Lockhart’s Lament

I thought I posted this a while back, but apparently I didn’t. This is perhaps the single best essay that I have ever read. The link above takes you to an MAA page with an intro by Keith Devlin, a link to the essay is near the bottom of the page.

Today is Bob Dylan’s birthday, he’s 70. Be sure to listen to your favorite Dylan song today! This is one of my favorite videos of him

My friend first told me about Sal Khan and the Khan Academy about a month ago. I must say, I was skeptical (and still am), but this definitely has potential. I think something like this is best used to supplement teaching that goes on in the classroom. I think I’ll mention this site to my students and see what they think.

Education: High school physics teacher has perfected the formula for inspiration – latimes.com

This is a really cool story. Amir Abo-Shaeer’s engineering academy was brand new when I was at Dos Pueblos. It wasn’t so much of an academy then, Mr. Abo-Shaeer was able to teach one robotics class. Now, seven years later, Dos Pueblos’ engineering academy has really taken off. Great work Mr. Abo-Shaeer!

One of my favorite songs by Simon and Garfunkel. It was also featured in one of my favorite movies. Happy April and happy spring everyone.

A couple of notes: Because of copyright stuff (blah blah) you can only watch the video on youtube. Also, did you know that Art Garfunkel has a masters degree in math from Columbia University? That’s pretty cool.

This was something I worked on a while back but never got around to posting. This is my rendition of Ansel Adams’ Rose and Driftwood.

I didn’t have any driftwood, so my desk had to suffice. Also, origami roses are kind of difficult to make. I was quite proud of how mine turned out. Here’s the real thing for comparison.

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 $\mathbb{Q}$ denote the additive group of rational numbers.

a) Prove that every finitely generated subgroup of $\mathbb{Q}$ is cyclic.

Proof. Let $H$ be a finitely generated subgroup of $\mathbb{Q}$ and let $\{h_1,\ldots,h_m\}$ be the generators of $H$. Pick $x\in H$. We want to show that $x$ can be generated by a single $h$ in $H$. Then $x=n(h_1+\ldots+h_m)$ for some positive integer $n$. But, $h_1+\ldots+h_m=\tilde{h}\in H$. Thus, $x=n\tilde{h}$. $\Box$

b) Prove that $\mathbb{Q}$ is not finitely generated.

Proof. Seeking a contradiction, suppose that $\mathbb{Q}$ is finitely generated. Note that $\mathbb{Q}$ is a subgroup of itself. Thus, $\mathbb{Q}$ is a finitely generated subgroup of $\mathbb{Q}$ and by part (a), this means that $\mathbb{Q}$ must be cyclic. This of course is a contradiction because $\mathbb{Q}$ is not cyclic. $\Box$

2. List, up to isomorphism, all abelian groups of order $3528=2^3\cdot3^2\cdot7^2$.

Solution. The partitions of $3$ are $3$, $2+1$, and $1+1+1$. The abelian groups associated with $2^3$ are:

• $\mathbb{Z}_8$
• $\mathbb{Z}_4 \times \mathbb{Z}_2$
• $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$

The partitions of $2$ are $2$ and $1+1$. The abelian groups associated with $3^2$ are:

• $\mathbb{Z}_9$
• $\mathbb{Z}_3 \times \mathbb{Z}_3$

The abelian groups associated with $7^2$ are:

• $\mathbb{Z}_{49}$
• $\mathbb{Z}_7 \times \mathbb{Z}_7$

By the fundamental theorem of finitely generated abelian groups, all abelian groups of order $3528$ are isomorphic to one of the groups listed below and none of the groups is isomorphic to another.

1. $\mathbb{Z}_8 \times \mathbb{Z}_9 \times \mathbb{Z}_{49}$
2. $\mathbb{Z}_8 \times \mathbb{Z}_9 \times \mathbb{Z}_7 \times \mathbb{Z}_7$
3. $\mathbb{Z}_8 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_{49}$
4. $\mathbb{Z}_8 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_7 \times \mathbb{Z}_7$
5. $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_9 \times \mathbb{Z}_{49}$
6. $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_9 \times \mathbb{Z}_7 \times \mathbb{Z}_7$
7. $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_{49}$
8. $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_7 \times \mathbb{Z}_7$
9. $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_9 \times \mathbb{Z}_{49}$
10. $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_9 \times \mathbb{Z}_7 \times \mathbb{Z}_7$
11. $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_{49}$
12. $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_7 \times \mathbb{Z}_7$ $\Box$

3. Let $V$ be a vector space over a field $F$ and let $T:V\to V$ be a linear operator such that $T^2=T$. Prove that $V=\text{ker}(T)\oplus\text{im}(T)$.

