2026 Spring Quals in Algebra

Problem 1.

How many elements of order are there in a simple group of order ?

Proof.

Let be a simple group of order . The elements of order in are exactly the nonidentity elements of the Sylow -subgroups of .

The number of Sylow -subgroups divides and satisfies , so . Because is simple, , since a unique Sylow subgroup would be normal. Therefore, . The intersection of two distinct subgroups of order must be trivial.

Consequently, the number of elements of order is

Problem 2.

Let be a finite group and let be a subgroup. Prove that the number of conjugates of in divides the index .

Proof.

Let denote the set of conjugates of in . Consider the action of on given by

This is a group action because

By the Orbit-Stabilizer Theorem, the size of the orbit of is equal to the index of the stabilizer of . Let

The stabilizer contains , since for every . Thus,

By Lagrange's Theorem, for some positive integer . Therefore, the number of conjugates of is

which divides

Problem 3.

Prove that the symmetric group has a normal subgroup such that .

Proof.

Let

This is a subgroup of . It is normal because it is a union of conjugacy classes: the identity class and the class consisting of all permutations of cycle type .

The quotient has order

Every group of order is isomorphic either to or to . No quotient of can contain an element of order , since the order of an element in a quotient divides the order of a preimage and has no element of order . Hence cannot be cyclic. Therefore,

Problem 4.

Let be a unique factorization domain, and let be an irreducible element. Let denote the field of fractions of . Prove that the subring

of is a principal ideal domain with a single irreducible element up to associates.

Proof.

Let and be two nonzero elements of , and let and be the powers of in the factorizations of and , respectively. Write

where and .

If , then divides in , because

and the second factor belongs to .

Now let be a nonzero ideal of . Choose a nonzero element for which the exponent of in is minimal. By the preceding divisibility observation, divides every element of . Hence , proving that is a principal ideal domain.

It remains to identify the irreducible elements. If and , then is a unit in , since its inverse also belongs to . If , then is a product of two nonunits and is not irreducible. Thus an irreducible element must have the form

where . Since is a unit of , this element is associate to . Therefore, up to associates, is the unique irreducible element of .

Problem 5.

Let be a primitive seventh root of unity. Find the minimal polynomial of over .

Proof.

Let . Since is Eisenstein at ,

Also,

The degree is divisible by both and , and hence is divisible by . Therefore,

is divisible by . On the other hand, satisfies , so this degree is at most . It is therefore equal to . Consequently, the minimal polynomial of over is

Problem 6.

Show that there is a real matrix whose minimal polynomial is and which is not similar to any matrix with rational entries.

Proof.

Let

The matrix is the companion matrix of , so its minimal polynomial is . The matrix is the real rotation matrix through the angle , and it satisfies .

Consider the block diagonal matrix

Then , so the minimal polynomial of divides . The minimal polynomial of a block diagonal matrix is the least common multiple of the minimal polynomials of its blocks. Since the minimal polynomial of the block is , the minimal polynomial of is exactly .

Finally, is not similar to a rational matrix. Indeed, trace is invariant under similarity, but

Every matrix with rational entries has rational trace, so no rational matrix can be similar to .

Problem 7.

Find the Galois group of the splitting field of the polynomial

up to isomorphism. You may use without proof that .

Proof.

Let be the splitting field of . Its roots satisfy

Thus the roots are

Set

Then . Moreover,

so .

By Eisenstein's criterion, applied at either or , the polynomial is irreducible over . Hence

The degree of over is at most , and the given fact shows that it is greater than . Therefore,

and

Thus has order . Because contains the nonnormal extension , the corresponding subgroup of is not normal. Among the groups of order , the only one having a nonnormal subgroup is the dihedral group of order . Therefore,

where denotes the dihedral group of order .

Problem 8.

If is a prime power, let denote the finite field of order . Find all polynomials such that

Proof.

A polynomial satisfies

if and only if is irreducible over and has degree .

A reducible quartic either has a linear factor or is a product of two irreducible quadratics. A quartic has a linear factor over precisely when it has a root in . The only irreducible quadratic over is , and

After eliminating the quartics having a root in and the square above, the irreducible monic quartics are

Therefore, these are exactly the polynomials for which

Problem 9.

Let be an integer. Prove that the polynomial

has no roots in with multiplicity greater than .

Proof.

A polynomial has a repeated root if and only if it has a nonconstant common divisor with its derivative. Any common divisor of and must also divide