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As mentioned in the previous post, here is the algebra qual with my solutions. To be honest, this is one of the easiest quals I’ve seen from them. The scores were quite high; I scored 24 out of 25 and there were a couple of perfect scores. I think my only mistake came on problem 5(a), I gave a pretty hand wavy argument. Oh well…

**1.** Let be a group.

**(a)** Let be defined by for all . Prove that is a homomorphism if and only if is abelian.

*Proof. *Suppose that is a homomorphism. Let . Then . Also, . Thus, . Multiplying on the left by and on the right by we have that .

Conversely, suppose that is abelian. Let . Then . Where the second equality holds because and commute.

**(b)** If is abelian and finite, show that is an automorphism if and only if has odd order.

*Proof. *Suppose that is an automorphism. Seeking a contradiction assume that does not have odd order. Then is a prime which divides , so by Cauchy’s theorem, has an element of order . Let be this element. Then , so is a nontrivial element in the kernel of . This, of course, is a contradiction because is injective and so must be the only element in the kernel of .

Conversely, suppose that has odd order. Since is abelian, by part (a) we know that is a homomorphism. Moreover, because and , being injective will imply being surjective. Thus, it suffices to show that is injective. We do this by showing that the kernel of is trivial. Suppose that is in the kernel of . Then . Thus must be or . But, the order of must divide the order of . Since has odd order, so . Thus, the kernel of is trivial.

**2. **Let be a group and let . Prove: If is a subgroup of and , then is a normal subgroup of .

*Proof. *Let and . We are given that is a subgroup so exists and it suffices to show that is normal. We have that . But so for some . Thus, because is closed.

**3. **Prove that in an integral domain every prime element is an irreducible.

*Proof. *Let be a prime element of . Then the ideal generated by , , is a prime ideal. Suppose that . Then and so or . Thus, or for some . By cancellation in an integral domain we see that or . In either case is a unit or is a unit which means that is an irreducible.

**4. **Find necessary and sufficient conditions on such that the matrix

is diagonalizable over .

*Solution. *Recall, an matrix is diagonalizable if and only if the sum of the dimensions of the eigenspaces is equal to . Let

First, we find the eigenvalues of . Computing the characteristic polynomial of , we obtain . The eigenvalues of are then and .

Next we find the corresponding eigenspaces.

For we solve the system given by and get that and . Thus,

is a basis for the eigenspace corresponding to .

For we solve the system given by and get that is free, , and .

If then and

would be our basis for the eigenspace corresponding to . In this case, would not be diagonalizable.

If then

would be our basis for the eigenspace corresponding to . In this case, would be diagonalizable.

In conclusion, is diagonalizable if and only if .

**5. **Let be the vector space of matrices with entries in and let and denote the set of real symmetric and skew-symmetric matrices, respectively.

**(a) **Show that the dimension of is . A brief justification is sufficient.

*Proof. *Let

be an element of . Then

.

Thus,

…and so on. By counting, we see that there must be elements in our basis for .

**(b) **Let be the linear transformation defined by for all . Prove that and .

*Proof. *First we show that . Let . Then and so . Thus, . Now, let . Then and so . Thus, .

Next we show that . Let . Then there exists such that . This implies that and so . Now, let . Then and so . Thus, .

**(c) **Compute the dimension of .

*Solution. *By the rank nullity theorem we have that

.

In particular,

.

Thus, .

I had my algebra qual yesterday. These are the problems along with my solutions. Hopefully they are correct

Some preliminary comments. First, I know my solution to 1(a) is incorrect. Second, I didn’t know how to prove problem 3. I literally wrote down garbage so I am expecting zero points for that one.

**1.** Let denote the additive group of rational numbers.

**a)** Prove that every finitely generated subgroup of is cyclic.

*Proof. *Let be a finitely generated subgroup of and let be the generators of . Pick . We want to show that can be generated by a single in . Then for some positive integer . But, . Thus, .

**b) **Prove that is not finitely generated.

*Proof.* Seeking a contradiction, suppose that is finitely generated. Note that is a subgroup of itself. Thus, is a finitely generated subgroup of and by part (a), this means that must be cyclic. This of course is a contradiction because is not cyclic.

**2. **List, up to isomorphism, all abelian groups of order .

*Solution.* The partitions of are , , and . The abelian groups associated with are:

The partitions of are and . The abelian groups associated with are:

The abelian groups associated with are:

By the fundamental theorem of finitely generated abelian groups, all abelian groups of order are isomorphic to one of the groups listed below and none of the groups is isomorphic to another.

**3. **Let be a vector space over a field and let be a linear operator such that . Prove that .

*Proof.* 😦

**4.** Let , be matrices with entries in a field and suppose that . Prove that if has an eigenspace of dimension one, then and share a common eigenvector.

