2018 Spring Qualifying Exam in Algebra (AI-generated)

Proof.

Let and let have order . By Sylow's theorems, the number of Sylow -subgroups satisfies

Thus , so the Sylow -subgroup is normal and cyclic. If is a Sylow -subgroup, then and , so

The homomorphism sends the generator of to an automorphism of order dividing . Since

there are exactly two possibilities: the trivial automorphism and inversion.

For the trivial action, . For inversion, is the dihedral group

Consequently, up to isomorphism, the two groups of order are

Proof.

(a) Let act on itself by conjugation. The class equation is

where the represent the noncentral conjugacy classes. Every index in the sum is a positive power of and is therefore divisible by . Hence divides . Since the identity lies in the center, this implies , so .

(b) We use induction on . The assertion is clear for . By part (a), contains a subgroup of order . It is central and therefore normal. The quotient is a smaller -group, so it is solvable by induction. Also is cyclic, hence abelian and solvable. An extension of a solvable group by a solvable group is solvable; therefore is solvable.

Proof.

Set . Maximality of makes a field, and the images of generate as a ring over the image of .

We first show that has positive characteristic. If , then embeds in . Since is a field, it contains . The generalized Zariski lemma says that if a field is finitely generated as an algebra over an integral domain , then must be a field after inverting finitely many elements. Applied to , this would imply that

is a field for some nonzero integer , which is impossible: a prime not dividing is not invertible in this ring. Thus for some prime .

It follows that is a field finitely generated as an algebra over the finite field . By Zariski's lemma, is a finite algebraic extension of . Hence

Therefore is a finite field.

Proof.

It is enough to prove that every nonzero ideal is principal. Let

be a nonzero finitely generated ideal. Define successively

The hypothesis implies

Indeed, divides and , while the Bezout expression for gives the reverse inclusion. Inductively,

Thus is principal. The zero ideal is also principal, so every ideal of is principal and is a PID.

Proof.

For every abelian group ,

If is finite and , equality of orders gives , so . Conversely, if , then , and the required isomorphism holds.

By the classification of finite abelian groups, the finite abelian groups annihilated by are precisely

where are arbitrary nonnegative integers. This includes the trivial group when .

Proof.

(a) Taking determinants gives

Because is nonsingular, , so in . Hence , and therefore

(b) For every integer , including negative because is invertible,

Taking traces yields

In characteristic , the element has multiplicative order , since and neither nor equals . Thus whenever , and so .

(c) Similar matrices have the same characteristic polynomial, so and do. Write

Then

Since and , comparison of coefficients gives

Thus and . Since the characteristic is , also . Therefore

with (in fact ).

Proof.

Let , and make a -module by defining

Because , the action factors through . Since is irreducible,

is a field, so becomes an -vector space. If , then the tower formula for vector-space dimensions gives

Hence .

Proof.

Because and have the same irreducible polynomial over , the -embedding

extends to an -automorphism of the splitting field . Therefore

for every positive integer .

(a) The group is finite, so has finite order, say . Then

and hence . Thus has positive characteristic, say .

(b) The same equality shows that divides the order of . The cyclic group therefore contains a subgroup of order . Let

Artin's fixed-field theorem gives

Proof.

The norm map

is surjective. Indeed, the multiplicative groups are cyclic, and the image has order

Choose with . Every root of is , where . All st roots of unity lie in because divides . Thus the polynomial splits over , and .

The polynomial is separable because is not divisible by the characteristic. It has distinct roots, so it cannot split in the field , which has only elements. Therefore . Since the only possibilities inside the degree-two extension are degrees and , we conclude that

Proof.

(a) The identity forms one class. The three double transpositions

form one conjugacy class. The eight -cycles split into two conjugacy classes of size in ; a representative choice is

Thus the class sizes are .

(b) The normal Klein four subgroup gives

Inflating the three irreducible characters of produces three one-dimensional characters of . The fourth irreducible character is the three-dimensional standard representation obtained from the permutation representation on four letters after removing its invariant line. Let . The character table is

The sum of the squares of the degrees is

so these are all the irreducible characters.