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From Ben:

We'll have the following people present in class:

  • 22 -
  • 23 -
  • 24 -
  • 26 -
  • 27 -
  • 28 -
  • 29 -

Problem 22 (Creating Cayley Graphs Of Simple Shift Permutations)

Let's make some Cayley graphs as they relate with the game of Scoring on simple shift permutations. To get Sage to make the graphs, you'll first have to write the simple shift permutation in disjoint cycle notation. As an example, if there are 6 letters in our alphabet $X=\{1,2,3,4,5,6\}$, then we could write the simple shift $\phi_2$ as $(2,4,6)(1,3,5)$. If player 1 chooses $\sigma_1=\phi_2$ and player 2 chooses $\sigma_2=\phi_3$, then the following Sage code visually provides a graph of $\text{span}(\{\sigma_1\})$ and then $\text{span}(\{\sigma_1,\sigma_2\})$. 

g1 =  PermutationGroup(["(2,4,6)(1,3,5)"])    
d1=g1.cayley_graph()                                                 
d1.show(color_by_label=True, vertex_size=0.03, vertex_labels=True)   
print(g1.list())                                                     

g2 = PermutationGroup(["(2,4,6)(1,3,5)","(1,4)(2,5)(3,6)"]) 
d2=g2.cayley_graph()
d2.show(color_by_label=True, vertex_size=0.03, vertex_labels=True)
print(g2.list())

By hand, my first graph would contain two unconnected triangles. My second graph would look similar to the graph above.

  1. Suppose the alphabet is $X=\{1,2,3,4,5,6\}$. Player 1 chooses $\phi_1$. Draw the associated graph.
  2. Suppose the alphabet is $X=\{1,2,\ldots,10\}$. Player 1 chooses $\phi_5$. Player 2 chooses $\phi_4$. Draw the two associated graphs.
  3. Suppose the alphabet is $X=\{1,2,\ldots,12\}$. Player 1 chooses $\phi_5$. Draw the associated graph.
  4. Suppose the alphabet is $X=\{1,2,\ldots,12\}$. Player 1 chooses $\phi_6$. Player 2 chooses $\phi_2$. Player 1 then chooses $\phi_3$. Draw the three associated graphs. The first graph should consist of 6 arrows between pairs of vertices (Sage will only show you the segment connnecting $()$ and $\phi_6$). Your second graph should have 2 hexagons with arrows crossing in the middle. The last graph should connect the edges of the hexagons.

Here's a blank Sage cell that you can use to perform some computations. You'll probably want to copy code from above and make some changes.

When we create graphs with Sage, it always creates a graph of the span of the permutations given. These graphs have a lot of really nice symmetry properties, and these properties are central to abstract algebra.

Problem 23 (Permutation Scoring On S 3)

Consider the game of permutation scoring on the set $S_3$, where $S_3$ represents the set of all permutations of $X=\{1,2,3\}$. We already know that there are 6 elements in $S_3$.

  1. Play the game of permutation scoring on $S_3$. Play it a few times.
  2. Can player 1 win by choosing a single permutation in $S_3$?
  3. Does player $1$ have a winning strategy? What is it?
  4. Pick an element in $S_3$ other than the identity, and call it $\sigma$. Let $S=\text{span}(\sigma)$. What is the span of $S$?
  5. Repeat part 4 with a different element of $S_3$. You can use Sage to make this part fast (first compute the span of $\sigma$ with Sage. The list below the graph gives you the elements of the span. Then create a graph using these elements.
  6. Make a conjecture.
Definition (A Closed Set Of Permutations)

We say that a set of permutations is closed if the set $S$ is equal to its span, so notionally we write $$S=\text{span}{(S)}.$$ In other words, we say that $S$ is closed if the set $S$ already contains every composition combination of permutations of elements in $S$.

Problem 24 (The Span Of A Set Of Permutations Is Closed)

Let $S$ be a set of permutations of $X$. Show that $\text{span}(S)$ is closed. In other words, show that once you form the span of a set $S$ of permutations of $X$, you cannot obtain any new permutations by spanning the span. For this reason, we'll often refer to $\text{span}(S)$ as the closure of $S$.

