2009 Fall Qualifying Exam in Algebra (AI-generated)

Proof.

Let have order . Sylow's theorems give

so the Sylow -subgroup is normal. By Schur--Zassenhaus,

where has order and is therefore or .

Now , so an action of has image of order at most . For each choice of , there is a trivial action and, up to automorphisms of , one nontrivial action onto , whose nonidentity element acts on by inversion. Hence the four groups are

and

The first two are abelian and the last two are nonabelian. Thus exactly two are nonabelian.

Proof.

Let have order . The number of Sylow -subgroups is or , and the number of Sylow -subgroups is or . If either number is , has a nontrivial proper normal subgroup.

Otherwise, the six Sylow -subgroups contribute

nonidentity elements, and the ten Sylow -subgroups contribute

different nonidentity elements. These two sets are disjoint, giving more than nonidentity elements, which is impossible. Hence one Sylow subgroup is normal and is not simple.

Proof.

(a) The additive group is infinite, and every element has finite order.

(b) Let the characteristic be . For , the equation is equivalent to

The cyclic group contains exactly

solutions to this latter equation, one of which is . The value solves the original equation exactly when

that is, when . Therefore the number of solutions is

Proof.

Suppose but neither nor lies in . By maximality, both larger ideals and meet . Choose

where . Then

But is multiplicative, so , contradicting . Thus or , and is prime.

Proof.

Consider the block matrix over

Using block row elimination with the upper-left identity block gives

Using the lower-right identity block instead gives

because is central. Equating the two expressions proves the identity.

Proof.

Suppose the displayed operator were zero. Then would be a module over

because

The cyclotomic polynomial is irreducible over and has degree , so this quotient is a field of degree over . Consequently

contradicting the assumption. Hence the operator is nonzero.

Proof.

By the primitive element theorem, write

with separable minimal polynomial . Then

Over , factor

into distinct monic irreducibles. The factors remain distinct because is separable. By the Chinese remainder theorem,

Each factor is a field, proving the result.

Proof.

(a) If is abelian, all operators commute. Over , they have a common eigenvector. The span of that vector is a nonzero invariant subspace, so irreducibility forces the entire representation to be one-dimensional.

(b) Every one-dimensional representation annihilates the commutator subgroup and therefore corresponds to a homomorphism

Conversely, every such homomorphism gives a one-dimensional representation. A finite abelian group has exactly as many complex characters as elements, so the number is

Proof.

Because is monic and separable, it is determined by its finite set of distinct roots. If contains a nonzero element, then is a finite multiplicatively closed subset of a group. Powers of each element eventually repeat, showing that this set contains and inverses; hence it is a finite subgroup of and therefore cyclic. Thus

for some not divisible by .

The root set may either omit or include . Therefore the nonconstant possibilities are

where and, in positive characteristic, the characteristic does not divide . These polynomials are monic and separable, and their root sets are closed under multiplication. If constant polynomials are allowed, is the additional empty-root-set case.

Proof.

The polynomial is Eisenstein at , so it is irreducible. Hence its Galois group acts transitively on five roots, and therefore divides . By Cauchy's theorem, contains a -cycle.

Modulo , one has

Both factors are irreducible over and are distinct, so this is a square-free factorization of type . The Frobenius cycle-type theorem implies that contains an element with cycle type . Its cube is a transposition.

Conjugating this transposition by powers of the -cycle produces transpositions whose associated graph on the five roots is connected; such transpositions generate . Consequently