2016 Fall Algebra

Problem 1.

Is every group of order 39 cyclic? Either prove this or construct a non-cyclic group of order 39.

Proof.

We recall that for distinct primes there is a non-abelian group of order if and only if . Since our goal is to construct the non-abelian group of order 39. The easiest way to do this is via semidirect product. By Cauchy's theorem a group of order 39 has a subgroup of order 3 and a subgroup of order 13.

We know that . By Cauchy's theorem again this group contains a unique subgroup of order 3. Thus, there is a nontrivial homomorphism from into , which implies that the semidirect product of and is not abelian.

Though it is not necessary to solve this problem, when and are primes with a presentation for the (unique up to isomorphism) non-abelian group of order is given by

where and .

Problem 2.

Recall that for a subgroup of , the normalizer is defined to be . Let be a prime and denote the symmetric group on letters. Let be a subgroup of of order . What is the order of ? Explain your answer.

Proof.

We will show that . Since a subgroup of order is a Sylow -subgroup. We recall that the index of in is the number of Sylow -subgroups of . Each such subgroup contains exactly elements of order and the intersection of any two such subgroups is trivial. The elements of order are exactly the -cycles, and there are of these. Therefore , so , and .

Problem 3.

Prove or disprove: The quotient ring is a field.

Proof.

Let be a field. We recall that is a field if and only if is irreducible. Therefore, we need only see whether is irreducible over . By substituting 0 and 1 we see that it has no linear factors. So, if this polynomial is not irreducible it must factor as the product of two quadratic polynomials. That is, we must have a solution:

where . Since we must have , which implies and , which is impossible.

Problem 4.

Does there exist a ring with 1 whose additive group is isomorphic to ? Prove your answer.

Proof.

Let be such a ring and be an isomorphism of additive groups. We first note that we cannot have in since this implies and the additive group of cannot be isomorphic to a non-trivial group.

Consider the image . This is equal to for some . Note that in so since is injective we must have in (where here we are considering multiplication by in terms of repeated addition since ).

Now consider . We have , so since is injective . But , so , and we find in , which is a contradiction.

Problem 5.

Let denote a commutative ring with 1 such that for every , we have for some integer . Prove that every prime ideal in is maximal.

Proof.

Solution 1: Let be a prime ideal of and assume there is an ideal of strictly containing . We will show that .

Let . By assumption there is an such that . For any we have

Since is a prime ideal . Since we must have , completing the proof.

Solution 2: Since is a prime ideal we know that is an integral domain. Our goal is to show that it is a field, which implies that is maximal. The condition on elements means that every element satisfies for some . Suppose in . This means in . Since is an integral domain, we must have , which implies that is invertible in .

Problem 6.

Let denote a primitive 7-th root of unity.

(a) True/False: Every element in can be expressed uniquely in the form

where . Briefly explain.

(b) Find the order of the element induced by . Briefly explain your answer.

Proof.

(a) False- the minimal polynomial for is the 7-th cyclotomic polynomial . We see for example that

which means that we do not have uniqueness of these representations of elements.

(b) For a primitive -th root of unity we know that is natually isomorphic to where the isomorphism is given by . Therefore, the order of this element is qual to the order of 2 in , which is 3. Note that

(c) We note that the element is fixed by the element of the previous problem. Let , a subgroup of . The degree of the fixed field of over is equal to the index and it contains the field . Note that is not in (if it were then the minimal polynomial of would have degree at most 4), so is the fixed field of , which has degree 2.

Problem 7.

Give an example of a matrix over with minimal polynomial which is not similar to any matrix with rational entries. Briefly explain your answer.

Proof.

Consider the -module

The linear transformation corresponding to multiplication by has minimal polynomial . The corresponding rational canonical form for this linear transformation is

This matrix cannot be similar to a matrix with rational coefficients, because any similar matrix has the above as its rational canonical form, but a matrix with rational coefficients has as its rational canonical form another matrix with rational coefficients.

Problem 8.

Let denote the dihedral group of order 10.

(a) Give an example of a non-trivial degree one representation .

(b) Give an example of an irreducible degree two representation . Prove that your representation is irreducible.

Proof.

(a) We consider the presentation of given by . Every element can be written uniquely as where and . We consider the sign representation which maps to . This is a non-trivial degree one representation.

(b) We want to find matrices and satisfying the relations and , then the map and extends uniquely to a homomorphism from to the matrix group generated by and . We can take

the matrix that corresponds to rotation by radians. This is irreducible as there are no -invariant 1-dimensional subspaces, since rotation by radians sends no line in to itself.

Problem 9.

True/False. For each of the following answer True or False and give a brief explanation.

(a) Every finite subgroup of is abelian.

(b) A finite extension of cannot have infinitely many distinct subfields.

Proof.

(a) False. We have embedded as a subgroup of via the permutation matrices- where a permutation is sent to the matrix where row has a single 1 in column .

(b) True. Let be a finite extension of and the normal closure of over . Suppose has order . A group of order has only finitely many subgroups, and the subfields of (which certainly contain the subfields of ) are in one-to-one correspondence with the subgroups of this group.

Problem 10.

For each of the following, either give an example or state that none exists. In either case, give a brief explanation.

(a) A non-zero zero divisor in .

(b) An injective group homomorphism .

Proof.

(a) We know that is a free -module with basis 1, . Therefore, is a free -module of rank 4 with basis , , , and . Note that . We have that

so , and is a non-zero zero divisor.

(b) This is not possible. is isomorphc to which has three elements of order 2, which has only one.