2002 Fall Qualifying Exam in Algebra (AI-generated)

Proof.

(a) Differentiation is nilpotent, and

form a basis. Thus is a cyclic vector and the Jordan form is one nilpotent block

(b) Set and . Then

For each , the subspace

is -invariant and gives one nilpotent Jordan chain of length . These subspaces form a direct sum of the whole polynomial space. Hence the Jordan form is

Proof.

(a) The conjugacy classes in have the following representatives and sizes:

and two distinct classes of -cycles, each of size , represented for example by

The -class of -cycles splits in because its centralizer lies in .

(b) If , , and , then . Thus, whenever contains one member of a conjugacy class, it contains the whole class.

(c) A normal subgroup of must contain the identity and be a union of some of the four nonidentity classes of sizes . Its order must also divide . Checking sums of with subsets of these four sizes shows that none is a divisor of except and . Therefore the only normal subgroups are and , so is simple.

Proof.

(a) The primary decomposition theorem gives

Thus there is exactly one abelian group of this order.

(b) The number of Sylow -subgroups divides and is congruent to modulo , so it is . The number divides and is congruent to modulo . If it were , then , contrary to the hypothesis. Thus it is also . Both Sylow subgroups are normal, intersect trivially, and commute, so

is abelian.

(c) Since , the eleven groups correspond to the partitions of :

and

(d) No. The unitriangular group

has order and is nonabelian.

Proof.

(a) No. The trace of every commutator is zero because

whereas over .

(b) Such matrices exist exactly when . The same trace argument proves necessity. For sufficiency when , act on

Let and let be multiplication by . Then the product rule gives

These are matrices over . If , take block-diagonal matrices consisting of copies of this pair. Their commutator is .

Proof.

(a) Since is normal, . Therefore

(b) Every unitary matrix is unitarily diagonalizable. If it has determinant , its eigenvalues are and . The diagonalizing matrix may be adjusted by a scalar to have determinant , so every element of is conjugate in to

Two such matrices are conjugate exactly when they have the same trace , so this list parametrizes all conjugacy classes.

(c) Let contain a noncentral element , where . Part (a), applied while the second element varies through conjugates near the identity, shows that contains rotations through all sufficiently small angles about one axis. Normality then gives the corresponding small rotations about every axis. Products of these elements form a neighborhood of the identity in . Hence is an open subgroup. Since is connected, it has no proper open subgroup, so . Therefore every proper normal subgroup is contained in the center , and the only nontrivial one is .

Normal subgroups of correspond under to normal subgroups of containing . The only such subgroups are the kernel and all of , whose images are and . Hence is simple.

Proof.

(a) The squares in the cyclic group are exactly its even powers .

(b) We have

Because is even, a power of is a square exactly when its exponent is even. This proves the assertion.

(c) As printed, this assertion is false because the congruences modulo the do not control the sign in quadratic reciprocity. Take , , and

Then

But is a nonsquare modulo each of , so

Consequently in . A condition controlling the primes modulo as well as modulo each would make the intended quadratic-reciprocity argument possible.

Proof.

(a) Here the number of variables and the degree are both . Since is not a square modulo , the equation

has only the solution . Thus the number of zeros is , not divisible by .

(b) The homogeneous polynomial

has degree , so its number of zeros is divisible by . There is at least the zero solution, and therefore there must be nonzero solutions.

We need one with . If is not a square, the only solution with is , so any other solution works. If is a square, there are exactly solutions with , a number not divisible by . Since the total is divisible by , some solution must have . Dividing its coordinates by gives

Proof.

(a) Irreducibility makes the Galois group transitive, so and Cauchy's theorem gives a -cycle in . Complex conjugation fixes the three real roots and interchanges the two nonreal roots, so also contains a transposition. The conjugates of this transposition by powers of the -cycle are transpositions along a connected -cycle graph, and such transpositions generate . Hence .

(b) Translate by :

Every nonleading coefficient is divisible by , while the constant term is not divisible by . Eisenstein's criterion at proves that , and hence , is irreducible over .

(c) Put . The derivative is

so the real critical points are and . Moreover,

Together with the behavior at , this shows that has exactly three real roots. Parts (a) and (b) now give

Proof.

(a) The conjugacy classes are

Thus representatives are . The center is

and, since while , the commutator subgroup is

(b) The subgroups are , , the three cyclic subgroups , , of order , and . The order- subgroups have index and hence are normal; the remaining cases are immediate.

(c) The abelianization is , so there are four one-dimensional representations. They all send to , and are obtained by independently choosing the signs assigned to and ; the value on is their product.

There is one further irreducible representation of degree , given by

with mapped to the product of these matrices. It is irreducible because the two displayed matrices have no common invariant line. The squared dimensions sum to

so this list is complete.