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Today is the birthday of my favorite Beatle, George Harrison. He would have been 68. Be sure to listen to your favorite George Harrison song today! This is one of mine

Theorem. A subgroup of a cyclic group is cyclic.

Proof. Let G be a cyclic group generated by a and let H\leq G. If H=\{e\} then H=\langle e \rangle is cyclic. If H\neq\{e\}, then a^n\in H for n\in\mathbb{Z}^+. Let m be the smallest integer in \mathbb{Z}^+ such that a^m\in H.

I claim that c=a^m generates H; that is, H=\langle a^m \rangle = \langle c \rangle. We must show that every b\in H is a power of c. Let b\in H. Then since H\leq G, b=a^n for some positive integer n. By the division algorithm for \mathbb{Z}, there exist q, r \in \mathbb{Z} such that n=mq+r where 0\leq r < m. Then a^n=a^{mq+r}=a^{mq}a^r so a^r=a^na^{-mq}. We must have that a^r\in H because H is a group and a^n, a^{-mq} \in H. Since m is the smallest positive integer such that a^m\in H and 0\leq r < m, it must be the case that r=0. Thus, n=mq, b=a^n=a^{mq}=c^q, and so b is a power of c. \Box

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 \varphi : G \to H is a homomorphism of groups, then \text{ker}(\varphi) is a normal subgroup of G and G/\text{ker}(\varphi)\cong\varphi(G).

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 G/\text{ker}(\varphi) to \varphi(G) 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 G/\text{ker}(\varphi) to \varphi(G). However, as we will see, the choice of such a function is quite natural.

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

Nonempty. Since \varphi is a homomorphism, it must map the identity in G to the identity in H. Thus, 1_G\in\text{ker}(\varphi) which shows that the kernel is nonempty.

Closure. Let a and b be in the kernel of \varphi. We want to show that ab^{-1} is in the kernel too. We compute that

\varphi(ab^{-1})=\varphi(a)\varphi(b)^{-1}=(1_H)(1_H)^{-1}=1_H.

Thus, ab^{-1} is indeed in the kernel of \varphi, which shows that \text{ker}(\varphi) is a subgroup of G.

Normal. Let g be an element of G and let k be an element in the kernel of \varphi. It suffices to show that gkg^{-1} is also in the kernel. We have that

\varphi(gkg^{-1})=\varphi(g)\varphi(k)\varphi(g)^{-1}=\varphi(g)1_H\varphi(g)^{-1}=1_H.

Thus, gkg^{-1} is in the kernel of \varphi as desired. This shows that \text{ker}(\varphi) is a normal subgroup of G.

Next, we will show that G/\text{ker}(\varphi) is isomorphic to \varphi(G). Define \psi : G/\text{ker}(\varphi) \to \varphi(G) by \psi(g\text{ker}(\varphi))=\varphi(g). I claim that \psi is a well-defined isomorphism. Indeed, we check the following:

Well-Defined. Suppose that a\text{ker}(\varphi)=b\text{ker}(\varphi). Then a=bk for some element k in the kernel of \varphi. We wish to show that \psi(a\text{ker}(\varphi))=\psi(b\text{ker}(\varphi)). Evaluating \psi(a\text{ker}(\varphi)) we see,

\psi(a\text{ker}(\varphi))=\varphi(a)=\varphi(bk)=\varphi(b)\varphi(k)=\varphi(b)1_H=\psi(b\text{ker}(\varphi)).

Thus, \psi is well defined.

Homomorphism. Let a\text{ker}(\varphi) and b\text{ker}(\varphi) be cosets in the factor group G/\text{ker}(\varphi). By multiplying cosets we observe that a\text{ker}(\varphi)b\text{ker}(\varphi)=ab\text{ker}(\varphi). Consequentially,

\psi(a\text{ker}(\varphi)b\text{ker}(\varphi))=\varphi(ab) =\varphi(a)\varphi(b) =\psi(a\text{ker}(\varphi))\psi(b\text{ker}(\varphi)).

Thus, \psi is a homomorphism.

Injective. Suppose that \psi(a\text{ker}(\varphi))=\psi(b\text{ker}(\varphi)). We wish to show that a\text{ker}(\varphi)=b\text{ker}(\varphi). Consider the following,

\varphi(a)=\psi(a\text{ker}(\varphi))=\psi(b\text{ker}(\varphi))=\varphi(b).

Since \varphi(b) is an element of H and H is a group, \varphi(b)^{-1} exists. Therefore,

\varphi(ab^{-1})=\varphi(a)\varphi(b)^{-1}=1_H,

which shows that ab^{-1} is in the kernel of \varphi. It follows that a\text{ker}(\varphi)=b\text{ker}(\varphi) and so \psi is injective.

Surjective. Let x be an element in the image of \varphi. Then there exists an element g in G such that \varphi(g)=x. Thus, \psi(g\text{ker}(\varphi))=\varphi(g)=x which shows that \psi is surjective.

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

We now state the analogous result for rings.

