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.

Back to Lesson 29

On to Lesson 30

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