*Proof.* Since has an eigenspace of dimension one, let be an eigenvalue of and let be the one eigenvector corresponding to . Then we have that and . Thus, which means that is an eigenvector of with corresponding eigenvalue . But is the only eigenvector corresponding to so . This shows that is an eigenvector of with corresponding eigenvalue .

**5. **Let be a principal ideal domain. Prove that every proper nontrivial prime ideal is maximal.

*Proof.* Let be a proper nontrivial prime ideal of . Let be an ideal of such that . Then for some in . Thus, . Since is prime, or .

Case 1: Suppose that . Then so .

Case 2: Suppose that . Then for some in . Thus, implies that and so is a unit. Thus, .

We have shown that or which means that is a maximal ideal of .

Each problem is worth 5 points, 15 is a passing score. Best case I think my score will be 2+5+0+5+5 which is 17 and a pass. Realistically, I think my score could be 1+5+0+4+5 which is 15 and a pass. But it could be 1+4+0+4+4 which is a 13 and not a pass. I feel very confidant about problems 2, 4, and 5; those should be five points each. But the qual committee can be very particular about grading. I also feel very confidant about 1(b); I just don’t know how many points that one is worth. All I can do now is wait…

**Definition. **A vector space is *finite dimensional* if it has a basis consisting of a finite number of vectors. The unique number of vectors in each basis for is called the *dimension* of and is denoted by .

**Definition. **Let and be vector spaces and let be linear. The *null space* (or kernel) of is the set of all vectors such that ; that is, . The *range* (or image) of is the subset of consisting of all images, under , of vectors in ; that is, .

**Definition. **Let and be vector spaces and let be linear. The *nullity *of is the dimension of denoted by . The* rank* of is the dimension of denoted by .

**Rank Nullity Theorem. **Let and be vector spaces and let be linear. If is finite dimensional, then .

*Proof.* Suppose that forms a basis for . We can extend this to form a basis for , . Since and , it suffices to show that . I claim that forms a basis for .

First we show that spans . Let . Then there exist unique scalars and such that

.

Thus,

.

Since each is in , we have that and so . This shows that spans .

Next we show that is linearly independent. Suppose that

for scalars . Since is linear, we have that

.

Thus, is in . Since span , there exist scalars such that

.

Which implies that

.

But, is a basis for which means that all the and must be zero. Thus, is linearly independent and is indeed a basis for . This proves that as desired.

While studying for the algebra qual, I stumbled upon the following problem:

Let be a group of order where is an odd prime. Prove that contains a nontrivial normal subgroup.

The problem is simple enough, however, it uses many results from group theory. We’ll investigate the problem in detail bellow.

**Cosests and Lagrange’s Theorem**

**Definition. **Let be a subgroup of a group . The subset of is the *left coset* of containing . The subset is the* right coset* of containing .

Cosets are important because they allow us to define normal subgroups, which in turn allow us to construct factor groups. We will need two important results concerning cosets.

If , then the set of left cosets of form a partition of . Indeed, let the relation be defined on by if and only if . It is easily verified that is an equivalence relation on with equivalence class . Moreover, if and only if . This result follows a simple set containment argument along with the definition of left coset. It should be noted that both of these results have an analogue for right cosets.

**Lagrange’s Theorem. **If is a finite group and is a subgroup of , then the order of divides the order of .

*Proof. *Let and . I claim that every left coset of has order . Define the function by . Then is surjective by the definition of as . Suppose that for and in . Then and by the cancellation law of a group, . Thus, is injective. Since is a bijection, we conclude that and have the same order and the claim is established. Since the left cosets of form a partition of , we have that

.

Thus,

.

Since the order of each left coset of is equal to , we have that

where is summed up times. That is, which shows that is a divisor of .

In our proof, to establish that every left coseet of has the same order as we showed that the function was a bijection. It turns out that every right coset of also has the same order as . To establish this, check that defined by is a bijection.

Lagrange’s theorem is perhaps one of the most useful results in group theory since it allows us to actually compute orders of subgroups. The converse of Lagrange’s theorem is not true. The classic counter example is the alternating group , which is the group of even permutations on elements. The order of is and is a divisor of , but has no subgroup of order . There are however some partial converses to Lagrange’s theorem. A partial converse which holds for arbitrary finite groups is the following result.

**Cauchy’s Theorem.** If is a finite group and is a prime dividing , then has an element of order .

Cauchy’s theorem will be useful for us in proving our original problem.

**Normal Subgroups**

**Definition. **A subgroup of a group is *normal* if for all . If is a normal subgroup of then we shall write .

The next theorem is of interest to us since it will be essential for proving our original problem.

**Theorem 1. **Let be a group of order . If is a subgroup of and , then .

*Proof. *Let . Then , , and . Since , we have that and . The elements of cosets are* *distinct because they from a partition of , thus, it must be the case that .

