2000 Fall Qualifying Exam in Algebra (AI-generated)

Proof.

(a) Let

This gives permutation matrices. The kernel is trivial because a permutation fixing every basis vector is the identity.

(b) No. The nonzero vector is fixed by every permutation.

(c) Take

and

Both are invariant and .

(d) The action on is the trivial matrix . For , use the basis

For the generators and , the matrices are

These determine the entire representation.

Proof.

(a) The proper nontrivial subgroups are and the three cyclic subgroups , , . The latter have index and hence are normal, while is central. Thus every subgroup is normal.

(b) The five conjugacy classes are

(c) False. In , let

and . Then and because is abelian, but is not normal in because conjugation permutes the three double transpositions.

Proof.

(a) Every nonleading coefficient is divisible by , and the constant term is not divisible by . Eisenstein's criterion at proves irreducibility.

(b) This polynomial is . Translate by :

Every nonleading coefficient is divisible by , while the constant term is , not divisible by . Thus the translated polynomial is Eisenstein at , and translation preserves irreducibility.

Proof.

(a) Let be the exponent of the finite abelian group . Every element is a root of , so . But divides , hence . A finite abelian group contains an element whose order is its exponent, so contains an element of order and is cyclic.

(b) In the product of all nonzero residues, pair each element with its inverse. Every pair contributes . The only self-inverse elements solve , hence are and . Therefore

The statement is also immediate for .

Proof.

The group consists of real orthogonal matrices of determinant , while

Every element of is conjugate to exactly one matrix

with conjugacy class determined by its trace .

If , it is a union of conjugacy classes. Moreover, for and ,

If contains a noncentral element , varying shows that these commutators contain rotations through all sufficiently small angles; conjugacy supplies all axes. Their products contain a neighborhood of the identity. Thus is open, and connectedness of forces . Therefore the only nontrivial proper normal subgroup is its center

The inverse image of a normal subgroup of is a normal subgroup of containing . The only possibilities are the kernel and all of , whose images are trivial and all of . Hence is simple.

Proof.

(a) The relation matrix is

It has rank . The gcd of all entries is , and the gcd of its minors is . Therefore its Smith normal form is

Consequently,

(b) Since , the -primary part is one of

and the -primary part is one of

All six groups are obtained by taking one direct product from each list, and these six are pairwise nonisomorphic.

Proof.

(a) Count the set

in two ways. Fixing first gives the left sum, while fixing first gives the right sum.

(b) Partition into orbits. For an orbit and any , orbit-stabilizer gives

Thus

Summing over the orbits and applying part (a) proves the formula.

Proof.

(a) The field . Every -automorphism of sends to another root of its minimal polynomial lying in , hence to a real root. Since is the unique real root, it must be fixed, and therefore the automorphism is the identity.

(b) The tower formula gives

For a proper nontrivial intermediate field, both factors exceed , so both equal . Hence .

(c) In , the subgroup fixing the root is a point stabilizer isomorphic to , and its fixed field is . Intermediate fields between and correspond to subgroups between and . A point stabilizer is a maximal subgroup of , so no proper intermediate subgroup, and hence no proper intermediate field, exists.

Proof.

(a) The conjugacy classes have sizes

They are the identity, the three double transpositions, and two separate classes of four -cycles. Thus

(b) Since , there are three one-dimensional representations. There are four conjugacy classes, so one further irreducible representation remains. The degree-sum formula gives

so . This is the standard rotation representation of the tetrahedron. Its character on the four classes is

(c) Tensoring multiplies character values. Every linear character is on the double transpositions and takes cube roots of unity on the two classes of -cycles. Since the three-dimensional character vanishes on both -cycle classes, the product character remains

Thus every tensor product is isomorphic to the same three-dimensional representation; no new representation appears.

Proof.

Since , the ring of integers is

with discriminant . The Minkowski bound for an imaginary quadratic field is

which lies between and . Therefore every ideal class has an integral ideal representative of norm at most .

The relevant rational primes are and . Since , it ramifies:

Modulo , is irreducible, so is inert and there is no ideal of norm . Ideals of norm introduce no new class because they are products involving or are principal.

Thus the class group is generated by , and

It is nontrivial: if were principal, there would be an algebraic integer of norm , requiring

which has no integer solution. Therefore