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December 2

On Monday last week, one of the problems we looked at was the following.

Problem 104 (A Direct Product Of Cyclic Groups Is Cyclic If And Only If The Groups Have Relatively Prime Orders)

Suppose that $G$ and $H$ are finite cyclic groups. Prove that $G\oplus H$ is cyclic if and only if $|G|$ and $|H|$ are relatively prime.

Your assignment for Monday is to extent this problem to work with the direct product of a finite number of cyclic groups.

Exercise (A Direct Product Of Cyclic Groups Is Cyclic If And Only If The Groups Have Relatively Prime Orders)

Use induction to extend the previous result to the external direct product of $n$ cyclic groups. So prove that if $G_1,G_2,\ldots,G_n$ are $n$ cyclic groups, then $G_1\oplus G_2\oplus \cdots\oplus G_n$ is cyclic if and only if $|G_i|$ and $|G_j|$ are relatively prime when $i\neq j$.

Click to see a hint.

This exercise require that you use induction on an "if $p$ then $q$" statement. In addition, the $q$ part of the statement involves an if and only if. You must be very careful how you tackle this problem.

Problem.Induction Mastery Two

Solve the previous exercise. Pay close attention to your logical order. What are you assuming at each stage? Be very careful with the order in which you place your assumptions, statements, and conclusions. This problem is extremely easy to mess up on.

Once you have a solution, your assignment is to type up your solution using a $\LaTeX$ processor that you have install on your computer, or used on a campus computer (there are a few in the computer lab that have LaTeX). Use this template (latex-template.tex) to type your file.

Here are some instructions for obtaining LaTeX.

  1. If you use Windows, Download MikTeX from http://miktex.org/download and install MikTeX.
    • This will give your compute the ability to build PDF documents from $\LaTeX$ files without needing access to the internet.
    • This will also install a program called TeXWorks where you edit your $\LaTeX$ code (generally saved in a .tex file).
  2. If you use Mac, then download MacTeX from http://tug.org/mactex/. There should be an option to install TeXWorks as part of your installation. Do this.
  3. If you use Linux (Caley), then I followed the information on http://askubuntu.com/questions/163682/how-do-i-install-the-latest-tex-live-2012 to install TeX. You'll then need an editor (like TeXWorks, see http://www.tug.org/texworks/).

Once you have a working copy of LaTeX, you should be able to download the following template (latex-template.tex), double click on the download to open TeXWorks, and then type your solution and build your PDF file.

December 4

This first problem is a repeat from Monday last week. Brennan showed much of this work in class. I would like to start Wednesday's class by reviewing this problem. Each step of the proof below should be quite short.

Problem (Introduction To Internal Direct Products)

Consider the group $G\times H$. Let $A=G\times \{e_H\}$ and $B=\{e_G\}\times H$. Prove the following.

  1. $A\cap B$ contains only the identity element of $G\times H$.
  2. The set product $AB$ equals the whole group $G\times H$.
  3. $A$ and $B$ are normal subgroups of $G\times H$.
  4. $G\approx A$ and $H\approx B$, which means $G\times H=AB\cong A\times B$.

Did you notice what we did above. We started with a group created by using an external direct product. We then found two subgroup $A$ and $B$ of the group, and showed that the group was equal to the set product $AB$ (a product done internally in the group) and at the same time isomorphic to the external direct product $A\times B$. Given any group $G$, we'd like to know when we can write the group as $G=HK\approx H\oplus K$. The first three conditions from the problem above are the key.

Definition (Internal Direct Product Of Two Subgroups)

Suppose that $G$ is a group. If $H$ and $K$ are normal subgroups of $G$ such that $H\cap K=\{e\}$ and $HK=G$, then we say that $G$ is the internal direct product of $H$ and $K$.

Let's now show that given any two subgroups $H$ and $K$ of a group $G$, that if these subgroups are both normal, their intersection is trivial, and their set product is all of $G$, then their external direct product is isomorphic to $G$.

Problem (Internal Direct Products Are Isomorphic To External Direct Products)

If $G$ is the internal direct product of $H$ and $K$, then prove that the internal direct product $G=HK$ is isomorphic to the external direct product $H\times K$.

Click to see a hint.

We need to build a map from one group to the other, and then prove that the map is an isomorphism (a surjective homomorphism with trivial kernel). The map that takes an element $(h,k)\in H\times K$ and sends it to the element $hk\in HK=G$ should be what we need. It should be clear that the map is surjective (why? Start with something in $G=HK$ and produce an element $(h,k)$ that maps to it.) To prove that this map is a homomorphism, we'll have to show that $hk=kh$. To prove that the map is injective, show that the kernel is trivial. The three properties of being an internal direct product will help.


Given any two finite subgroups $H$ and $K$, there is a simple way to compute the size of their set product. You'll see a similar theorem in combinatorics or probability courses. This is counting principles, and your job is to show that if $x=hk$, then for each $y\in H\cap K$, there is another way to write $x=h_yk_y$, and these are the only ways you can create $x$.

Problem (The Number Of Elements In A Set Product)

Suppose that $G$ is a group and that $H$ and $K$ are finite subgroups of $G$. Prove that $|HK|=|H||K|/|H\cap K|$.


The next problem has you practice working with orders of elements in external direct products, and recognizing when two groups are, or are not, isomorphic. Much of what we'll be doing in the next two weeks has to do with determining when two groups are, or are not, isomorphic.

Problem (Which Groups Of Order 60 Are Isomorphic)

Prove or disprove each of the following. Either build an isomorphism, or show that no such isomorphism exists.

  1. $\mathbb{Z}_4\oplus Z_{15}\approx \mathbb{Z}_{6}\oplus \mathbb{Z}_{10}$
  2. $\mathbb{Z}_4\oplus \mathbb{Z}_{15}\approx \mathbb{Z}_{20}\oplus \mathbb{Z}_{3}$
  3. $D_{20}\oplus \mathbb{Z}_{3}\approx D_{60}$ (Recall that $D_{2n}$ is the automorphisms of a regular $n$-gon.)
  4. $D_{20}\oplus \mathbb{Z}_{3}\approx \mathbb{Z}_{12}\oplus \mathbb{Z}_5$

Click to see a hint.

Consider orders of elements. What's the largest possible order in each group? How many elements of order 2 does each group have? If you find a mismatch between groups, they cannot be isomorphic.


Problem 101 (Homomorphisms Preserve Normal Subgroups)

Suppose that $f:G\to H$ is a homomorphism. Use The Normal Subgroup Test to prove the following:

  1. If $N$ is normal in $G$, then $f(N)$ is normal in $f(G)$.
  2. If $B$ is normal in $H$, then $f^{-1}(B)$ is normal in $G$.

At this point, we've proved quite a large collection of facts about homomorphisms and what they preserve. I strongly suggest that you look at pages 202-204 of your text to see a list of these properties. Then read the remarks on page 204.

The following two facts follow immediately from using the properties of homomorphisms. The first fact shows that any time you have a subgroup that contains half the elements of the group, that subgroup must be a normal subgroup.

Problem 102 (Subgroups Of Index 2 Are Normal)

Suppose that $H$ is a subgroup of $G$ with index $|G:H|=2$. Recall that the index of $H$ in $G$ is the number of distinct cosets of $H$ in $G$.

  1. Build a surjective homomorphism from $G$ to $\mathbb{Z}_2$.
  2. Show that $H$ is normal in $G$.

This next fact shows that any time you have subgroup in a factor group $G/N$, it corresponds to a subgroup $H$ in $G$ that contains $N$. So if we know that $G/N$ has a subgroup of

Problem 103 (Subgroups Of A Quotient Group Correspond To Subgroups Of The Original Group)

Suppose that $N$ is a normal subgroup of $G$ and that $B$ is a subgroup of $G/N$. Prove the following:

  1. There exists a subgroup $H\leq G$ such that $H/N=B$.
  2. If we know that $n=|N|$ and $m=|B|$ (so $n$ is the order of $N$ in $G$, and $m$ is the order of $B$ in $G/N$), then $G$ must have a subgroup $H$ of order $nm$.

