explicitly, every element of the subgroup generated by lies in

But

does not lie in , since there is no integer such that

Thus cannot generate all of .

Therefore is not finitely generated as a -module.

Solution.

Let

The splitting field of over is

The roots of are

But

because

Therefore the splitting field of

is still

Now is irreducible over by Eisenstein's criterion at , so

Also , while , so

Thus

The Galois group acts faithfully on the three roots of , so it embeds into . Since the splitting field has degree , the Galois group has order . Hence it is all of :

Therefore the Galois group of over is

Solution.

For part (a), let . Since is Galois, it is normal. Let be the minimal polynomial of over . Since is normal, splits completely over .

Because has rational coefficients, its coefficients are fixed by complex conjugation. Therefore

So is another root of . Since all roots of lie in , we get

Thus

Applying the same argument to gives the reverse inclusion, so

For part (b), suppose for contradiction that . By part (a), complex conjugation restricts to an automorphism of over . Let

Since and

the automorphism is nontrivial. Also , so has order .

Now has a unique subgroup of order , namely . Hence by the Galois correspondence, has a unique intermediate field of degree over , namely

But since , the field

is an intermediate field of degree over . Therefore uniqueness forces

This is impossible, because every element of is fixed by complex conjugation and hence is real, while is not contained in .

This contradiction shows that

Solution.

For part (a), no such example exists.

Every finite abelian group decomposes as a direct sum of cyclic groups:

If is nontrivial, then some . Since tensor product distributes over direct sums,

contains

as a direct summand. But

which is nontrivial. Hence cannot be trivial.

Therefore no nontrivial finite abelian group satisfies

For part (b), no such example exists.

A finite field has order for some prime and integer . Its underlying additive group is

Thus if two finite fields are isomorphic as additive groups, then they have the same characteristic and the same dimension over . Therefore they have the same cardinality

But finite fields of the same cardinality are isomorphic as rings.

Therefore no two finite fields are isomorphic as additive groups but not isomorphic as rings.