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We'll be presenting in small groups. Please head to your respective board and start working.

 FrontRightBackLeft 
 BryceAustinAlexanderJustin 
 CarlaBrennanKimberlyKinsey 
 CassieCaleyLauraLi 
 JoeTerryMeganLilia 
 SamTimRichNick 
      
1.1CarlaTimKimberlyJustinCan We Use Division
1.2BryceCaleyMeganLiliaOrder of Z_n and U(n) (just show how to get the orders in Z_7 and U(7)
1.3TimTimRichRichGCD Theorem (see chapter 0)
1.4JoeJoeLiliaLiliaCenter is a subgroup
1.5CarlaCarlaLauraLauraPowers of products in an abelian group
      
2CassieTimKimberlyLiSubgroup Generated by an element
3CaleyCaleyAlexanderAlexanderIntersection of two subgroups
4JoeJoeKinseyKinseyProperties of <a>
5CarlaBrennanRichLiliaThe center of a dihedral Group.

Problem (Problems from the past)

We already have people ready to present these problems. If you have not yet completed them, please spend 10 minutes tops with each. Don't let yourself get bogged down with completing each of these before class, but make sure that you at least have spent 10 minutes with each (otherwise the discussion in class is pointless).

Click to see the problems.

The presenters are Justin, Kimberly, Carla, Tim

Problem 47 (Can We Use Division To Create A Group)

Let $G=\mathbb{R}$ and $H=\mathbb{R}\setminus \{0\}$.

  1. Show that division $a\div b$ is not a
    binary operation
    Definition (Binary Operation)

    Let $G$ be a set. A binary operation on $G$ is a way of combining two elements of $G$ to obtain a new element in $G$. Formally, we just say that a binary operation $*$ is function $*:G\times G\to G$, and we use the notation $a*b$ to represent the function notation $*(a,b)$.

    on $G$.
  2. Show that division $a\div b$ is a binary operation on $H$.
  3. Since division is a binary operation on $H$, determine if $(H,\div)$ is a
    group
    Definition (Group)

    Let $G$ be a nonempty set, and let $*$ be a binary operation on $G$, which means for every $x,y\in G$ we have $x*y\in G$ $\textbf{[Closure]}$. The structure $\mathbb{G} = (G,*)$ is called a $\textdef{group}$ if the following hold.

    1. $\textbf{[Associativity]}$ For all $x,y,z\in G$ we have $(x* y)* z = x* (y* z)$.
    2. $\textbf{[Identity]}$ There is an $e\in G$ such that for all $x\in G$ we have $x * e = e* x = x$.
    3. $\textbf{[Inverses]}$ For all $x\in G$ there is a $y\in G$ such that $x* y = y* x = e$.

    We usually simply write $G$ when referring to the entire structure $\mathbb{G}=(G,*)$. The element $e$ from the second point is called the $\textdef{identity}$. The element $y$ from the third point is called the $\textdef{inverse}$ of $x$ and is usually denoted $x^{-1}$. One often simply writes $xy$ in place of $x*y$, and for every positive integer $n$, we'll write $x^n$ as shorthand for $x* x* \cdots * x$ ($n$ times).

    .
    • Does $e=1$ satisfy the property of being an identity?
    • If $x\in H$, find an inverse $x^{-1}\in H$ or explain why none exists.
    • Is the operation $\div$ associative?


The presenters are Caley, Bryce, Lilia, Megan

Problem 48 (Orders Of $\mathbb{Z}_n$ And $U(n)$ And Their Elements)

We've already shown that $\mathbb{Z}_n$ under addition mod $n$ and $U(n)$ under multiplication mod $n$ are groups.

  1. For each $n$ between 2 and 10, compute the order of $Z_n$ and the order of each element of $\mathbb{Z}_n$. Organize your work into a table where you first make a list of the elements, and then underneath each element state the order. For example, if $n=6$ then our table would look like the one below. $$\begin{array}{|c|c|c|c|c|c|c|} \hline \text{Element}&0&1&2&3&4&5\\\hline \text{Order}&1&6&3&2&3&6\\\hline \end{array}$$
  2. For each $n$ between 2 and 10, compute the order of $U(n)$ (state $\varphi(n)$) and then compute the order of each element of $U(n)$. You can check your work with the sage code below. I kept the numbers under 10 because you can do these by hand (or in your head) fairly quickly. Make sure you do enough by hand that you feel comfortable with this process.
  3. You should have noticed that $U(8)$ and $U(10)$ both have order 4. Is there a 1-1 correspondence between these two groups that matches elements with the same order?

Here's some Sage code you can use to check your computations with $U(n)$.

for n in (2..20):
 Zn = Integers(n)
 Un = [x for x in Zn if gcd(ZZ(x), n) == 1]  #This creates Un
 show(table(["U("+str(n)+r") has order $\varphi($"+str(n)+"$)=$"+str(len(Un))+
             ".The elements, with orders below them, are listed below."]))
 orders=[x.multiplicative_order() for x in Un] #This computes the multiplicative order of each element.
 show(table([Un,orders])) #This creates a table of elements (top row) and their orders (bottom row).