As a suggestion, consider the homomorphism $f:G\to G/N$ given by $f(g)=Ng$ and use some properties of homomorphisms.


December 6

Hi all. There are no new prep problems for today. Please work on Wednesday's problems. If you are stuck on them, then please instead see the "Helpful Practice Problems" section below.

As a class, let me tell you that you are doing amazing collectively. Even the weakest students have learned so much, and have demonstrated it. I would say that this course has had a higher comprehension of theory than any proof type course I have ever taught. I am dead serious here. And yet, I think many of you don't realize this. You don't realize how far you have come, and how much you have learned. I must apologize for not helping you see the growth you have made in yourselves. We need something to help you realize how far you have come.

  • On Friday, I plan to have short interviews, 3-5 minutes, with every student in the class. I'll have them in my office, and they'll start before class starts. I want an opportunity with each of you to let you know the good I've seen in you this semester.
  • During class time on Friday, we may have some presentations from students, but a good chunk of class time will be devoted to building your confidence in what you have learned. This is where your book comes in handy. I'll have a collection of problems from the text that I want you to work through together in small groups. Bring your textbook, as we'll need it.
  • For the remainder of the semester, right now I'm thinking that we'll have two different tracks of work that you can follow. There will be new prep problems that lead us to the Fundamental Theorem of Finitely Generated Abelian Groups and the Sylow Theorems. The other track will be to instead work on problems (from the text) to build confidence (I'll point you to problems that should do this). You may choose which track you want to follow. I'll find a way to allow both tracks to carry on meaningful discussions in class.
Helpful Practice problems

Please open up your abstract algebra text. Turn to any of chapters 6-10, and look at the homework problems at the end of the chapter. You might want to start in chapter 6 or 7 (do you feel more confident with isomorphisms or cosets - whichever you feel more confident with, start there). Do 5-10 of the homework problems near the beginning of the homework list. Do ones that look interesting to you. Feel free to swap to a different chapters (I'd move to chapter 10 before going to chapters 8 and 9) and do some from each chapter. I think you'll find that you can do much more than you may have thought. You have learned so much as a class and individually, but some of you don't feel like you have. As your teacher, I've been trying to stretch you each day to help you grow. It's worked, but I haven't done a good enough job helping you see that growth. I'm sorry.

Here's a list of some of the problems to work on in each chapter. I sorted them by my perceived difficulty. Some of the "basic" problems are very computational in nature. If you understand how to do the computations, I wouldn't waste time doing them. If you don't know how to do the computations, then do them. Don't forget that there are solutions to many of these problems (or at least partial solutions) in the back of the book.

Chapter

Basic

Intermediate

Advanced

5 (Permutations)

1-6, 9, 11 - 14, 17-18, 20, 21, 36, 38, 39,

7, 8, 15, 16, 19, 22-24, 27-30, 32-35, 37,

10, 26, 31,

6 (Isomorphisms)

1, 3, 4, 5, 6, 7, 8, 18, 19, 20, 21, 22, 25, 26, 29, 30, 38, 39, 40,

2, 10, 13, 14, 15, 16, 17, 24, 28, 31, 35,

9, 11, 12, 23, 27, 32, 33, 34, 36, 37, 41-47

7 (Cosets)

1-10 (do 10), 12-14, 20, 22, 27-29, 39, 43, 47, 49,

11, 15, 17-19, 21, 25, 26, 30, 31, 37, 38, 40-42,

16, 23, 24, 32-36, 44-46, 48, 50

8 (Direct Products)

1-11, 13-14, 15 (DO THIS ONE - IT'S the crux of the induction mastery II assignment.), 16-18, 20, 21, 23 (Matrices), 24, 26-32, 39, 46-48,

12, 19, 22, 25(this shows up in vector subspaces, topology, and much more), 34 (fun), 35-38, 40, 43-45, 49, 50, 53, 56, 57,

33, 41, 42, 51, 52, 54, 55, 58-75

5-8 Mixed together. See page 174.

9 (Normal Subgroups and Factor Groups)

1-9, 12-14, 16-18, 20, 22-24, 27, 28, 31, 34, 52 (Fun, Everyone should do), 53, 54, 56, 65 (we did in class), 67

3, 10, 11, 15, 19, 21, 26, 32, 37 (key problem), 38, 39, 42, 43, 49, 55, 57, 59, 60, 62, 64, 67, 74, 75

29, 30, 33, 35, 36, 40, 41, 44-48 (Use G/Z theorem to make easy), 50, 51, 58, 63, 66, 69-73,

10 (Homomorphisms)

1-6, 8, 9, 10, 12-18, 21, 24, 26-30, 31-34, 35, 36, 38, 42, 44, 45, 46, 48 (linear algebra application), 54 (orthogonal matrix application),

7, 11, 19, 20, 22 (done in class), 23, 25, 37, 43 (what are the normal subgroups?), 49, 50, 52, 55 (fun use of preimages of normal subgroups), 59, 60,

39, 40, 41, 47, 51, 53, 56, 57, 58 (Introduction to counting the number of elements in $\text{Hom}(A, B)$ and $\text{Hom}(A)$), 61, 62.

8.22, 23, 27, 27, 28

December 9

To prove that $H$ is a subgroup of $G$, we have been using two different options, namely The Subgroup Test and The Finite Subgroup Test. Our version of the subgroup test requires that you show (1) that $H$ is nonempty, (2) that $H$ is closed under the operation, and (3) that $H$ is closed under taking inverse. There is another test that can often result in shorter proofs, though not always easier to use. It's called the "One-Step Subgroup Test" and requires you show that $H$ is nonempty and that $ab^{-1}\in H$ whenever $a$ and $b$ are in $H$.

Problem (One Step Subgroup Test)

Suppose that $H$ is a nonempty subset of a group $G$. Suppose that $ab^{-1}\in H$ whenever $a,b\in H$. Prove that $H$ is a subgroup of $G$.


Many of the ideas we'll be doing today hinge up on the notation of an equivalence relation. It's way of grouping together common objects. Given a huge collection of things, we like to sort them into piles of common things. Each object gets put in exactly one pile, and no piles overlap. This is what we call a partition. We say that things in the same pile are equivalent. Many of you have seen the following definition in your past.

(:include :)

Definition (Equivalence Relation)

Let $S$ be set. Let $\cong$ be a relation on $S$, meaning $\cong$ is a collection $\mathscr{C}$ of ordered pairs of $S$. We way that $A\cong B$ if and only if the ordered pair $(A,B)$ is an element of $\mathscr{C}$. We say that $\cong$ is an equivalence relation if and only if

  1. (Reflexive) For every $A\in S$, we know that $A\cong A$ (so the ordered pair $(A,A)$ is always in $\mathscr{C}$).
  2. (Symmetric) If $A\cong B$, then $B\cong A.$
  3. (Transitive) If $A\cong B$ and $B\cong C$, then $A\cong C$.

When we created cosets of a subgroup $H$, we created an equivalence relation on the elements in a group by saying that two elements in the group are "equivalent" if and only if they are in the same coset. We know that this relation is reflexive because for each $a\in G$, we know that $a$ and $a$ are in the same coset, namely the coset $Ha$. We also know that if $a$ and $b$ are in the same coset, then $b$ and $a$ are in the same coset. Finally, we know that if $a$ and $b$ are in the same coset, and $b$ and $c$ are in the same coset, then $a$ and $c$ must be in the same coset. This partitions the group up into groups of "equivalent" elements. We created the factor group of $G$ by $H$ by just using the set product on these cosets.

The same principle applies to comparing groups. We have way of determining when two groups are "the same". We call the groups isomorphic. The next problem has you show that the notation of isomorphisms creates an equivalence relation on the set of all groups.

