2018 Fall Qualifying Exam in Algebra (AI-generated)

Proof.

By Sylow's theorems, the number of Sylow -subgroups satisfies

Hence . Let be the unique Sylow -subgroup; then .

To produce a subgroup of order , let be a Sylow -subgroup and let act by conjugation on the set of Sylow -subgroups. Their number is either or . In either case it is odd, so the involution group has a fixed point. Thus normalizes some Sylow -subgroup . Therefore

is a subgroup and .

Since and are relatively prime, . Also , so . Hence

(equivalently, , depending on the semidirect-product convention).

Finally, is a normal Hall subgroup of . The Schur--Zassenhaus theorem says that complements to a normal Hall subgroup exist and are all conjugate. Every subgroup of order is a complement to , so any two such subgroups are conjugate in .

Proof.

We prove the stronger statement for every finite -group acting linearly on a nonzero finite-dimensional vector space over .

Let act on the underlying finite set of . Every nontrivial orbit has size divisible by , so

where is the common fixed subspace. Since is divisible by and , the number is divisible by ; hence contains a nonzero vector .

The induced action on is again an action of a -group. Induction on produces an -stable complete flag

such that acts trivially on each quotient . Choose a basis adapted to this flag. Every then has an upper-triangular matrix, and triviality on each one-dimensional quotient makes every diagonal entry .

A Sylow -subgroup of is a finite -group, so the conclusion applies to the given .

Proof.

We use induction on . The assertion is clear for the trivial group. If is a nontrivial finite -group, then its center is nontrivial. The quotient is a smaller -group and is nilpotent by induction. Therefore, for some ,

The lower central series is compatible with quotients, so this means

Consequently,

Thus the lower central series of terminates, and is nilpotent.

Proof.

The ring is the localization

Ideals of a localization correspond to ideals of disjoint from , after saturation. Since every ideal of is principal, every nonzero ideal of is principal.

More explicitly, if an integer is written

then and are units in , so . Hence the ideals of are

where ranges over positive integers relatively prime to . This description is unique: if are both relatively prime to and , then is a unit of , forcing .

Therefore is a PID.

Proof.

(a) Use the norm

For with , choose a Gaussian integer by rounding the real and imaginary parts of to nearest integers. Then

With , this gives

Thus is Euclidean.

(b) A rational prime congruent to modulo remains prime in . Since , the element is already irreducible. Thus, up to multiplication by a unit, its irreducible factorization is simply

(c) Since ,

Each factor has norm , a rational prime, so both factors are irreducible.

Proof.

(a) If lies in the -eigenspace of , then

Thus lies in the same eigenspace.

(b) Choose an eigenvalue of and its nonzero eigenspace . By part (a), is invariant under all the other operators. Restrict to and choose a nonzero eigenspace . Continue. After finitely many steps, every nonzero vector in is an eigenvector for all .

(c) By part (b), choose a common eigenvector and set . The commuting operators induce commuting operators on . By induction on , the quotient has a complete common invariant flag. Taking inverse images of that flag under the quotient map produces

with the required dimensions and invariance. Together with , this is the desired chain.

Proof.

Let the nonconstant invariant factors of over be

The invariant factors of are

where coefficients are conjugated. Every root of every lies in the open upper half-plane, while every root of every lies in the open lower half-plane. Hence

for all .

The -module associated with is the direct sum of those associated with and . By the Chinese remainder theorem,

Therefore the invariant factors of are

Each polynomial has real coefficients. The rational canonical matrix whose companion blocks correspond to these invariant factors is therefore a real matrix. Since two complex matrices with the same invariant factors are similar over , is similar to that real matrix.

Proof.

Over , let and let be a primitive sixth root of unity. The polynomial is Eisenstein at , so

Its splitting field is

The field is real but is not, so and . Define

and let be complex conjugation. Then has order , has order , and

These automorphisms account for all elements, so

where the nontrivial element of acts by inversion.

Over , the polynomial is separable because its derivative is

which has no common root with . The group is cyclic of order . If is a generator, then is one of or because it has order in . In either case the equation has a solution in . Moreover, all sixth roots of unity lie in because . Hence splits in .

It does not have a root in , since for one has , and is not a square modulo . Thus its splitting field is exactly , and

generated by the Frobenius automorphism .

Proof.

Let

By the fundamental theorem of Galois theory, intermediate fields correspond bijectively to subgroups of . Therefore has no subgroup other than and .

Choose . The subgroup is nontrivial, so it must equal ; hence is cyclic. If were composite, Cauchy's theorem (or the subgroup structure of a cyclic group) would give a nontrivial proper subgroup of , a contradiction. Thus is prime. Finally,

so is prime.

Proof.

(a) The elements and are central, so each forms its own conjugacy class. Direct calculation shows that each noncentral element is conjugate to its negative and to no other elements. Thus the conjugacy classes are

(b) Since

there are four one-dimensional characters. The remaining irreducible character has degree because

It is afforded, for example, by the usual faithful two-dimensional complex representation of the quaternion group. With columns ordered by the classes above, the table is

The rows are pairwise orthonormal under the character inner product, confirming that this is the complete irreducible character table.