The presenters are Rich and Tim

Problem 42 (The GCD Theorem Proof)

Prove the GCD Theorem.

Theorem (The GCD Theorem)

Let $a$ and $b$ be nonzero integers. The greatest common divisor of $a$ and $b$ is an integer linear combination of $a$ and $b$, or symbolically we have $\gcd(a,b)\in\text{span}(a,b)$. In addition, the smallest positive integer linear combination of $a$ and $b$ is $\gcd(a,b)$.

Hint: If you intersect $\text{span}(a,b)$ with the natural numbers, then you can apply the well ordering principle to get the smallest positive element $d$ of this intersection. You then just have to show that this smallest positive element $d$ is the greatest common divisor. This requires that you show

  1. that $d$ is a common divisor, and
  2. that $d$ is the greatest common divisor.

To show that $d$ divides both $a$ and $b$, use the division algorithm to obtain $a=qd+r$, and explain why the remainder must be zero. Then you must show that if $c$ is any other common divisor, then $d$ is greater, so $d\geq c$.


Joe and Lilia will present the following problem.

Problem 56 (The Center Of Group Is A Subgroup)

Prove that the center $Z(G)$ of a group $G$ is a subgroup of $G$. If $G$ is Abelian, then what is $Z(G)$?



Carla and Laura Will present the following problem.

Problem 57 (Powers Of Products In An Abelian Group)

Suppose $G$ is an Abelian group. Prove that if $a,b\in G$, then $(ab)^2=a^2b^2$. Then use induction to prove that if $a,b\in G$, then $(ab)^n=a^nb^n$ for each $n\in \mathbb{N}$.



We've been working with groups since the first day of the semester. The set of simple shift permutations on an alphabet is a group. In the problem
Simple Shift Repetition

Problem 9(Simple Shift Repetition)

Let's now devise a way to not only encrypt a message, but also keep track of who has seen the message? There are several ways to do this. Let's look at an example that involves repeated application of the same encrpytion key. For this example, let's use the encryption key $\phi_4:S\to S$ (the simple shift permutation that shifts right 4).

A group of military commanders decide to send messages using a message chain. The message passes sequentially from one person to the next, where at each stage they apply the encryption key to the ciphertext before sending it on. If I want to send the plain text message $attackatdawn$, then I'd send the message $exxegoexhear$ to the person after me in the chain. This person would decipher the text using $\phi_4^{-1}$ (shift left 4), and then apply $\phi_4$ to the cipher text $exxegoexhear$ and then send the ciphertext $ibbiksibliev$ to the person after them in the chain. This would repeat until the message made it to every commander.

  1. You receive the cipher text $skkzgzkomnz$. What is the plain text message? How many people have seen this message?
  2. How many commanders can we include in this encryption scheme and still tell who sent the message?
  3. If we use $\phi_5$ instead of $\phi_4$, how many commanders can we include in this encryption scheme and still tell who sent the message?
  4. Is there an encryption key $\phi_n$ so that $\phi_n^2=\phi_0$. In other words, is there an encryption key $\phi_n$ that will only allow up to two commands to be in the chain?

, we looked at what happens if we repeatedly apply the same simple shift many times. We noticed that eventually the message would cycle back to the identity message. We will now show that this is true for any finite group. First, we need some notation that works for arbitrary groups. Instead of using the word "span" to talk about repeatedly applying a function, we'll now use the notation $\langle a\rangle$.
Definition (Subgroup Generated By An Element)

Let $G$ be a group and $a\in G$. Then the subgroup generated by $a$ is the set $$\langle a\rangle = \{a^n\mid n\in\mathbb{Z}\},$$ where we define $a^0=e$ and $a^{-n}=(a^{-1})^n$.

Problem 58 (The Subgroup Generated By An Element Is Actually A Subgroup)

Let $G$ be a group with $a\in G$. Show that $\langle a\rangle$, using the definition above, is a subgroup of $G$.



Problem 59 (The Intersection Of Two Subgroups Of $\mathbb{Z}$)

Let $G=\mathbb{Z}$. Because the group operation is addition, remember that $a^2$ actually means $a+a$, and $a^5=a+a+a+a+a=5a$. Beware of this issue, as $a^n$ actually means $na$ because the group operation is addition when working in $\mathbb{Z}$.

  1. What is $\langle 2\rangle$? What is $\langle 3\rangle$? Convince yourself that $\langle 2\rangle\cap \langle 3\rangle = \langle 6\rangle$.
  2. What is $\langle 4\rangle$? What is $\langle 6\rangle$? Find an integer $c$ so that $\langle c\rangle=\langle 4\rangle\cap \langle 6\rangle$? Prove that your result is true.
  3. If $a,b\in \mathbb{Z}$, then conjecture what $c$ should equal so that $\langle c\rangle=\langle a\rangle\cap \langle b\rangle$. You don't have to prove this result (unless you'd rather prove this result than proving part 2).