Problem (Isomorphisms Yield An Equivalence Relation On The Set Of All Groups)

We say that $A$ is isomorphic to $B$ and write $A\approx B$ if and only if there exists a bijection from $A$ to $B$. In this problem, you'll prove that $\approx$ is an equivalence relation on the set of all groups.

  1. Let $A$ be a group. Show that $A$ is isomorphic to $A$ by building an isomorphism from $A$ to $A$.
  2. Suppose that $A$ is isomorphic to $B$. Prove that $B$ is isomorphic to $A$.
  3. Suppose that $A$ is isomorphic to $B$ and suppose that $B$ is isomorphic to $C$. Use this to produce an isomorphism from $A$ to $C$.

We'll now be preparing ourselves to prove the first Sylow theorem. When we study normal subgroups, we look at the product $x^{-1}ax$ quite often. You also studied this idea in linear algebra when you were studying changing bases. We called two matrices $A$ and $B$ similar in linear algebra if $B=P^{-1}AP$ for some invertible matrix $P$. When two matrices were similar, they represented the exact same linear transformation, just using a different basis. One of the biggest ideas in linear algebra is that if you use the eigenvectors as your basis vectors (the columns of $P$), then the matrix $A$ is similar to a diagonal matrix. Studying the products $P^{-1}AP$ had a huge payoff in linear algebra. It does in group theory as well.

Definition (Conjugacy Class Of A)

Let $G$ be a group. Suppose that $a$ and $b$ are elements of $G$. We say that $a$ and $b$ are conjugate in $G$ if $b=x^{-1}ax$ for some $x\in G$, and we say that $b$ is a conjugate of $a$. The set of conjugates of $a$ is called the conjugacy class of $a$ which we denote by $\text{cl}(a)=\{x^{-1}ax\mid x\in G\}$.


Exercise (Conjugacy Classes Of The Automorphisms Of The Square)

Let $G = D_8$, the automorphisms of the square. Compute the conjugacy classes of $G$.

Click to see a solution.

The conjugacy classes are $$ \begin{align} \text{cl}(R_0)&=\{R_0\}\\ \text{cl}(R_{180})&=\{R_{180}\}\\ \text{cl}(R_{90})&=\{R_{90},R_{270}\}\\ \text{cl}(R_{270})&=\{R_{90},R_{270}\}\\ \text{cl}(H)&=\{H,V\}\\ \text{cl}(V)&=\{H,V\}\\ \text{cl}(D)&=\{D,D'\}\\ \text{cl}(D-)&=\{D,D'\}. \end{align} $$ Notice how this partitions the group. The next problem has you show that conjugacy is an equivalence relation.


You've seen conjugacy classes before when you looked at similar matrices in linear algebra. If two matrices are similar, then lots of things are known about the matrices, in particular that they have the same eigenvalues, and the same dimensions of eigenspaces. There's an entire world of facts to explore about similar matrices, which is a specific example of a conjugacy class.

For now, let's show that conjugacy classes partition the group $G$. As a side note, every idea we learn here about conjugacy classes will immediately apply to matrix groups. After we have shown that conjugacy classes partition the group $G$, we can then count the number of elements of $G$ by instead counting the number of elements in each distinct conjugacy class. This means we have $G = \bigcup \cl(a)$ and hence $|G|=\sum |\text{cl}(a)|$ where the sum only includes each conjugacy class once.

Problem (Conjugacy Is An Equivalence Relation)

Show that conjugacy is an equivalence relation on $G$. In other words, show the following three things.

  1. For each $a\in G$, we know that $a$ is a conjugate of $a$.
  2. If $a$ is a conjugate of $b$, show that $b$ is a conjugate of $a$.
  3. If $a$ is a conjugate of $b$ and $b$ is a conjugate of $c$, show that $a$ is a conjugate of $c$.

Let's now connect conjugacy to centralizers. The center of a group is the set of all $x$ that commute with every element in the groups. The centralize of an element is the collection of all $x$ that commute with $a$.

Definition (Centralizer Of An Element)

Let $a\in G$. The centralizer of $a$ is the collection of all $x\in G$ such that $ax=xa$, or alternately $x^{-1}ax=a$. We write this set symbolically as $$C(a)=\{x\in G\mid ax=xa\} = \{x\in G\mid x^{-1}ax=a\} .$$


Exercise (Centralizers Of The Automorphisms Of The Square)

Let $G = D_8$, the automorphisms of the square. Compute the centralizes $C(a)$ for each $a\in G$.

Click to see a solution.

The centralizers are $$ \begin{align} C(R_0)&=\{R_0,R_{90}, R_{180}, R_{270}, H, V, D, D' \}\\ C(R_{180})&=\{R_0,R_{90}, R_{180}, R_{270}, H, V, D, D' \}\\ C(R_{90})&=\{R_0,R_{90}, R_{180}, R_{270}\}\\ C(R_{270})&=\{R_0,R_{90}, R_{180}, R_{270}\}\\ C(H)&=\{R_0, R_{180}, H, V\}\\ C(V)&=\{R_0, R_{180}, H, V\}\\ C(D)&=\{R_0, R_{180}, D, D'\}\\ C(D-)&=\{R_0, R_{180}, D, D'\}\\. \end{align} $$ Compare this with the exercise about the conjugacy classes of $G$. Do you notice that a large centralizer means a small conjugacy class, and vice versa. The order of the centralizer and the order of the conjugacy class are inversely related. Their product is always the order of the group.


If an element is in the centralizer, then we know that we can use that element to show that $a$ is a conjugate of itself. On your midterm you showed that the centralizer of $a$ is a subgroup of $G$. The next problem has you connect centralizers and conjugacy classes.

Problem (The Number Of Conjugates Of A)

Let $G$ be a group and let $a\in G$. Prove that the number of conjugates of $a$ equals the index of the centralizer of $a$ in $G$. In symbols, we write this as $|\text{cl}(a)|=|G:C(a)|$.

Hint: Recall that the index of $H$ in $G$ is the number of distinct cosets of $H$ in $G$.

Build a map from $\text{cl}(a)$ to the collection of cosets of $C(a)$ in $G$. Try sending $x^{-1}ax\in \text{cl}(a)$ to the coset $C(a)x$. You'll have to show that this map is well defined, meaning that if $x^{-1}ax=y^{-1}ay$, then you must have $C(a)x=C(a)y$. You'll need to then show that this map is a bijection. I would suggest you look up the properties of cosets to see what is equivalent to saying $Hx=Hy$.

One of the easiest ways to show that two sets have the same number of elements is to produce a bijection from one the others.

On a side note, the map you create is NOT a homomorphism. You won't be able to prove that it is. Luckily, you don't need to.

December 11

Inquiry Based Prep Problems

To prove that $H$ is a subgroup of $G$, we have been using two different options, namely The Subgroup Test and The Finite Subgroup Test. Our version of the subgroup test requires that you show (1) that $H$ is nonempty, (2) that $H$ is closed under the operation, and (3) that $H$ is closed under taking inverse. There is another test that can often result in shorter proofs, though not always easier to use. It's called the "One-Step Subgroup Test" and requires you show that $H$ is nonempty and that $ab^{-1}\in H$ whenever $a$ and $b$ are in $H$.

Problem (One Step Subgroup Test)

Suppose that $H$ is a nonempty subset of a group $G$. Suppose that $ab^{-1}\in H$ whenever $a,b\in H$. Prove that $H$ is a subgroup of $G$.


Many of the ideas we'll be doing today hinge up on the notation of an equivalence relation. It's way of grouping together common objects. Given a huge collection of things, we like to sort them into piles of common things. Each object gets put in exactly one pile, and no piles overlap. This is what we call a partition. We say that things in the same pile are equivalent. Many of you have seen the following definition in your past.