**The Original Problem**

Let be a group of order where is an odd prime. Prove that contains a nontrivial normal subgroup.

*Proof.* By Cauchy’s theorem, there exists an element such that . Then the cyclic subgroup is nontrivial and has order . By theorem 1, is a normal subgroup of .

The proof is quite elegant, but as you have seen, we needed to recall some important results in order achieve such a simple proof.

**Theorem.** Let be a commutative ring with unity. The ideal of is a maximal ideal if and only if the quotient ring is a field.

*Proof. *Suppose that is a maximal ideal of . Let . It suffices to show that has a multiplicative inverse. Consider the set . I claim that is an ideal of that properly contains . Indeed, we check the following:

*Nonempty. *Since contains the element , we have that is nonempty.

*Subtraction.* Let . Then we have that

.

Since is a ring and is an ideal, and . Thus, .

*Multiplication.* Let and . Since is commutative, it suffices to show that . We have that

.

Since is a ring and is an ideal, and . Thus,

*Properly contains .* By definition of , . Moreover, notice that and by assumption, . Thus, is a proper subset of

Therefore, is as claimed. Now, since is maximal, we must have that . Thus , say, . Then,

We conclude that is the multiplicative inverse of ; hence, is a field.

Conversely, suppose that is a field. Let be an ideal of that properly contains (i.e. ) and let . Then is a nonzero element of and so there exists an element such that , the multiplicative identity of . Since , , and is an ideal, we have that . Moreover, since

,

we have that . Thus, and since contains it must be that . We conclude that is a maximal ideal of .

**Theorem.** A subgroup of a cyclic group is cyclic.

*Proof.* Let be a cyclic group generated by and let . If then is cyclic. If , then for . Let be the smallest integer in such that .

I claim that generates ; that is, . We must show that every is a power of . Let . Then since , for some positive integer . By the division algorithm for , there exist such that where . Then so . We must have that because is a group and . Since is the smallest positive integer such that and , it must be the case that . Thus, , , and so is a power of .

The first isomorphism theorem is probably the most important result one will learn in a first course on algebra. We will state and prove the theorem for groups. There is, of course, an analogous result for rings. We will state the theorem for rings (the proof is identical to that for groups) and then look at an example using the theorem.

**Theorem. **If is a homomorphism of groups, then is a normal subgroup of and .

The proof is not terribly complicated; in fact, it is rather “computational”, meaning, we just have to verify a few definitions. It is just a matter of checking that the kernel satisfies the definition of normal and then defining a function from to and checking that the function satisfies the definition of an isomorphism. The “hardest” part of the proof, I think, is actually defining a function from to . However, as we will see, the choice of such a function is quite natural.

*Proof.* First, we will show that the kernel of is a normal subgroup of *. *We check the following:

*Nonempty. *Since is a homomorphism, it must map the identity in to the identity in . Thus, which shows that the kernel is nonempty.* *

*Closure. *Let and be in the kernel of . We want to show that is in the kernel too. We compute that

Thus, is indeed in the kernel of , which shows that is a subgroup of .

*Normal.* Let be an element of and let be an element in the kernel of . It suffices to show that is also in the kernel. We have that

Thus, is in the kernel of as desired. This shows that is a normal subgroup of .

Next, we will show that is isomorphic to . Define by . I claim that is a well-defined isomorphism. Indeed, we check the following:* *

*Well-Defined. *Suppose that . Then for some element in the kernel of . We wish to show that Evaluating we see,

Thus, is well defined.* *

*Homomorphism. *Let and be cosets in the factor group . By multiplying cosets we observe that Consequentially,

Thus, is a homomorphism.* *

*Injective. *Suppose that . We wish to show that . Consider the following,

Since is an element of and is a group, exists. Therefore,

which shows that is in the kernel of . It follows that and so is injective.

*Surjective. *Let be an element in the image of . Then there exists an element in such that . Thus, which shows that is surjective.

We conclude that is a bijective homomorphism, hence an isomorphism.

We now state the analogous result for rings.

**Theorem. **If is a homomorphism of rings, then is an ideal of and .

**Example.** Let be a commutative ring with unity. Prove that

Define the function to be the evaluation homomorphism for any in . By the first isomorphism theorem, it suffices to check that is a surjective homomorphism with . We check the following:

*Homomorphism. *Let and be in . We compute that

.

Thus, is indeed a ring homomorphism.

*Surjective. *Let be an element in . Then is in and which shows that is surjective.

*Kernel.* Let be in the kernel of . Then which means that is a factor of and so is in the ideal . Therefore, . Now, let be in the ideal . Then for some in . Thus, which shows that is in the kernel of . Therefore, . We conclude that .

By the first isomorphism theorem we have that