Problem 26 (Powers With Modular Arithmetic)

  1. We know that $481\mod {483} =(-2)\mod {483}$. Use this fact to quickly compute $481^2\mod{483}$ without a calculator. [How can you use the $-2$ to simplify this?]
  2. Using the same idea, what is $481^{3}\mod{483}$ and $481^4\mod{483}$? You should be giving a positive integer less than 483 for all your answers.
  3. Explain why $12\cdot 16 \mod 17$ is the same as $(-5)(-1)\mod 17$.
  4. Now compute $14^k \pmod {17}$ without a calculator for as many values of $k$ as needed until you find a repeating pattern. See the hint below.

Let me give you a suggestion. Rather than trying to compute $14^4$, instead first compute $14^2$ by considering $(-3)^2$. To compute $(14)^3$, instead use the product $(14)^2(14)=(9)(-3)=(-8)(-3)$. Use your work from the previous power to figure out the next power. If you try to do this with brute force, it gets outrageously large really fast. If you swap from positive to negative values when the numbers get large (above 8), you should be able to do this without ever needing long division. You should find that no product ever gets above 24.

The following Sage code will compute powers with modular arithmetic. You are welcome to check your answers with this, but the whole point to the problem is to learn how to simplify the computations by hand. To compute $a^m\mod n$ you would write "power_mod($a$,$m$,$n$)." 

power_mod(481, 4, 483)

Problem 27 (Matrix Encryption Mod 5)

In the Problem (Simple Matrix Encryption), we took a plain text message written in columns of a matrix $M$ and a matrix $A$ and used the product $AM$ to encode a message. Let's now combine matrix encryption with modular arithmetic to obtain a new encryption scheme. Once we have done this, we'll be one step closer to understanding modern encryption techniques. To simplify our work, let's use a small alphabet consisting of 5 letters. Instead of using the set $S=\{a,b,c,d,e\}$, let's just use numbers for our letters and we'll consider the alphabet $S=\{0,1,2,3,4\}=\mathbb{Z}_5$. Given the plain text "bad bed" or "10 31 43", we can encode it in the columns of the matrix $$M=\begin{bmatrix}1&3&4\\0&1&3\end{bmatrix}.$$ An 2 by 2 encryption key using coefficients mod 5 is a 2 by 2 matrix $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ where $a,b,c,d\in \mathbb{Z}_5$. Here are the steps to our encryption scheme.

  • First convert the plain text to a matrix $M$ where the letters are place in the matrix top to bottom, column by column.
  • Take the encryption key $A$ and compute the product $AM$.
  • Reduce all the entries in $AM$ mod 5. So the cipher text only consists of the numbers in $\mathbb{Z}_5$.

Answer the following questions.

  1. Suppose we let $A=\begin{bmatrix}4&2\\1&3\end{bmatrix}$. Why is this matrix not suitable as an encryption key? (Hint: what is the determinant, and why is that a problem?)
  2. Suppose we let $A=\begin{bmatrix}2&1\\4&3\end{bmatrix}$. The inverse of $A$ if we ignore modular arithmetic is $\frac{1}{2}\begin{bmatrix}3&-1\\-4&2\end{bmatrix}$. We know that we can replace $-4$ with $1$ when working modular 5. Similarly we can replace $-1$ with a positive number. We need to find the multiplicative inverse of 2 when working modular 5. Find a value $b\in \mathbb{Z}_5$ so that $2b\mod 5=1$. This number $b$ is the inverse of $2$ modular 5.
  3. State a matrix $B$ with coefficients in $\mathbb{Z}_5$ so that $AB$ is the identity matrix after reducing each entry mod 5. Actually perform the product by hand, and show how this matrix product reduces to the identity.
Definition (Modular Multiplicative Inverse and $U(n)$)