Definition (Equivalence Relation)

Let $S$ be set. Let $\cong$ be a relation on $S$, meaning $\cong$ is a collection $\mathscr{C}$ of ordered pairs of $S$. We way that $A\cong B$ if and only if the ordered pair $(A,B)$ is an element of $\mathscr{C}$. We say that $\cong$ is an equivalence relation if and only if

  1. (Reflexive) For every $A\in S$, we know that $A\cong A$ (so the ordered pair $(A,A)$ is always in $\mathscr{C}$).
  2. (Symmetric) If $A\cong B$, then $B\cong A.$
  3. (Transitive) If $A\cong B$ and $B\cong C$, then $A\cong C$.

When we created cosets of a subgroup $H$, we created an equivalence relation on the elements in a group by saying that two elements in the group are "equivalent" if and only if they are in the same coset. We know that this relation is reflexive because for each $a\in G$, we know that $a$ and $a$ are in the same coset, namely the coset $Ha$. We also know that if $a$ and $b$ are in the same coset, then $b$ and $a$ are in the same coset. Finally, we know that if $a$ and $b$ are in the same coset, and $b$ and $c$ are in the same coset, then $a$ and $c$ must be in the same coset. This partitions the group up into groups of "equivalent" elements. We created the factor group of $G$ by $H$ by just using the set product on these cosets.

The same principle applies to comparing groups. We have way of determining when two groups are "the same". We call the groups isomorphic. The next problem has you show that the notation of isomorphisms creates an equivalence relation on the set of all groups.

Problem (Isomorphisms Yield An Equivalence Relation On The Set Of All Groups)

We say that $A$ is isomorphic to $B$ and write $A\approx B$ if and only if there exists a bijection from $A$ to $B$. In this problem, you'll prove that $\approx$ is an equivalence relation on the set of all groups.

  1. Let $A$ be a group. Show that $A$ is isomorphic to $A$ by building an isomorphism from $A$ to $A$.
  2. Suppose that $A$ is isomorphic to $B$. Prove that $B$ is isomorphic to $A$.
  3. Suppose that $A$ is isomorphic to $B$ and suppose that $B$ is isomorphic to $C$. Use this to produce an isomorphism from $A$ to $C$.

We'll now be preparing ourselves to prove the first Sylow theorem. When we study normal subgroups, we look at the product $x^{-1}ax$ quite often. You also studied this idea in linear algebra when you were studying changing bases. We called two matrices $A$ and $B$ similar in linear algebra if $B=P^{-1}AP$ for some invertible matrix $P$. When two matrices were similar, they represented the exact same linear transformation, just using a different basis. One of the biggest ideas in linear algebra is that if you use the eigenvectors as your basis vectors (the columns of $P$), then the matrix $A$ is similar to a diagonal matrix. Studying the products $P^{-1}AP$ had a huge payoff in linear algebra. It does in group theory as well.

Definition (Conjugacy Class Of A)

Let $G$ be a group. Suppose that $a$ and $b$ are elements of $G$. We say that $a$ and $b$ are conjugate in $G$ if $b=x^{-1}ax$ for some $x\in G$, and we say that $b$ is a conjugate of $a$. The set of conjugates of $a$ is called the conjugacy class of $a$ which we denote by $\text{cl}(a)=\{x^{-1}ax\mid x\in G\}$.


Exercise (Conjugacy Classes Of The Automorphisms Of The Square)

Let $G = D_8$, the automorphisms of the square. Compute the conjugacy classes of $G$.

Click to see a solution.

The conjugacy classes are $$ \begin{align} \text{cl}(R_0)&=\{R_0\}\\ \text{cl}(R_{180})&=\{R_{180}\}\\ \text{cl}(R_{90})&=\{R_{90},R_{270}\}\\ \text{cl}(R_{270})&=\{R_{90},R_{270}\}\\ \text{cl}(H)&=\{H,V\}\\ \text{cl}(V)&=\{H,V\}\\ \text{cl}(D)&=\{D,D'\}\\ \text{cl}(D-)&=\{D,D'\}. \end{align} $$ Notice how this partitions the group. The next problem has you show that conjugacy is an equivalence relation.


You've seen conjugacy classes before when you looked at similar matrices in linear algebra. If two matrices are similar, then lots of things are known about the matrices, in particular that they have the same eigenvalues, and the same dimensions of eigenspaces. There's an entire world of facts to explore about similar matrices, which is a specific example of a conjugacy class.

For now, let's show that conjugacy classes partition the group $G$. As a side note, every idea we learn here about conjugacy classes will immediately apply to matrix groups. After we have shown that conjugacy classes partition the group $G$, we can then count the number of elements of $G$ by instead counting the number of elements in each distinct conjugacy class. This means we have $G = \bigcup \cl(a)$ and hence $|G|=\sum |\text{cl}(a)|$ where the sum only includes each conjugacy class once.

Problem (Conjugacy Is An Equivalence Relation)

Show that conjugacy is an equivalence relation on $G$. In other words, show the following three things.

  1. For each $a\in G$, we know that $a$ is a conjugate of $a$.
  2. If $a$ is a conjugate of $b$, show that $b$ is a conjugate of $a$.
  3. If $a$ is a conjugate of $b$ and $b$ is a conjugate of $c$, show that $a$ is a conjugate of $c$.

Let's now connect conjugacy to centralizers. The center of a group is the set of all $x$ that commute with every element in the groups. The centralize of an element is the collection of all $x$ that commute with $a$.

Definition (Centralizer Of An Element)

Let $a\in G$. The centralizer of $a$ is the collection of all $x\in G$ such that $ax=xa$, or alternately $x^{-1}ax=a$. We write this set symbolically as $$C(a)=\{x\in G\mid ax=xa\} = \{x\in G\mid x^{-1}ax=a\} .$$


Exercise (Centralizers Of The Automorphisms Of The Square)

Let $G = D_8$, the automorphisms of the square. Compute the centralizes $C(a)$ for each $a\in G$.

Click to see a solution.

The centralizers are $$ \begin{align} C(R_0)&=\{R_0,R_{90}, R_{180}, R_{270}, H, V, D, D' \}\\ C(R_{180})&=\{R_0,R_{90}, R_{180}, R_{270}, H, V, D, D' \}\\ C(R_{90})&=\{R_0,R_{90}, R_{180}, R_{270}\}\\ C(R_{270})&=\{R_0,R_{90}, R_{180}, R_{270}\}\\ C(H)&=\{R_0, R_{180}, H, V\}\\ C(V)&=\{R_0, R_{180}, H, V\}\\ C(D)&=\{R_0, R_{180}, D, D'\}\\ C(D-)&=\{R_0, R_{180}, D, D'\}\\. \end{align} $$ Compare this with the exercise about the conjugacy classes of $G$. Do you notice that a large centralizer means a small conjugacy class, and vice versa. The order of the centralizer and the order of the conjugacy class are inversely related. Their product is always the order of the group.


If an element is in the centralizer, then we know that we can use that element to show that $a$ is a conjugate of itself. On your midterm you showed that the centralizer of $a$ is a subgroup of $G$. The previous exercise showed that the order of the centralizer is inversely proportional to the cardinality of the conjugacy class. The next problem has you connect centralizers and conjugacy classes, namely that $$|\text{cl}(a)||C(a)|=|G|\quad\quad \text{or}\quad\quad |\text{cl}(a)|=|G:C(a)|.$$

Problem (The Number Of Conjugates Of A)

Let $G$ be a group and let $a\in G$. Prove that the number of conjugates of $a$ equals the index of the centralizer of $a$ in $G$. In symbols, we write this as $|\text{cl}(a)|=|G:C(a)|$.

Hint: Recall that the index of $H$ in $G$ is the number of distinct cosets of $H$ in $G$.

Build a map from $\text{cl}(a)$ to the collection of cosets of $C(a)$ in $G$. Try sending $x^{-1}ax\in \text{cl}(a)$ to the coset $C(a)x$. You'll have to show that this map is well defined, meaning that if $x^{-1}ax=y^{-1}ay$, then you must have $C(a)x=C(a)y$. You'll need to then show that this map is a bijection. I would suggest you look up the properties of cosets to see what is equivalent to saying $Hx=Hy$.

