Grade 9/10 Math Circles
An Introduction to Group Theory Part 2 - Solutions

Exercise 1 Solution

The complete list of permutations of \(X\) is as follows: \[(1,2,3,4) \quad (1,2,4,3) \quad (1,3,2,4) \quad (1,3,4,2) \quad (1,4,2,3) \quad (1,4,3,2)\] \[(2,1,3,4) \quad (2,1,4,3) \quad (2,3,1,4) \quad (2,3,4,1) \quad (2,4,1,3) \quad (2,4,3,1)\] \[(3,1,2,4) \quad (3,1,4,2) \quad (3,2,1,4) \quad (3,2,4,1) \quad (3,4,1,2) \quad (3,4,2,1)\] \[(4,1,2,3) \quad (4,1,3,2) \quad (4,2,1,3) \quad (4,2,3,1) \quad (4,3,1,2) \quad (4,3,2,1)\]

Exercise 2 Solution

We compute that \[\sigma\circ\beta= \begin{pmatrix} 1 & 2 & 3 & 4 & 5\\ 1 & 5 & 3 & 2 & 4 \end{pmatrix}\] and \[\beta\circ\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5\\ 5 & 2 & 1 & 4 & 3 \end{pmatrix}.\]

Exercise 3 Solution

We need to convince ourselves that the 3 group axioms hold for \(S_3\) with composition. Let’s go through each axiom:

Axiom 1: For all \(\sigma,\beta,\alpha\in S_3\), we need \((\sigma\circ\beta)\circ \alpha\) to be the same permutation as \(\sigma\circ(\beta\circ\alpha)\). Showing this in a quick way is a bit tough given our current mathematical tools. So, let’s do a concrete example to see why this axiom holds. Consider the following permutations \[\sigma=\begin{pmatrix} 1 & 2 & 3\\ 1 & 3 & 2 \end{pmatrix},\quad \beta=\begin{pmatrix} 1 & 2 & 3\\ 3 & 2 & 1 \end{pmatrix},\quad \alpha = \begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix}.\] We compute that \[\sigma\circ\beta= \begin{pmatrix} 1 & 2 & 3\\ 2 & 3 & 1 \end{pmatrix} \quad \text{and} \quad \beta\circ\alpha = \begin{pmatrix} 1 & 2 & 3\\ 1 & 3 & 2 \end{pmatrix}.\] So \[(\sigma\circ\beta)\circ\alpha = \begin{pmatrix} 1 & 2 & 3\\ 2 & 3 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix} =\begin{pmatrix} 1 & 2 & 3\\ 1 & 2 & 3 \end{pmatrix}\] and \[\sigma\circ(\beta\circ\alpha) = \begin{pmatrix} 1 & 2 & 3\\ 1 & 3 & 2 \end{pmatrix}\circ\begin{pmatrix} 1 & 2 & 3\\ 1 & 3 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3\\ 1 & 2 & 3 \end{pmatrix}.\] We see that \((\sigma\circ\beta)\circ\alpha = \sigma\circ(\beta\circ\alpha)\), as desired.

Axiom 2: The identity element in \(S_3\) is the permutation \[\begin{pmatrix} 1 & 2 & 3\\ 1 & 2 & 3 \end{pmatrix}.\] Given how composition works, if you compose this element with any other element \(\sigma\in S_3\), we just get \(\sigma\) back. So indeed, this is the identity element \(\text{id}_{S_3}\). For example, we see that \[\begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix}\circ \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix}=\begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3\\ 1 & 2 & 3 \end{pmatrix}\circ \begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix}.\]

Axiom 3: We need to show that every element in \(S_3\) has an inverse. Let’s show how to construct the inverse of any element in \(S_3\) by using a concrete example. Consider \[\sigma= \begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix}.\] The inverse \(\sigma^{-1}\) of \(\sigma\) is obtained by switching the rows of \(\sigma\) and then reordering the columns so that the first row is \(1,2,3\). That is, \[\sigma=\begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix}\xrightarrow{\text{switch rows}} \begin{pmatrix} 3 & 1 & 2\\ 1 & 2 & 3 \end{pmatrix}\xrightarrow{\text{reorder columns}} \begin{pmatrix} 1 & 2 & 3\\ 2 & 3 & 1 \end{pmatrix}=\sigma^{-1}.\] And indeed, we see that \(\sigma\circ\sigma^{-1} = \text{id}_{S_3} =\sigma^{-1}\circ\sigma\) as desired. We can do this with any element in \(S_3\) to get it’s inverse!

Exercise 4 Solution

In lesson 2 we saw that every symmetry of the equilateral triangle can be identified with a permutation of \(\{1,2,3\}\). In the exact same way, we can identity every symmetry of the square with a permutation of \(\{1,2,3,4\}\). For example, consider the symmetry counter-clockwise rotation by 90 degrees:

We see that the 1 moves to where 2 was, 2 moves to where 3 was, 3 moves to where 4 was, and 4 moves to where 1 was. Given this, we identify this symmetry with the permutation \[\begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 3 & 4 & 1 \end{pmatrix}.\] In a similar way, we can identify every symmetry of the square with an element of \(S_3\). However, not every permutation of \(\{1,2,3,4\}\) can be identified with a symmetry of the square. One way to see this is to notice that there are more permutations of \(\{1,2,3,4\}\) than symmetries of the square. As seen in the solution for Exercise 1, there are 24 permutations of \(\{1,2,3,4\}\). However, there are only 8 symmetries of the square (see Problem 7 Solution in the solutions for Lesson 1). Since the number of symmetries of the square differs from the number of elements in \(S_3\), the symmetry group of the square can not be the same as the symmetric group on \(\{1,2,3,4\}\).

Exercise 5 Solution

The answer to both questions is yes! Let’s see why. Recall that our current situation is as follows:

The key idea is to introduce two beings who have never switched minds yet. In the episode they introduce the two characters Sweet and Bubblegum:

We first get Farnsworth’s body and Bubblegum’s body to switch minds:

Then we get Amy’s body and Sweet’s body to switch minds:

Note that Amy’s body has switched minds with Farnsworth’s body an Sweet’s body. So the only switch Amy’s body can do is with Bubblegum’s body. Similarly, the only switch Farnsworth’s body can do is with Sweet’s body. These switches result in:

We see that Amy’s mind is back in Amy’s body, and the same is true for Farnsworth. Since Sweet and Bubblegum’s bodies have not switched minds yet, we can do this switch to get everyone’s mind back into their own body:

We have successfully answered (1) and (2)! Note that this approach can be generalized to any number of mind switches.

Source: All character photos in this solution set are from: Futurama Fandom. Futurama wiki characters [Online]. Available from: https://futurama.fandom.com/wiki/Characters [Accessed March 24 2023].