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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.* ðŸ˜¦

**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 . Â This shows that spans .

Next we show that is linearly independent. Suppose that

for scalars . Since is linear, we have that

.

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

.

Which implies that

.

But,Â 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.

I Googled math jokes the other day (because I’m a nerd) and I came across this:

I took this from loveallthis. A few of comments. First, I really love this. Second, I don’t understand tumblr. Third, as much as I love this, the flowchart is incorrect.

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.

Exciting news, I’ll be teaching calculus next quarter!

…kind of.

There is a bit of a catch. I’ll be teaching calculus for business and economics. Commonly referred to as business calc, or (as the TA’s like to call it) busy calc. The catch is that there is no trig (how can you do anything fun in calculus without trig?) and there will be a lot of applications to “business problems”. Regardless, it is still calculus and I am excited!

Keith Devlin is a professor at Stanford but probably better known as the NPR “math guy”. I really like his latest piece on *Weekend Edition.*

The Way You Learned Math Is So Old School : NPR

Remember going through your multiplication tables in elementary school? Or crunching through long division? Yeah, it sucked and that’s because it’s lame and useless, literally useless. Especially in today’s society where everyone uses a calculator or a computer to carry out arithmetic calculations. Computers do arithmetic for us, Devlin says, but making computers do the things we want them to do requires algebraic thinking. Elementary schools are starting to notice this and it’s changing the way arithmetic is taught. The emphasis in teaching mathematics today is on getting people to be sophisticated, algebraic thinkers. This doesn’t mean that we should stop teaching arithmetic in school; arithmetic is the foundation for strong algebraic thinking, it is not however, the end goal.

The next time you slice a bagel you might want to try this! You can slice a bagel in a single cut so that the result is two equal halves linked together. Don’t believe it? George W. Hart, sculptor and mathematician, has instructions you can follow here.

The two halves turn out to be mobius strips. For a better visual check out the video.

**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 .