One of the easiest ways to show that two sets have the same number of elements is to produce a bijection from one the others.

On a side note, the map you create is NOT a homomorphism. You won't be able to prove that it is. Luckily, you don't need to.

Problem (The Class Equation)

Prove the following corollaries of The number of conjugates of $a$

  1. For any group $G$, we have $$|G|=\sum|G:C(a)|$$ where the sum runs over one element from each conjugacy class of $G$.
  2. The center of any group consists precisely of the elements $a\in G$ such that $\text{cl}(a)=\{a\}$. In symbols, we have $a\in Z(G)$ if and only if $|\text{cl}(a)|=1$.
  3. For any group $G$, we know that $$|G|=|Z(G)|+\sum|G:C(a)|$$ where the sum runs over one element from each conjugacy class of $G$ that has more than one element.
  4. If $G$ has order $p^k$ for some prime $p$ and $k\in \mathbb{N}$, then $p$ divides $|Z(G)|$, and so the center is nontrivial.

This problem carries over from last week.

Given any two finite subgroups $H$ and $K$, there is a simple way to compute the size of their set product. You'll see a similar theorem in combinatorics or probability courses. This is counting principles, and your job is to show that if $x=hk$, then for each $y\in H\cap K$, there is another way to write $x=h_yk_y$, and these are the only ways you can create $x$.

Problem (The Number Of Elements In A Set Product)

Suppose that $G$ is a group and that $H$ and $K$ are finite subgroups of $G$. Prove that $|HK|=|H||K|/|H\cap K|$.


Problem (The First Sylow Theorem Proof)

Prove the First Sylow Theorem.

Theorem (The First Sylow Theorem)

Suppose $G$ is a finite group and $p^k$ divides $|G|$ for some prime $p$ and integer $k$. Then $G$ contains a subgroup of order $p^k$.

Click to see a hint.

Use induction on on the order of the group. If $|G|=1$, then what do you know? If $|G|=2$, then what do you know. Assume for some $n\in \mathbb{N}$ that the first Sylow theorem is true for all groups of order $k<n$. Then show that the theorem is true if $|G|=n$.

At some point in your work, you'll need to use two problems that we have already shown to be true.

  1. Cauchy's Theorem for Abelian Groups - If $p$ divides $|G|$, then $G$ has an element of order $p$. This applies to the center of every group, so if $G$ is not Abelian you can still apply the theorem to the center $Z(G)$ of $G$.
  2. If $G/N$ has a subgroup $B$, then $G$ has a subgroup $H$ with $H/N=B$. If $G$ is finite, then we know $|H|/|N|=|B|$.

What do you do? If $G$ has order $p^k$, then you're done. If $G$ has a proper subgroup of order $p^k$, then you're done. So suppose it does not have a proper subgroup of order $p^k$. What does this mean about $|C(a)|$? In particular, what does this mean about $|G:C(a)|$ which equals $|\text{cl}|$? Can you prove that $p$ must divide this index for each $a\notin Z(G)$? You may have to use the induction hypothesis your work here. If you then use the class equation, you should be able to show that the center of $G$ contains an element $x$ of order $p$. Once you do that, you can create the factor group $G/\left<x\right>$, which has order less than $G$. Here's where you get to use the induction hypothesis again. Problem 7 from Wednesday last week should help you to get to a subgroup in $G$ from knowing things about subgroups in $G/\left<x\right>$. I believe Nick presented this in class on Monday (at the very end).


Practice with the past

If you have decided to practice problems from the past, then, please use the table below to pick problems from the text. Try problems from multiple sections each day. If you hit a wall with one topic, try simpler problems with a different topic. The last thing you want is to get to next week on Wednesday and finally discover that you are completely stuck in the last 2 chapters. You want to find that out ASAP, and come get help.

Problem. What to Report

Please tell me which problems you worked on (at least 10 of them). Also, tell me 3 of them that you would be willing to teach to your peers. When you come to class, I'll group you together with some peers and then ask you to share these problems with each other.

Chapter

Basic

Intermediate

Advanced

5 (Permutations)

1-6, 9, 11 - 14, 17-18, 20, 21, 36, 38, 39,

7, 8, 15, 16, 19, 22-24, 27-30, 32-35, 37,

10, 26, 31,

6 (Isomorphisms)

1, 3, 4, 5, 6, 7, 8, 18, 19, 20, 21, 22, 25, 26, 29, 30, 38, 39, 40,

2, 10, 13, 14, 15, 16, 17, 24, 28, 31, 35,

9, 11, 12, 23, 27, 32, 33, 34, 36, 37, 41-47

7 (Cosets)

1-10 (do 10), 12-14, 20, 22, 27-29, 39, 43, 47, 49,

11, 15, 17-19, 21, 25, 26, 30, 31, 37, 38, 40-42,

16, 23, 24, 32-36, 44-46, 48, 50

8 (Direct Products)

1-11, 13-14, 15 (DO THIS ONE - IT'S the crux of the induction mastery II assignment.), 16-18, 20, 21, 23 (Matrices), 24, 26-32, 39, 46-48,

12, 19, 22, 25(this shows up in vector subspaces, topology, and much more), 34 (fun), 35-38, 40, 43-45, 49, 50, 53, 56, 57,

33, 41, 42, 51, 52, 54, 55, 58-75

5-8 Mixed together. See page 174.

9 (Normal Subgroups and Factor Groups)

1-9, 12-14, 16-18, 20, 22-24, 27, 28, 31, 34, 52 (Fun, Everyone should do), 53, 54, 56, 65 (we did in class), 67

3, 10, 11, 15, 19, 21, 26, 32, 37 (key problem), 38, 39, 42, 43, 49, 55, 57, 59, 60, 62, 64, 67, 74, 75

29, 30, 33, 35, 36, 40, 41, 44-48 (Use G/Z theorem to make easy), 50, 51, 58, 63, 66, 69-73,

10 (Homomorphisms)

1-6, 8, 9, 10, 12-18, 21, 24, 26-30, 31-34, 35, 36, 38, 42, 44, 45, 46, 48 (linear algebra application), 54 (orthogonal matrix application),

7, 11, 19, 20, 22 (done in class), 23, 25, 37, 43 (what are the normal subgroups?), 49, 50, 52, 55 (fun use of preimages of normal subgroups), 59, 60,

39, 40, 41, 47, 51, 53, 56, 57, 58 (Introduction to counting the number of elements in $\text{Hom}(A, B)$ and $\text{Hom}(A)$), 61, 62.

Don't forget that there are solutions (or rather sketches of solutions) to many of the odd problems in the text.

December 13

Inquiry Based Prep Problems

Pick up where you left off in class. I think you got partway through 5. The big idea is the Sylow theorem. The problems that come after it are other applications related to the ideas that built up to the Sylow theorem. If you are not sure where to go with the Sylow theorem, then work on the other problems first. The Sylow theorem is the longest one. Some of the problems below are extremely short.

Problem (The First Sylow Theorem Proof)

Prove the First Sylow Theorem.

Theorem (The First Sylow Theorem)

Suppose $G$ is a finite group and $p^k$ divides $|G|$ for some prime $p$ and integer $k$. Then $G$ contains a subgroup of order $p^k$.

Click to see a hint.

Use induction on on the order of the group. If $|G|=1$, then what do you know? If $|G|=2$, then what do you know. Assume for some $n\in \mathbb{N}$ that the first Sylow theorem is true for all groups of order $k<n$. Then show that the theorem is true if $|G|=n$.

At some point in your work, you'll need to use two problems that we have already shown to be true.

  1. Cauchy's Theorem for Abelian Groups - If $p$ divides $|G|$, then $G$ has an element of order $p$. This applies to the center of every group, so if $G$ is not Abelian you can still apply the theorem to the center $Z(G)$ of $G$.
  2. If $G/N$ has a subgroup $B$, then $G$ has a subgroup $H$ with $H/N=B$. If $G$ is finite, then we know $|H|/|N|=|B|$.