Definition (The Subgroup $n\mathbb{Z}$)

For each $n\in \mathbb{N}$ we define $n\mathbb{Z}$ to be the subgroup $\langle n\rangle = \{kn\mid k\in \mathbb{Z}\}$ of the group $\mathbb{Z}$ under addition. This is just the multiples of $n$.


When we were working with simple shift permutations of the standard 26 letter alphabet, we showed that the set of all simple shift permutations, which we denoted $H=\{\phi_n\mid n\in\mathbb{Z}\}$, was actually a set that consisted of just 26 elements, namely we showed that $H=\{\phi_0,\phi_1, \ldots,\phi_{25}\}$. Each of these simple shifts can be generated using $\phi_1$, and the order of $\phi_1$ is $|\phi_1|=26$. Using the language of groups, we showed that $$H =\langle \phi_1\rangle = \{\phi_0,\phi_1, \ldots,\phi_{25}\}.$$ We showed this was true because if $n\in \mathbb{Z}$, then we could use the division algorithm to write $n=26q+r$ where $0\leq r<26$. Then we knew that $$\phi_1^{n}=\phi_1^{26q+r}=\phi_1^{26q}\circ \phi_1^r = \phi_0\circ \phi_r = \phi_r.$$ We knew that any repeated shift of 26 letters would take us back to the identity. We now generalize this problem to show that if an element of a group has order $n$, then $\langle a\rangle$ must always consist of the elements $\{e,a,a^2,\ldots, a^{n-1}\}$.

Problem 60 (Properties Of $\langle a \rangle$ When $a$ Has Finite Order)

Let $G$ be a group with $a\in G$. Suppose that the order of $a$ is $|a|=n$. Prove the following:

  1. We have $\langle a\rangle = \{e,a,a^2,\ldots, a^{n-1}\}$. (You are showing two sets are equal.)
  2. We have $a^i=a^j$ if and only if $i-j$ is a multiple of $n$.
  3. The order of an element equals the order of the subgroup generated by that element, namely $|a|=|\langle a\rangle|$. (How can you combine 1 and 2 to get this.)

If you didn't read the paragraph above this proof, then please do. You are just generalizing a result we already showed true for simple shift permutations. The key is the division algorithm.
Exercise (If $a^k=e$, Then The Order Of $a$ Divides $k$)

Suppose that $a$ is a group element with order $n$. If $a^k=e$, prove that $k$ is a multiple of $n$.

Click to see a solution.

Suppose $a^k=e$. This means $a^k=a^0$, which from the previous problem is true if and only if $k-0$ is a multiple of $n$. This shows that $k=k-0$ is a multiple of $n$.

Exercise (Properties Of An Element With Infinite Order)

Let $G$ be a group with $a\in G$. Suppose that the order of $a$ is infinite. Show that $a^i=a^j$ if and only if $i=j$, and state the order of $\langle a\rangle$.

Click to see a solution.

Suppose the order of $a$ is infinite. Then by definition this means that $a^n\neq e$ unless $n=0$. Clearly if $i=j$ then $a^i=a^j$. We only need to prove the converse of this statement. Suppose that $a^i=a^j$, and assume without loss of generality that $i\geq j$. Then $a^{i-j}=e$ (just multiply both sides on the right by $a^{-j}$). Because $a$ has infinite order, then we know $a^k\neq e$ for any positive integer $k$. Since $i-j\geq 0$ and $a^{i-j}=e$, we must have $i-j=0$. This means $i=j$.


The next problem asks you to compute the center $Z(G)$ of a nonabelian group, and the result is not trivial group. The automorphisms of the square gave us a group with 8 elements. This group consisted of 4 rotations and 4 reflections. Similarly, for any regular $n$-gon we can construct the automorphism group which will have order $2n$ and consist of $n$ rotations and $n$ reflections.

Definition (The Dihedral Groups $D_{n}$)

For each integer $n\geq 2$, we define the dihedral group on $n$ vertices, written $D_{n}$, to be automorphism group of the regular $n$-gon. This group consists of $n$ rotations and $n$ reflections, so has order $2n$.

In some texts, the author uses $D_{2n}$ instead of $D_{n}$, at which point they'll call it the dihedral group of order $2n$ instead of the dihedral group on $n$ vertices.

Problem 61 (The Center Of A Dihedral Group)

Let $G=D_4$, the automorphism group of the square. Recall that $Z(G)$ is the center of the group, or the set of elements that commute with every element of the group.

  1. What is $\langle R_{90} \rangle$? What is $\langle R_{180} \rangle$? What is $\langle R_{270} \rangle$? What is $\langle H \rangle$?
  2. Does $R_{90}\in Z(G)$? Explain. (Does $R_{90}$ commute with every element in $G$? In particular, does $R_{90}H=HR_{90}$?)
  3. Compute the center $Z(G)$ and show that it consists of more than just $R_0$. Make sure you can explain why each element is either in $Z(G)$, or not in $Z(G)$.


For more problems, see AllProblems