Proof. 😦

4. Let $A$, $B$ be $n\times n$ matrices with entries in a field $F$ and suppose that $AB=BA$. Prove that if $B$ has an eigenspace of dimension one, then $A$ and $B$ share a common eigenvector.

Proof. Since $B$ has an eigenspace of dimension one, let $\lambda$ be an eigenvalue of $B$ and let $v$ be the one eigenvector corresponding to $\lambda$. Then we have that $ABv=BAv$ and $ABv=A\lambda v=\lambda Av$. Thus, $BAv=\lambda Av$ which means that $Av$ is an eigenvector of $B$ with corresponding eigenvalue $\lambda$. But $v$ is the only eigenvector corresponding to $\lambda$ so $Av=v$. This shows that $v$ is an eigenvector of $A$ with corresponding eigenvalue $1$. $\Box$

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

Proof. Let $(p)$ be a proper nontrivial prime ideal of $D$. Let $(a)$ be an ideal of $D$ such that $(p)\subseteq(a)$. Then $p=ab$ for some $b$ in $D$. Thus, $ab\in(p)$. Since $(p)$ is prime, $a\in(p)$ or $b\in(p)$.

Case 1: Suppose that $a\in(p)$. Then $(a)\subseteq(p)$ so $(p)=(a)$.

Case 2: Suppose that $b\in(p)$. Then $b=pu$ for some $u$ in $D$. Thus, $p=apu$ implies that $1=au$ and so $a$ is a unit. Thus, $(a)=D$.

We have shown that $(p)=(a)$ or $(a)=D$ which means that $(p)$ is a maximal ideal of $D$. $\Box$

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 $V$ is finite dimensional if it has a basis consisting of a finite number of vectors. The unique number of vectors in each basis for $V$ is called the dimension of $V$ and is denoted by $\text{dim}(V)$.

Definition. Let $V$ and $W$ be vector spaces and let $T:V\to W$ be linear. The null space (or kernel) $N(T)$ of $T$ is the set of all vectors $x\in V$ such that $T(x)=0$; that is, $N(T)=\{x\in V:T(x)=0\}$. The range (or image) $R(T)$ of $T$ is the subset of $W$ consisting of all images, under $T$, of vectors in $V$; that is, $R(T)=\{T(x):x\in V\}$.

Definition. Let $V$ and $W$ be vector spaces and let $T:V\to W$ be linear. The nullity of $T$ is the dimension of $N(T)$ denoted by $\text{nullity}(T)$. The rank of $T$ is the dimension of $R(T)$ denoted by $\text{rank}(T)$.

Rank Nullity Theorem. Let $V$ and $W$ be vector spaces and let $T:V\to W$ be linear. If $V$ is finite dimensional, then $\text{dim}(V)=\text{rank}(T)+\text{nullity}(T)$.

Proof. Suppose that $\{u_1,\ldots,u_m\}$ forms a basis for $N(T)$. We can extend this to form a basis for $V$, $\{u_1,\ldots,u_m,w_1,\ldots,w_n\}$. Since $\text{nullity}(T)=m$ and $\text{dim}(V)=m+n$, it suffices to show that $\text{rank}(T)=n$. I claim that $S=\{Tw_1,\ldots,Tw_n\}$ forms a basis for $R(T)$.

First we show that $S$ spans $R(T)$. Let $v\in V$. Then there exist unique scalars $a_1,\ldots,a_m$ and $b_1,\ldots,b_n$ such that

$v=a_1u_1+\ldots+a_mu_m+b_1w_1+\ldots+b_nw_n$.

Thus,

$Tv=a_1Tu_1+\ldots+a_mTu_m+b_1Tw_1+\ldots+b_nTw_n$.

Since each $u_i$ is in $N(T)$, we have that $Tu_i=0$ and so $Tv=b_1Tw_1+\ldots+b_nTw_n$.  This shows that $S$ spans $R(T)$.

Next we show that $S$ is linearly independent. Suppose that

$c_1Tw_1+\ldots+c_nTw_n=0$

for scalars $c_i$. Since $T$ is linear, we have that

$T(c_1w_1+\ldots+c_nw_n)=0$.