What do you do? If $G$ has order $p^k$, then you're done. If $G$ has a proper subgroup of order $p^k$, then you're done. So suppose it does not have a proper subgroup of order $p^k$. What does this mean about $|C(a)|$? In particular, what does this mean about $|G:C(a)|$ which equals $|\text{cl}|$? Can you prove that $p$ must divide this index for each $a\notin Z(G)$? You may have to use the induction hypothesis your work here. If you then use the class equation, you should be able to show that the center of $G$ contains an element $x$ of order $p$. Once you do that, you can create the factor group $G/\left<x\right>$, which has order less than $G$. Here's where you get to use the induction hypothesis again. Problem 7 from Wednesday last week should help you to get to a subgroup in $G$ from knowing things about subgroups in $G/\left<x\right>$. I believe Nick presented this in class on Monday (at the very end).


Problem (If The Factor Group Of G By The Center Is Cyclic Then G Is Abelian)

Suppose that $G$ is a group and $Z(G)$ is the center of $G$. Prove that if $G/Z(G)$ is cyclic, then $G$ is Abelian.

Problem (Groups Of Order $pq$)

Suppose that $G$ has order $pq$ for primes $p$ and $q$. If $G$ is not Abelian, then what does the previous problem tell you about the center of $G$?


Problem (Groups Of Order $p^2$ Are Abelian)

Prove that every group of order $p^2$ is Abelian.


Definition (Automorphisms And Inner Automorphisms)

Let $G$ be a group.

  • An automorphism of $G$ is an isomorphism from $G$ to $G$.
  • We write $\text{Aut}(G)$ to represent the set of all automorphisms of $G$.
  • The function $\phi_g:G\to G$ defined by $\phi_g(x)=gxg^{-1}$ is called an inner automorphism of $G$.
  • We write $\text{Inn}(G)$ to represent the set of all inner automorphisms of $G$.

Problem ($\text{Inn}(G)$ is a normal subgroup of $\text{Aut}(G)$)

Let $G$ be a group. Prove the following:

  1. $\text{Aut}(G)$ is a group,
  2. $\text{Inn}(G)$ is a subgroup of $\text{Aut}(G)$, and
  3. $\text{Inn}(G)$ is a normal subgroup of $\text{Aut}(G)$.

Problem ($G/Z(G)$ is isomorphic to $\text{Inn}(G)$)

Suppose that $G$ is a group. Let $f:G \to \text{Inn}(G)$ be defined by $f(x)=\phi_x$.

  1. Show that $f$ is a homomorphism with kernel $Z(G)$.
  2. Then prove that $G/Z(G)$ is isomorphic to $\text{Inn}(G)$.
  3. Compute $\text{Inn}(G)$ for any Abelian group $G$ and then for $G=D_8$.
  4. Why is $D_{10}$ isomorphic to $\text{Inn}(D_{10})$?

December 16

For the remaining two days of class, we will be working through the next 9 problems. Everyone should work on problems 1-4. These problems come from the text, and I want you to complete as them. We won't have people present their work on problems 1 and 4, but we will have someone share what they did on 2 and 3. I'll put some solutions up for 1 and 4. Don't go back to tackle problems 5 and 6 from Friday, until after you have tackled the ones below.

Problem. Factor Group Practice

Please open your text to chapter 9 and complete problems 14, 15, 17, 18, 19, 24, 25, 27, 28. Do as many of these as are valuable to your continued learning. Check your work with the solutions below.

Click For Some Solutions.

The odd numbered problems have partial solutions in the back of the text.

Exercise 14

First, remember that the identity in a factor group $G/H$ is the coset $H$. So we need to know how many times we have to add the coset $14+\left<8\right>$ to itself before we obtain $\left<8\right>$. This is equivalent to asking, "how many times must we add 14 to itself before we reach a multiple of 8?" We could just notice that the multiples of 14 are 14, 28, 42, 56, and so the 4th one gets us to a multiple of 8. This means the order is 4.

We could also simplify the problem first, before computing the order. Since $14+\left<8\right>=6+8+\left<8\right>$, and because $8+\left<8\right> = \left<8\right>$ (cosets absorb their own elements) we can just write $14+\left<8\right>=6+\left<8\right>$. To find the order, we just need to ask how many times we have to add 6 to itself until we arrive at a multiple of 8. The least common multiple is 24, which means the order of the coset is 4.

Exercise 15

Remember that $U_5(105)$ is the collection of elements $x\in U(105)$ such that $x\mod 5=1$. We need to find the order of $4U_5(105)$. Since $4^2=16$, and we know that $16\mod 5=1$, then $(4U_5(105))^2 = 16U_5(105)=U_5(105)$. This shows that the order of $4U_5(105)$ is 2.

Exercise 17

Let $G=Z/\left<20\right>$ and $H=\left<4\right>/\left<20\right>$. Then we have $$H = \{4+\left<20\right>, \quad 8+\left<20\right>, \quad 12+\left<20\right>, \quad 16+\left<20\right>, \quad 0+\left<20\right>\}.$$ The elements of $G/H$ are $$H, \quad (1+\left<20\right>)+H, \quad (2+\left<20\right>)+H, \quad (3+\left<20\right>)+H\}.$$

Exercise 18

There are 60 elements in $\mathbb{Z}_60$. The subgroup generated by 15 has 4 elements. So there are 60/4=15 elements in the factor group.

Exercise 19

There are 10 elements in $\mathbb{Z}_{10}$ and 4 elements in $U(10)$. This means that their external direct product has 40 elements. We know 2 has order 5 in $\mathbb{Z}_{10}$, and we know $9^2=81$ which equals 1 mod 10, so this means that 9 has order 2 in $U(10)$. The order of $(2,9)$ in the direct product is the least common multiple of the orders of the individual elements, which means we have $$|(2,9)|=\text{lcm}(|2|,|9|) = \text{lcm}(5,2)=10.$$ We are creating a quotient group of a group of order 40, using a subgroup of order 10. This means the factor group has order $40/10 = 4$.

Exercise 24

Let's start by listing the cosets of $H = \left<(2,2\right>$ in the group $G=\mathbb{Z}_{4}\oplus \mathbb{Z}_{12}$. The cosets are $$ \begin{align} H &= \left<(2,2\right> =\{(0,0),(2,2),(0,4),(2,6),(0,8),(2,10)\} \\ (1,0)+H &= \{(1,0),(3,2),(1,4),(3,6),(1,8),(3,10)\} \\ (2,0)+H &= \{(2,0),(0,2),(2,4),(0,6),(2,8),(0,10)\} \\ (3,0)+H &= \{(3,0),(1,2),(3,4),(1,6),(3,8),(1,10)\} \\ (0.1)+H &= \{(0,1),(2,3),(0,5),(2,7),(0,9),(2,11)\} \\ (1.1)+H &= \{(1,1),(3,3),(1,5),(3,7),(1,9),(3,11)\} \\ (2.1)+H &= \{(2,1),(0,3),(2,5),(0,7),(2,9),(0,11)\} \\ (3.1)+H &= \{(3,1),(1,3),(3,5),(1,7),(3,9),(1,11)\}. \end{align} $$ If you just ignore the plus $H$ on the end of each coset, you should see that this group is basically the same as $\mathbb{Z}_4\oplus \mathbb{Z}_2$. There are clearly no elements of order 8, which means is not isomorphic to $\mathbb{Z}_8$. There are elements of order $4$, which means it's not isomorphic to $\mathbb{Z}_2\oplus \mathbb{Z}_2\oplus \mathbb{Z}_2$.