Let $n$ be a positive integer, and let $a\in \mathbb{Z}_n$. Then the modular multiplicative inverse of $a \mod n$ is an integer $b\in \mathbb{Z}_n$ such that $(ab)\mod n=1$. If such an integer exists, then we'll write $a^{-1}\mod n$ to represent the modular multiplicative inverse. If the context is clear, we may just refer to this as the inverse instead of the modular multiplicative inverse. We'll use the notation $U(n)$ to represent the set of elements in $\mathbb{Z}$ that have a modular multiplicative inverse. Notationaly, we write $$U(n)=\{a\in\mathbb{Z}_n\mid \text{ there exists } b\in \mathbb{Z}_n \text{ such that } (ab)\mod n=1\}.$$

The Sage code "inverse_mod(a,n)" will compute the modular multiplicative inverse of $a\mod n$ or return an error. 

inverse_mod(7, 11)

Problem 28 (Computing Multiplicative Inverses For Small N)

To use modular arithmetic in matrix encryption, we must be able to find the modular multiplicative inverse of $a \mod n$. For each $n\leq 13$ and $a\in \mathbb{Z}_n$, we now compute the inverse of $a\mod n$ or explain why it does not have one.

  1. Why does $0\mod n$ never have an inverse?
  2. What is the multiplicative inverse of $1\mod n$?
  3. For $n=3$, what is the multiplicative inverse of $a=2$.
  4. For $n=4$, show that 2 does not have an inverse by computing $2\cdot 0$, $2\cdot 1$, $2\cdot 2$, and $2\cdot 3$ all modulo $4$. What is $3^{-1}\mod 4$?
  5. For each $n$ between 5 and 13, create a table that shows the inverse of each $a\in \mathbb{Z}_n$. List the elements of $\mathbb{Z}_n$ along the top row, and beneath each element list the multiplicative inverse, or cross off the number if there is no multiplicative inverse.

When you are done you should have a list of the elements of $U(n)$ for each $n$ up to 13, as well as the inverse of each element in $U(n)$. Do you see any patterns?

For $n=9$, you should obtain a table somewhat similar to the following table.

$a$012345678
$a^{-1}$x15x72x48
Definition (Cayley Graph Of A Span Of Permutations)

Let $X$ be a set. Let $S$ be a set of permutations of $X$. Then a Cayley graph of $H=\text{span} (S)$ generated by $S$, which we'll write as $(H,S)$, is a colored directed graph that satisfies the the following three properties:

  1. The vertex set is $\text{span} (S)$. Each vertex corresponds to a permutation.
  2. Each element $s \in S$ is assigned a unique color which we'll denote by $c_s$.
  3. For each color $c_s$, and each vertex $\sigma$, we draw the colored arrow $(\sigma ,s \circ \sigma)$.

Most of the time we assume that $S$ does not contain the identity. However, if it does contain the identity, then we just draw a colored loop $(\sigma,\sigma)$ at each vertex.

Problem 29 (Cayley Graph Patterns)

In each scenario below, you should draw a Cayley graph.

  1. Let $X=\mathbb{Z}$, an infinite set. Consider the permutation $\phi_1:\mathbb{Z}\to \mathbb{Z}$ defined by $\phi_1(x)=x+1$. This permutation shifts elements in $\mathbb{Z}$ right one, with no wrap around. If we let $S=\{\phi_1\}$, give a rough sketch of the Cayley graph of $\text{span}(S)$. It's a rough sketch of the graph because there are infinitely many vertices.
  2. How would you change your graph in part 1 if instead you used $S=\{\phi_2,\phi_3\}$, where these are shifts right 2 and 3 respectively?
  3. Suppose that $X$ is a set, and that $S$ contains a single automorphism $s$ of $X$. Suppose also that we know $s^6$ is the identity (so following the arrow colored $c_s$ 6 times will return us to where we started). Construct a Cayley graph of $\text{span}(S)$. Is there more than one right answer to this?
  4. Suppose that we know $S$ contains two elements $a$ and $b$. We also know that $a^2$ and $b^4$ are both the identity. In addition, we know that $a\circ b=b\circ a$, so if we start somewhere and follow the arrows colored $c_a$ and then $c_b$, then we should end up at the same place if we followed the arrows colored $c_b$ and then $c_a$. Construct a graph that satisfies these relationships. We'll often write this as $$\left<a,b\mid a^2=id, b^4=id, a\circ b=b\circ a\right>.$$

For more problems, see AllProblems