Thus, $c_1w_1+\ldots+c_nw_n$ is in $N(T)$. Since $u_i$ span $N(T)$, there exist scalars $d_i$ such that

$c_1w_1+\ldots+c_nw_n=d_1u_1+\ldots+d_mu_m$.

Which implies that

$c_1w_1+\ldots+c_nw_n-(d_1u_1+\ldots+d_mu_m)=0$.

But, $\{u_1,\ldots,u_m,w_1,\ldots,w_n\}$ is a basis for $V$ which means that all the $c_i$ and $d_i$ must be zero. Thus, $S$ is linearly independent and is indeed a basis for $R(T)$. This proves that $\text{rank}(T)=n$ as desired. $\Box$

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 $G$ be a group of order $2p$ where $p$ is an odd prime. Prove that $G$ 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 $H$ be a subgroup of a group $G$. The subset $aH=\{ah:h\in H\}$ of $G$ is the left coset of $H$ containing $a$. The subset $Ha=\{ha:h\in H\}$ is the right coset of $H$ containing $a$.

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 $H\leq G$, then the set of left cosets of $H$ form a partition of $G$. Indeed, let the relation $\sim_L$ be defined on $G$ by $a\sim_L b$ if and only if $a^{-1}b\in H$. It is easily verified that $\sim_L$ is an equivalence relation on $G$ with equivalence class $aH$. Moreover, $aH=H$ if and only if $a\in H$. 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 $G$ is a finite group and $H$ is a subgroup of $G$, then the order of $H$ divides the order of $G$.

Proof. Let $|G|=m$ and $|H|=n$. I claim that every left coset of $H$ has order $n$. Define the function $\varphi:H\to gH$ by $\varphi(h)=gh$. Then $\varphi$ is surjective by the definition of $gH$ as $\{gH:h\in H\}$. Suppose that $\varphi(h_1)=\varphi(h_2)$ for $h_1$ and $h_2$ in $H$. Then $gh_1=gh_2$ and by the cancellation law of a group, $h_1=h_2$. Thus, $\varphi$ is injective. Since $\varphi$ is a bijection, we conclude that $H$ and $gH$ have the same order and the claim is established. Since the left cosets of $H$ form a partition of $G$, we have that

$G = g_1H \cup g_2H \cup \cdots \cup g_rH$.

Thus,

$|G| = |g_1H| + |g_2H| + \ldots + |g_rH|$.

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

$|G| = m = n + n + \ldots + n$

where $n$ is summed up $r$ times. That is, $m=rn$ which shows that $n$ is a divisor of $m$. $\Box$

In our proof, to establish that every left coseet of $H$ has the same order as $H$ we showed that the function $\varphi(h)=gh$ was a bijection. It turns out that every right coset of $H$ also has the same order as $H$. To establish this, check that $\varphi:H\to Hg$ defined by $\varphi(h)=hg$ 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 $A_4$, which is the group of even permutations on $4$ elements. The order of $A_4$ is $12$ and $6$ is a divisor of $12$, but $A_4$ has no subgroup of order $6$. 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 $G$ is a finite group and $p$ is a prime dividing $|G|$, then $G$ has an element of order $p$.

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

Normal Subgroups

Definition. A subgroup $N$ of a group $G$ is normal if $gN=Ng$ for all $g\in G$. If $N$ is a normal subgroup of $G$ then we shall write $N\unlhd G$.

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

Theorem 1. Let $G$ be a group of order $2n$. If $H$ is a subgroup of $G$ and $|H|=n$, then $H\unlhd G$.

Proof. Let $a\in G\setminus H$. Then $aH\neq H$, $Ha\neq H$, and $|aH|=|Ha|=n$. Since $|G|=2n$, we have that $G=H\cup aH$ and $G=H\cup Ha$. The elements of cosets are distinct because they from a partition of $G$, thus, it must be the case that $aH=Ha$. $\Box$

The Original Problem

Let $G$ be a group of order $2p$ where $p$ is an odd prime. Prove that $G$ contains a nontrivial normal subgroup.

Proof. By Cauchy’s theorem, there exists an element $a\in G$ such that $|a|=p$. Then the cyclic subgroup $\langle a \rangle$ is nontrivial and has order $p$. By theorem 1, $\langle a \rangle$ is a normal subgroup of $G$. $\Box$

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