Lesson 29 Solutions

Exercise 1) Show that $r_3$ and $r_4$ can both be expressed as a product of $r_1$ with some rotation $\rho_i$ . Namely, find such a product for each reflection.

Solution: Recall that $r_3$ is the reflection that acts on the corners by swapping corners 1 and 2, and swapping corners 3 and 4, which we denote by $1\leftrightarrow 2, 3\leftrightarrow 4$ . Additionally, we have that $r_1$ acts as $1\leftrightarrow 3$ . There is more than one way that we can express $r_3$ as a product of $r_1$ with some rotation, but let’s find one such that we first act with $r_1$ , i.e., such that $r_1$ will show up on the right hand side of the rotation (since we apply it first). So, we want to find some $i$ such that $\rho_i\cdot r_1=r_3$ . Let’s take this one step at a time. Since applying $r_1$ 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 $\rho_3$ , so it better be the case that $\rho_3\cdot r_1=r_3$ , otherwise it will be impossible to find such a product. We know that we have $1\rightarrow 2$ like we need, so let’s check the others. Recall that $\rho_3$ acts as $1\rightarrow 4\rightarrow 3\rightarrow 2 \rightarrow 1$ as it is a rotation of 270 degrees. Where does $\rho_3\cdot r_1$ send corner 2 then? Well, $r_1$ doesn’t touch corner 2, and $\rho_3$ sends 2 to 1. Thus, combining our results, we have that $1\leftrightarrow 2$ , as we hoped. Let’s now see where corner 3 goes. We need it to go to 4, and sure enough $r_1$ send 3 to 1, and then $\rho_3$ sends 1 to 4, thus giving the desired action. The last thing we need to check is that 4 goes to 3. Sure enough, $r_1$ doesn’t touch corner 4, and then $\rho_3$ sends 4 to 3. We’ve therefore established that $\rho_3\cdot r_1=r_3$ . Note, if we wanted a product that had $r_1$ as the second (left-most) factor, then we’d have $r_1\cdot \rho_1=r_3$ , as the reader can check for herself.

Now we need to find a similar expression for $r_4$ , which acts as $1\leftrightarrow 4$ , $2\leftrightarrow 3$ . Again, we’ll look for a product that has $r_1$ acting first, though there will also exist a product with $r_1$ acting second, and we’ll simply state what this is after doing the first case. If we first act with $r_1$ , then we’ll have sent $1\rightarrow 3$ , so we need a rotation that now sends 3 to 4. The only rotation that does this is $\rho_1$ , 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 $\rho_1\cdot r_1$ sends 1 to 4, so let’s see where it sends 4. $r_1$ doesn’t touch corner 4, and then $\rho_1$ rotates 90 degrees so that 4 goes to 1, thus the net action of $\rho_1\cdot r_1$ is that 4 goes to 1, so we indeed have $1\leftrightarrow 4$ so far. Let’s now check the other corners. $r_1$ sends 3 to 1, and then $\rho_1$ takes 1 to 2, so that’s good. And sure enough $r_1$ doesn’t touch corner 2 while $\rho_1$ takes 2 to 3, so we also indeed have the action (on the relevant corners) of $\rho_1\cdot r_1$ being $2\leftrightarrow 3$ . Thus, the entire action of $\rho_1\cdot r_1$ is $1\leftrightarrow 4, 2\leftrightarrow 3$ , which is precisely $r_4$ , and we’ve therefore established that $\rho_1\cdot r_1=r_4$ , as desired. This establishes the fact that given $r_1$ alone, and all of the rotations, we can always get to any other reflection by simply composing $r_1$ with rotations (as we covered the $r_2$ case in the lesson).

Exercise 2) What is the subgroup containing at least one rotation as well as $r_3$ , and which isn’t the whole group $D_4$ ?

