Exercise 1) Show that and can both be expressed as a product of with some rotation . Namely, find such a product for each reflection.
Solution: Recall that is the reflection that acts on the corners by swapping corners 1 and 2, and swapping corners 3 and 4, which we denote by . Additionally, we have that acts as . There is more than one way that we can express as a product of with some rotation, but let’s find one such that we first act with , i.e., such that will show up on the right hand side of the rotation (since we apply it first). So, we want to find some such that . Let’s take this one step at a time. Since applying sends 1 to 3, we need a rotation that sends 3 to 2 so that the net action is that corner 1 goes to corner 2. The only rotation that does this is , so it better be the case that , otherwise it will be impossible to find such a product. We know that we have like we need, so let’s check the others. Recall that acts as as it is a rotation of 270 degrees. Where does send corner 2 then? Well, doesn’t touch corner 2, and sends 2 to 1. Thus, combining our results, we have that , as we hoped. Let’s now see where corner 3 goes. We need it to go to 4, and sure enough send 3 to 1, and then sends 1 to 4, thus giving the desired action. The last thing we need to check is that 4 goes to 3. Sure enough, doesn’t touch corner 4, and then sends 4 to 3. We’ve therefore established that . Note, if we wanted a product that had as the second (left-most) factor, then we’d have , as the reader can check for herself.
Now we need to find a similar expression for , which acts as , . Again, we’ll look for a product that has acting first, though there will also exist a product with acting second, and we’ll simply state what this is after doing the first case. If we first act with , then we’ll have sent , so we need a rotation that now sends 3 to 4. The only rotation that does this is , so let’s hope that this is indeed the right rotation for all the other corners too (spoiler alert, it will be). We now know that sends 1 to 4, so let’s see where it sends 4. doesn’t touch corner 4, and then rotates 90 degrees so that 4 goes to 1, thus the net action of is that 4 goes to 1, so we indeed have so far. Let’s now check the other corners. sends 3 to 1, and then takes 1 to 2, so that’s good. And sure enough doesn’t touch corner 2 while takes 2 to 3, so we also indeed have the action (on the relevant corners) of being . Thus, the entire action of is , which is precisely , and we’ve therefore established that , as desired. This establishes the fact that given alone, and all of the rotations, we can always get to any other reflection by simply composing with rotations (as we covered the case in the lesson).
Exercise 2) What is the subgroup containing at least one rotation as well as , and which isn’t the whole group ?
Solution: As we saw in the lesson, the only rotation that we can put in our subgroup is , so we have so far. Each element so far is its own inverse. We now need to find out what and are. Let’s do the first. acts as . If we then act with which acts as , we have the net action being that (as from and from ), , , and . Thus, the net action of is , which is precisely . Similarly, we would find that , so that is indeed the last element we need to add to our subset to make it closed under multiplication. Therefore, the subgroup that we’re after is .
Exercise 3) Show that any rotation in the dihedral group can be expressed as a product of . I.e., show that by multiplying to itself over and over again, we cycle through all possible rotations before coming back to the identity. This is an example of the special nature of dihedral groups of prime-gons discussed in the final paragraph.
Solution: This problem is straightforward, but let’s us see an example of how the subgroup of rotations behaves when the dihedral group is such that is prime. We note that the rotations for are where we include the identity as a rotation, and where is the counterclockwise rotation by degrees, just as was the counterclockwise rotation of degrees when , i.e. when we were considering the group . We note that is approximately 51.4 degrees. Additionally, , and the rest of the rotations are generated from in the obvious way, and finally we’ll adopt the convention that , , etc.
With this final convention, it is hopefully clear that , where the addition in the subscripts is addition modulo 7, so that for example in the obvious way, and that and , where we’ve taken advantage of the new convention we’ve adopted. Finally, we use the notation to mean composed with itself times, as we have before. For example, , etc. We’re now ready to address the question.
We first consider . We have . Using this result, we can calculate . We have . Using this result, we can now calculate . We have . We then have . Similarly, we get . Similarly, we get . Thus, after 7 powers of , we’ve cycled back through to the identity, meaning that the power of will only bring us back to the first power, and the ninth power will bring us back to the second power, etc. Moreover, we’ve hit every single rotation by only considering powers of , which means that if we wanted to construct a subgroup of containing , we’d have to also include all of the other rotations in order for the subset to be closed under multiplication. The reader can check, or just convince herself of the fact that the same will hold for any of the other rotations in , and indeed for any of the rotations in whenever is prime. We can, and will, prove this more rigorously in the future, but for now let’s just notice that it’s true.