Let
The complete list of permutations of
Let
We compute that
Convince yourself that
We need to convince ourselves that the 3 group axioms hold for
Axiom 1: For all
Axiom 2: The identity element in
Axiom 3: We need to show that every element in
Show that the symmetry group of the square (i.e., the regular polygon
with 4 sides) is not the same as the symmetric group on
In lesson 2 we saw that every symmetry of the equilateral triangle
can be identified with a permutation of
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
Consider the original situation where only Amy and Professor Farnsworth have switched minds (see Figure 3).
Can we get Amy’s mind back into Amy’s body and Farnsworth’s mind back into Farnsworth’s body?
Further, can (1) be done in such a way that everyone who switches minds ends up with their own mind in the end?
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].
Compute the following compositions:
Solution: We compute that:
Compute the inverses of the following permutations:
Solution: As illustrated in the solution to Problem 1, we
see that the inverse of (a) is itself. The inverse of (b) is also
itself. And the inverse of (c) is
Let
hint: how many elements are in
Solution: The argument for Exercise 4 also holds here. In
other words, one reason to see why the symmetry group of