Solution: As we saw in the lesson, the only rotation that we can put in our subgroup is $\rho_2$ , so we have $\{e, \rho_2, r_3, ...\}$ so far. Each element so far is its own inverse. We now need to find out what $\rho_2\cdot r_3$ and $r_3\cdot \rho_2$ are. Let’s do the first. $r_3$ acts as $1\leftrightarrow 2, 3\leftrightarrow 4$ . If we then act with $\rho_2$ which acts as $1\leftrightarrow 3, 2\leftrightarrow 4$ , we have the net action being that $1\rightarrow 4$ (as $1\rightarrow 2$ from $r_1$ and $2\rightarrow 4$ from $\rho_2$ ), $4\rightarrow 1$ , $2\rightarrow 3$ , and $3\rightarrow 2$ . Thus, the net action of $\rho_2\cdot r_3$ is $1\leftrightarrow 4, 2\leftrightarrow 3$ , which is precisely $r_4$ . Similarly, we would find that $\rho_2\cdot r_3=r_4$ , so that $r_4$ 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 $\{e, \rho_2, r_3, r_4\}$ .

Exercise 3) Show that any rotation $\rho_i$ in the dihedral group $D_7$ can be expressed as a product of $\rho_4$ . I.e., show that by multiplying $\rho_4$ 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 $D_n$ is such that $n$ is prime. We note that the rotations for $D_7$ are $\{e, \rho_1, \rho_2, \rho_3, \rho_4, \rho_5, \rho_6\}$ where we include the identity as a rotation, and where $\rho_1$ is the counterclockwise rotation by $\frac{360}{7}$ degrees, just as $\rho_1$ was the counterclockwise rotation of $\frac{360}{4}=90$ degrees when $n=4$ , i.e. when we were considering the group $D_4$ . We note that $\frac{360}{7}$ is approximately 51.4 degrees. Additionally, $\rho_2=\rho_1\cdot\rho_1$ , and the rest of the rotations are generated from $\rho_1$ in the obvious way, and finally we’ll adopt the convention that $\rho_7=e$ , $\rho_8=\rho_1$ , etc.

With this final convention, it is hopefully clear that $\rho_i\cdot \rho_j=\rho_{i+j}$ , where the addition in the subscripts is addition modulo 7, so that for example $\rho_2\cdot \rho_3=\rho_5$ in the obvious way, and that $\rho_3\cdot \rho_4=\rho_7=e$ and $\rho_5\cdot \rho_6=\rho_{11}=\rho_7\cdot \rho_4=e\cdot \rho_4=\rho_4$ , where we’ve taken advantage of the new convention we’ve adopted. Finally, we use the notation $\rho_4^n$ to mean $\rho_4$ composed with itself $n$ times, as we have before. For example, $\rho_4^3=\rho_4\cdot \rho_4\cdot \rho_4$ , etc. We’re now ready to address the question.

We first consider $\rho_4^2$ . We have $\rho_4^2=\rho_4\cdot \rho_4=\rho_8=\rho_1$ . Using this result, we can calculate $\rho_4^3$ . We have $\rho_4^3=\rho_4^2\cdot \rho_4=\rho_1\cdot \rho_4=\rho_5$ . Using this result, we can now calculate $\rho_4^4$ . We have $\rho_4^4=\rho_4^3\cdot \rho_4=\rho_5\cdot \rho_4=\rho_9=\rho_2$ . We then have $\rho_4^5=\rho_4^4\cdot \rho_4=\rho_2\cdot \rho_4=\rho_6$ . Similarly, we get $\rho_4^6=\rho_4^5\cdot \rho_4=\rho_6\cdot \rho_4=\rho_{10}=\rho_3$ . Similarly, we get $\rho_4^7=\rho_4^6\cdot \rho_4=\rho_3\cdot \rho_4=\rho_7=e$ . Thus, after 7 powers of $\rho_4$ , we’ve cycled back through to the identity, meaning that the $8^{th}$ power of $\rho_4$ 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 $\rho_4$ , which means that if we wanted to construct a subgroup of $D_7$ containing $\rho_4$ , 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 $D_7$ , and indeed for any of the rotations in $D_n$ whenever $n$ is prime. We can, and will, prove this more rigorously in the future, but for now let’s just notice that it’s true.