Theorem. If \varphi : R \to S is a homomorphism of rings, then \text{ker}(\varphi) is an ideal of R and R/\text{ker}(\varphi)\cong\varphi(R).

Example. Let R be a commutative ring with unity. Prove that R[x] / \langle x \rangle \cong R.

Define the function \varphi:R[x]\to R to be the evaluation homomorphism \varphi(f(x))=f(0) for any f(x) in R[x]. By the first isomorphism theorem, it suffices to check that \varphi is a surjective homomorphism with \text{ker}\varphi=\langle x \rangle. We check the following:

Homomorphism. Let f(x) and g(x) be in R[x]. We compute that

\varphi((f+g)(x))=(f+g)(0)=f(0)+g(0)=\varphi(f(x))+\varphi(g(x))

\varphi((fg)(x))=(fg)(0)=f(0)g(0)=\varphi(f(x))\varphi(g(x)).

Thus, \varphi is indeed a ring homomorphism.

Surjective. Let r be an element in R. Then f(x)=x+r is in R[x] and \varphi(f(x))=f(0)=0+r=r which shows that \varphi is surjective.

Kernel. Let f(x) be in the kernel of \varphi. Then f(0)=0 which means that x is a factor of f(x) and so f(x) is in the ideal \langle x \rangle. Therefore, \text{ker}\varphi\subseteq\langle x \rangle. Now, let g(x) be in the ideal \langle x \rangle. Then g(x)=xf(x) for some f(x) in R[x]. Thus, \varphi(g(x))=g(0)=0f(0)=0 which shows that g(x) is in the kernel of \varphi. Therefore, \langle x \rangle\subseteq\text{ker}\varphi. We conclude that \text{ker}\varphi=\langle x \rangle.

By the first isomorphism theorem we have that R[x] / \langle x \rangle \cong R. \Box

My blog doesn’t really have a theme; it’s just a random collection of things that are going on in my life that I want to share. I keep the topics pretty broad so more people will be inclined to read. As a result, the math posts tend to be low. I don’t want my blog to be a “math blog” (you can find a list of such blogs here, or just google “math blogs”). It should be noted that online collaborative mathematics has become quite popular recently. Sites like Math Overflow and Terence Tao’s blog are great examples.

Having said that, I’ve decided that my next few posts will be more math related; in particular, I’ll be posting a couple of algebra theorems each week for the next few weeks. Here’s why. The algebra qualifying exam is in about a month and I’ve been studying really hard for it. I know that some of the other TA’s are studying really hard for it too; and since some of them occasionally glance at my blog (I know, I can’t believe it either), I think it’s worth while to share some of the key algebra theorems.

For the mathematically inclined, please feel free to comment and critique any of the results you see posted. A lot of the proofs are my own, which means there might (will) be errors. Enjoy.

The NBA All-Star Game was this past weekend, which means, slam dunk contest. If you didn’t hear by now, Blake Griffin jumped over a car and won the contest (you can find videos of it all over the internet). But you probably don’t know how the alley-oop was invented. This video should enlighten you.

This is a broadcast from last year that NPR ran for Valentine’s Day. It’s too good to pass up. Be sure to listen to the story, don’t just read the article.

Radiolab: Carl Sagan And Ann Druyan’s Ultimate Mix Tape Of The Human Experience : NPR.

I’ve recently been on an origami craze. Even though origami is an ancient Japanese art, a lot has been written about the mathematics of origami and paper folding (as a simple Google search will show). Origami, and paper folding in general, has many interesting algebraic, and of course, geometric properties. The mathematics in many of the articles I’ve seen is pretty complicated and I won’t even try to pretend that I fully understand it. One of the AMS Notices from last year had an article that “use[d] origami as a tool to exhibit explicit solutions to some systems of partial differential equations”. If you are at all interested, you can find the article here.

As far as my interests in origami go, yeah there’s some pretty deep mathematics, but I think it’s a cool hobby to have and I can make some cool decorations for my desk.

A stereographic projection is a mapping that projects a sphere onto a plane. Stereographic projections occur in many areas of math. In fact, this past summer when I was studying minimal surfaces, we used a stereographic projection in the proof of Berstein’s Theorem (a major result in minimal surface theory). More recently, in the topology class I am currently enrolled in, a stereographic projection is used to show that the unit circle is the one point compactification of the real line.

Stereographic projections have a wide variety of uses in the math world, so what? Well, it turns out there are some pretty cool applications of them outside the pure math world. The most practical example is in the making of world maps. Since the Earth is a sphere, in order to project the Earth onto a flat piece of paper, to make a map, a stereographic projection is used.

But this is what’s really cool…

We can go the other way, project a plane onto a sphere. So imagine this, you take a picture, which is flat. We can project that image onto a sphere to get something that looks like this

photo courtesy of wikipedia

A fellow math peep introduced me to a really cool website, thesixtyone.com. You can friend me if you want, I’m rkneufeld. But I think the friending is done through facebook, which I don’t have. Regardless, it’s a great site for discovering new music, much better than pandora. Check it out!