2016 Spring Algebra

Problem 1.

(a) Prove that every subgroup of a cyclic group is cyclic.

(b) Is the automorphism group of a cyclic group necessarily cyclic? Explain.

Proof.

(a) Let and let . If we are done, so suppose . So for some . Let be the smallest positive integer such that . Let . Since , for some . By the division algorithm, where . So . We see that , and since is the smallest positive integer with , . Therefore, every element in is in the cyclic group generated by , so is cyclic.

(b) No, the automorphism group of is isomorphic to . You can see this by noting that an automorphism of a cyclic group is determined by where it sends a generator, and (since is surjective) and each of these automorphisms are of order 2.

Problem 2.

Let , the cyclic group of order 25.

(a) Can be given the structure of a -module? Explain your answer.

(b) Can be given the structure of a -module? Explain your answer.

Proof.

(a) We need only determine how an element acts on any . By the module axioms we must have ( times), so we need only determine . By the axioms, we have , so we need the action to give a homomorphism from to which is multiplication by -1 when we do it twice. So, we need where is a square root of -1. Since , we see that is a square, and checking small numbers we see . So, we can let , giving the structure of a -module.

(b) No: is a field and a module over a field is a vector space over that field. In particular, for any module we must have and so , which is not the case in .

Problem 3.

Prove there is no simple group of order 520.

Proof.

We start be factoring . Let be the number of Sylow -subgroups of . Our goal is to show that either , , or is 1, since then that unique Sylow -subgroup would have to be normal.

Since and if we write and if we write with , we have . Then

Suppose none of these are 1. Then and since each pair of these groups of order 13 intersect trivially we get 12 elements of order 13 in each. This gives elements of order 13 in . Similarly, we get 4 elements of order 5 from each Sylow 5-subgroup, so implies there are elements of order 5. Since we see that this is impossible and either or .

Problem 4.

Let be a finite group acting transitively on a set with .

(a) State the Orbit-Stabilizer Theorem.

(b) Show that there is some element of which fixes no element of .

Proof.

(a) Orbit-Stabilizer Theorem

(b) The starting point of this problem is Burnside's Lemma (which follows from the Orbit-Stabilizer Theorem plus some counting). For , let be the set of fixed points of . Burnside's Lemma says

where the 1 is because the action is transitive. Now , so one of the summands in this average is larger than 1. That means that one of the must be less than 1 (since the average is 1) and since it is a nonnegative integer, it must be 0.

Problem 5.

Let be a field and let be an algebraic closure of . Assume have degree 2 and 3 over , respectively.

(a) Can have degree 5 over ? Either give an example or prove this is impossible.

(b) Can have degree 6 over ? Either give an example or prove this is impossible.

Proof.

(a) This is impossible. By multiplicativity of the degree of field extensions we have . Since the degree of over must divide 6.

(b) This is possible. Let and . Since we see that . So, we need only see that , . Since and , we're done.

Problem 6.

Let denote a Galois extension with Galois group isomorphic to .

(a) Does there exist a quadratic extension contained in ? Prove your answer.

(b) Does there exist a degree 4 polynomial in with splitting field equal to ? Prove your answer.

Proof.

(a) By the Galois Correspondence Theorem this is equivalent to asking whether contains a subgroup of index 2. Suppose were a subgroup of of order 6. There must be a 3-cycle not in since they are 8 total 3-cycles in . Let be such a 3-cycle. Since there are only two cosets of we must have or , both of which are impossible.

(b) This is equivalent to asking whether there is a non-Galois quartic extension of contained in , which is the same as asking whether contains a non-normal index 4 subgroup. It does contain such a subgroup, for example taking the subgroup generated by any 3-cycle.

Problem 7.

Let be a linear transformation of a vector space over the field which satisfies the relation . Show that the dimension is divisible by 3.

Proof.

We assume is finite dimensional.

We consider the subspaces . Let . Since and are vector spaces, we must have

By assumption, . We shall prove that both dimensions of and are divisible by .

Let be a basis of . Since is not linear dependent to , we can find an such that replacing by , the set is still a basis of . Similarly, since has no rational solution, can not be represented by a linear combination of and . This we canb replacing some of () with and the set is still a basis. By induction, we shall find a basis of so that the basis can be given by , and hence .

Problem 8.

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

(a) If , are fields and is a ring homomorphism such that , then is injective.

(b) The unit group of is isomorphic to the additive group of .

Proof.

(a) This is true -- since is an ideal of so it is 0 or all of .

(b) This is false -- the additive group is never isomorphic to its multiplicative group. Alternatively, these groups have different number of elements of order 2.

Problem 9.

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

(a) An element such that .

(b) A tower of field extensions such that and are Galois extension but is not Galois.

Proof.

(a) The Primitive Element Theorem says that such an element must exist. For an explicit example, take .

(b) Consider , , and . We see that is not Galois since it does not contain the other roots of , but each of and is Galois since they are quadratic.

Problem 10.

Let be all pairwise non-isomorphic complex irreducible representations of a group of order 12. What are the possible values for their dimensions ? For each of the possible answers of the form give an example of which has such irreducible representations.

Proof.

We know that and that each divides 12 (we don't actually need this). This implies that each . Also, the number of is equal to the index of , the commutator subgroup in .

When is abelian we have for all , so we get . If is not abelian, then the number of is at most 6 (and is always at least 1). In particular, the sum of squares of the bigger than 1 must be at least 6.

The only possibilities left are and . The first set of numbers correspond to the irreducible representations of . You can see this by noting that the commutator subgroup has order 4. This is because the commutator subgroup is the largest normal subgroup such that the quotient is abelian, and has a normal subgroup of order 4.

The second set correspond to the irreducible representations of , which we can see by the fact that the dimensions of irreducible representations of a direct product are pairwise products of the dimensions of the irreducible representations of the direct factors. The dihedral group of order 12 also works here.