Exercise 25

If we list out the cosets of $H=\{1,31\}$ in the group $U(32) = \{1,3,5,7,9,11,13,1,17,19,21,23,25,27,29,31\}$, we obtain (noting that 31 and $-1$ are equivalent mod 32 to simply computations) $$ \begin{align} H &= \{1,31\}\\ 3H &= \{3,29\}\\ 5H &= \{5,37\}\\ 7H &= \{7,25\}\\ 9H &= \{9,23\}\\ 11H &= \{11,21\}\\ 13H &= \{13,19\}\\ 15H &= \{15,17\}. \end{align} $$ We now need to compute the order of these cosets. The only possible orders are 1, 2, 4, and 8, because the order of the factor group is 8 and each element must have an order that divides the order of the group. The powers of 3, taken mod 32, are 3, 9, 27=-5, -15=17, and so on. Notice that this is enough to rule out 1,2, and 4 as the order of $3H$. This means that $3H$ has order 8, and as such generates the entire group. This means that the group $G/H$ must be isomorphic to $\mathbb{Z}_8$.

Exercise 27

We know $H$ and $K$ are isomorphic, because they both have prime order 2, and hence are both cyclic. See the book for the reason why the factor groups are not isomorphic.

Exercise 28

Similar to the previous. The point to these last two problems is to show you that even if you know $H$ and $K$ are isomorphic, you do not know that $G/H$ and $G/K$ are isomorphic. They might be, but they don't have to be.

You'll want to use the The First Isomorphism Theorem to prove both of the following.

Theorem (The First Isomorphism Theorem)

Let $f:G\to H$ be a surjective homomorphism with kernel $K$. Because we know that $f(x)=f(y)$ for any $y\in Kx$ (elements in the same coset of the kernel have the same image under $f$), then we can define a map $\phi:G/K\to H$ by defining $\phi(Kg)=f(g)$. This map $\phi$ is always an isomorphism.

Problem. The Second and Third Isomorphism Theorems

Use The First Isomorphism Theorem to prove each of the following. In both cases, you just need to define a surjective map, compute the kernel, and then quote the theorem.

  • (Second Isomorphism Theorem) Suppose that $K\leq G$ and $N\trianglelefteq G$. Prove that $K/(K\cap N)\approx KN/N$.
  • (Third Isomorphism Theorem) Suppose that $M$ and $N$ are normal subgroups of $G$ with $N\leq M$. Prove that $(G/N)/(M/N)\approx G/M$.
Note that for the first problem, you only need to define a map from $K$ to $KN/N$, as the $K\cap N$ part should automatically show up when you compute the kernel and use the first isomorphism theorem. What map should you try? How about sending $k$ to the coset $Nk$?
For the third isomorphism theorem, you could define a map from $G$ to $(G/N)/(M/N)$ or from $G/N$ to $G/M$, and then in either case show that the kernel of your map is the missing denominator.
Theorem. Fundamental Theorem of Finite Abelian Groups

Every finite Abelian group $G$ is the direct product of cyclic groups. In particular, this direct product can be expressed as a direct product of cyclic groups with prime-power order, which means $$G\approx \mathbb{Z}_{p_1^{n_1}}\oplus \mathbb{Z}_{p_2^{n_2}}\oplus\cdots \oplus \mathbb{Z}_{p_k^{n_k}} .$$ When expressed in this manner, the number of terms in the direct product and the orders of the cyclic groups is uniquely determined by the group.

Problem.Isomorphism Classes of Abelian Groups of various orders

Start by reading pages 218-220 in your text. In each situation below, you are given an integer $n$. State all the different isomorphism classes of Abelian groups of the given order.

  1. $n=8=2^3$
  2. $n=48 =2^43 $
  3. $n=500 = 2^25^3$
  4. How many (don't list them all) isomorphism classes are there for a group of order $n=2^2\cdot 3^3\cdot 5\cdot 7^5\cdot 11^4$?

When $n=8=2^3$, the isomorphism classes are $$ \mathbb{Z}_2\oplus \mathbb{Z}_2\oplus\mathbb{Z}_2, \quad \mathbb{Z}_4\oplus \mathbb{Z}_2, \quad \text{ and }\quad \mathbb{Z}_8. $$

Problem.Practice with The Fundamental Theorem

Please complete problems 1-10 in chapter 11 of your text.

Click for some solutions.

Problem 1

The smallest such integer is $n=4$. The two groups are $\mathbb{Z}_4$ and $\mathbb{Z}_2\oplus\mathbb{Z}_2$.

Problem 2

The smallest such integer is $n=8$. The three groups are $\mathbb{Z}_8$, $\mathbb{Z}_4\oplus\mathbb{Z}_2$, and $\mathbb{Z}_3\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$.

Problem 3

The smallest such integer is $n=2^2\cdot 3^2 = 36$. The four groups are $\mathbb{Z}_4\oplus \mathbb{Z}_9$, $\mathbb{Z}_4\oplus\mathbb{Z}_3\oplus \mathbb{Z}_3$, $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus \mathbb{Z}_9$, and $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_3\oplus \mathbb{Z}_3$. You have to pick different primes to obtain this one.

Problem 4

In $\mathbb{Z}_{16}$ there is only one element of order 2, namely the integer 8. There are 2 elements of order 4, namely the integers 4 and 12.

In $\mathbb{Z}_{8}\oplus \mathbb{Z}_2$ we can obtain an element of order 2 by picking an element $(x,y)$ such that either $x$ or $y$ has order 2 and the other has order 1 or 2. There is one element in each group that has order 2. The elements of order 2 are $(4,0)$, $(4,1)$, and $(0,1)$. There are three such elements. To compute the elements of order 4, we must pick an element of the form $(x,y)$ where either $x$ or $y$ has order $4$, and the other has order 1, 2, or 4. The elements of order 4 are hence $(2,0),(6,0),(2,1),(6,1)$. That's it.

In $\mathbb{Z}_{4}\oplus \mathbb{Z}_4$ we can obtain an element of order 2 by picking an element $(x,y)$ such that either $x$ or $y$ has order 2 and the other has order 1 or 2. There is one element in each group that has order 2. The elements of order 2 are $(2,0)$, $(2,2)$, and $(0,2)$. There are three such elements. To compute the elements of order 4, we must pick an element of the form $(x,y)$ where either $x$ or $y$ has order $4$, and the other has order 1, 2, or 4. The elements of order 4 where we require the first component to have order 4 are $(1,0),(3,0),(1,1),(3,1),(1,2),(3,2),(1,3),(3,3),$. We also must count the elements where the first component does not have order 4, but the second does. This gives us the additional elements $(0,1),(0,3),(2,1),(2,3)$. This gives us 12 elements of order 4. A simpler computation could have just noticed that since there is 1 element of order 1 and 3 elements of order 2, and no element has order 8 or 16, then the remaining 12 elements must have order 4.

In $\mathbb{Z}_{4}\oplus \mathbb{Z}_2\oplus \mathbb{Z}_2$ we can obtain an element of order 2 by picking an element $(x,y,z)$ such that either $x$, $y$, or $z$ has order 2 and the others have orders 1 or 2. There is one element in each group that has order 2. There are 4 elements of the form $(2,x,y)$ that have order 2 (as there are two choices for $y$ and $z$). There are two choices of the form $(0,2,z)$, and one choice of the form $(0,0,2)$. This gives a total of 7 elements of order 2. To obtain an element of order 4, we must choose $x=1,3$ and then allow $y$ or $z$ to be anything. This gives $8=2\cdot 2\cdot 2$ elements of order 4. Again, we could have noticed that with 1 element of order 1 and 7 elements of order 2, there must be 8 elements of order 4.

Problem 5

Since $45=5\cdot 9$, there are two isomorphism classes of Abelian groups of order 45. They are $\mathbb{Z}_9\oplus \mathbb{Z}_5\approx \mathbb{Z}_{45}$ and $\mathbb{Z}_3\oplus\mathbb{Z}_3\oplus \mathbb{Z}_5\approx \mathbb{Z}_{15}\oplus \mathbb{Z}_3$. Remember that we can combine cyclic groups when the orders are relatively prime and obtain a cyclic group. Notice that in each case, we can obtain an element of order 15, namely $3\ in \mathbb{Z}_{45}$ and $(1,0)\in \mathbb{Z}_{15}\oplus \mathbb{Z}_3$.

Problem 6

Notice that $n=108=54\cdot 2 = 27\cdot 2^2 = 2^2\cdot 3^3$. There are two isomorphism classes for groups of order $2^2$, and three isomorphism classes for groups of order $3^3$. So there are a total of 6 isomorphism classes of groups of order 108. The cyclic group $\mathbb{Z}_{108}\approx \mathbb{Z}_{4}\oplus \mathbb{Z}_{27}$ has exactly one subgroup of order $3$. Since we are after subgroups of order 3, we can do whatever we want to the part related to the prime number 2, and it won't change the answer, as any element that includes something nonzero from any factors not divisible by 2 will force the element to have even order (recall that the order of an element is the least common multiple of the orders or the elements in each factor). This means the group $\mathbb{Z}_2\oplus \mathbb{Z}_2\oplus \mathbb{Z}_{27}$ has a single subgroup of order 3.

Problem 7

Continuing from problem 6, we notice that $\mathbb{Z}_{4}\oplus \mathbb{Z}_{9}\oplus \mathbb{Z}_{3}$ has several subgroups of order 3. Because any any subgroup of order 3 is cyclic, all we have to do is count the number of elements of order 3. So let $(x,y,z)$ be an element of $\mathbb{Z}_{4}\oplus \mathbb{Z}_{9}\oplus \mathbb{Z}_{3}$. Then $x=0$, otherwise the order is divisible by 2. The orders of $y$ and $z$ must be 1 or 3, with at least 1 of them being 3. So if $x=3$ or $x=6$, then $y\in\{0,1,2\}$. This gives 6 elements of order 3. If $x=0$, then $y\in \{1,2\}$. which gives another 2 elements of order 3. There are a total of 8 elements of order 3. Each subgroup of order 3 contains 2 of these elements, and as such there are 4 subgroups of order 3. A similar computation for $\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\oplus \mathbb{Z}_{9}\oplus \mathbb{Z}_{3}$ gives the same result.

Problem 8

We now examine the groups $\mathbb{Z}_{4}\oplus \mathbb{Z}_{3}\oplus\mathbb{Z}_{3}\oplus \mathbb{Z}_{3}$ and $\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\oplus \mathbb{Z}_{3}\oplus\mathbb{Z}_{3}\oplus \mathbb{Z}_{3}$. We just need to count the number of elements of order in $\mathbb{Z}_{3}\oplus\mathbb{Z}_{3}\oplus \mathbb{Z}_{3}$. This is simple, as every element other than the identity must have order 3, so there are 26 elements of order 3. This means there are 13 subgroups of order 3 in either group.

Problem 9

We know that $120=2^3\cdot 3\cdot 5$. This means that $G$ is isomorphic to one of the following groups: $$\mathbb{Z}_8\oplus\mathbb{Z}_3\oplus\mathbb{Z}_5, \mathbb{Z}_4\oplus\mathbb{Z}_2\oplus\mathbb{Z}_3\oplus\mathbb{Z}_5, \mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_3\oplus\mathbb{Z}_5.$$ The first group has one element of order 2. The last group has 7 elements of order 2. This leaves the middle group as the only possible answer.

Problem 10

We first compute $36\cdot 10 = 2^2\cdot 3^2\cdot 2\cdot 5=2^3\cdot 3^2\cdot 5$. There are 6 possible isomorphism classes. See page 220 in your text for the possible answers, where they list the possible isomorphism classes of groups of order $2^3\cdot 7^2\cdot 3$.

In all the work below, we'll be working with external and/or internal direct products. Recall that we say $G$ is the internal direct product of two subgroups $H$ and $K$ if
  1. we know that $G=HK$ as a set product,
  2. we know that $H\cap K = \{e\}$, and
  3. we know that both $H$ and $K$ are normal.

Because we are working with Abelian groups, the third fact is immediate anytime we have subgroups. Anytime we know that $G$ is an internal direct product of $H$ and $K$, then we know that $$G=HK\approx H\times K = H\oplus K.$$ Some authors will just write $G=H\times K$ instead of $G\approx H\times K$, because the external direct product is equal to the internal direct product $HK$. I'll be sticking with the use of $\approx$, intead of just saying that $G$ equals its external direct product.

The next theorem shows that we can take any Abelian group and separate the group appart as $G=H_{p_1}\times H_{p_2}\times\cdots H_{p_k}$ where $p_1, p_2, \ldots, p_k$ are distinct primes, and every element in $H_{p_i}$ has a prime-powered order.

Problem.Splitting an Abelian Group into $H\times K$ where all elements of prime $p$ powered order are in $H$

Suppose that $G$ is an Abelian group with order $p^km$ where $p$ does not divide $m$. Let $H=\{x\in G\mid x^{p^k}=e\}$. Let $K=\{x\in G\mid x^m=e\}$.

  1. Prove that both $H$ and $K$ are normal subgroups of $G$, that $H\cap K=\{e\}$, and that $G=HK$. In other words, prove that $G$ is the internal direct product of $H$ and $K$, which means that $G\approx H\times K$.
  2. Further more, show that $|H|=p^k$ and $|K|=m$.

Now that we've split up the group $G$ into a direct sum of prime powered subgroups, we need to show that each prime powered subgroup can be written as a direct product of cyclic groups. To do this, we'll show that we can write $G$ in the form $G\approx \left<a\right>\times$ where $a$ has maximal order in $G$ and $K$ is a subgroup of $G$. Once we have shown that we can strip off one cyclic group from $G$, we just repeat this process until we obtain $$G\approx \left<a_1\right>\times\left<a_2\right>\times \left<a_3\right>\times\cdots\times \left<a_k\right> \times \left<e\right>$$ where the orders satisfy $|a_1|\geq |a_2|\geq |a_3|\geq \cdots \geq |a_k|>1$.

Problem.Obtaining two cyclic subgroups with trivial intersection in a prime-powered Abelian group, one of which has maximal order

Suppose that $G$ has order $p^n$ for some prime $p$ and positive integer $n$, and suppose that $G$ is not cyclic. Let $a$ be an element of $G$ with maximal order. Let $b\in G$ be an element of smallest order such that $b\notin \left<a\right>.$ Prove that $\left<a\right>\cap \left<b\right> = \{e\}$.

Problem.The Greedy Algorithm for a prime-powered Abelian group

Suppose that $G$ has order $p^n$ for some prime $p$ and positive integer $n$. Let $a$ be an element of $G$ with maximal order. Prove that $G\approx \left<a\right>\times K$ for some subgroup $K$ of $G$.

You'll need to use induction on the order of a group to complete this problem. The key is to use the subgroup $\left<b\right>$ from the previous problem and then examine the factor group $G/\left<b\right>$ which has a smaller order than $G$. You'll need to prove that the order of $a$ in $G$ is the same as the order of $\left<b\right>a$ in $G/ \left<b\right>$.

You'll also need to use the fact that subgroups of $G/ \left<b\right>$ correspond to subgroups of $G$ in a very nice way (something that we showed in the last two weeks).

Problem.A finite Abelian Group of prime power order is a direct product of cyclic groups

Suppose that $G$ has order $p^n$ for a prime $p$ and positive integer $n$. Prove that $G$ can be written as a direct product of cyclic groups.

Problem.The Uniqueness of a Direct Product Decomposition for prime-powered groups

Suppose that $G$ is a group of prime power order, and that $G$ has been written in two ways as the product of cyclic groups, namely that $G\approx H_1\times H_2\times \cdots \times H_n$ and $G\approx K_1\times K_2\times \cdots \times K_m$, where the subgroups are written so that their orders are nonincreasing, and none of the factors are trivial. Prove that $n=m$ and that $H_i\approx K_i$ for each $i$ from 1 to $n$.

December 18

Look at the schedule on Monday. We'll just be picking up where we left of. If you haven't done the exercises in chapters 9 and 11 yet, make sure you do them.

December 20

The final exam is available in the testing center starting Wednesday, and available through Friday.


For more problems, see AllProblems