 Welcome back to our lecture series Math 42-20, Abstract Algebra I for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misseldine. In lecture 31, we're going to talk about the idea of a factor group, sometimes called a quotient group. But we're going to do that in the next video properly. In this video, I want to continue our discussion of conjugates and consciously classes that we introduced at the end of lecture 30. In our last video, I actually have proven that the conjugates in the symmetric group occur exactly when two permutations have the same cycle structure. We want to give in this video some sufficient and necessary conditions to know when two elements of the dihedral group are in fact conjugates of each other. Let Dn be the order two-n dihedral group. That is, it's the symmetry group of the regular n-gon. When it comes to the dihedral group, basically we can classify the elements in one of two families. There's the rotations of the n-gon. We're going to call those R. R is a single rotation counterclockwise. You have the rotations in there, which will include the identity, which you think of as a zero degree rotation. Then there's also going to be the reflections for what you can reflect across the x-axis, y-axis, or various oblique lines passing through your polygon there. How does the consciously classes of these elements, how are they determined? Well, it turns out for rotations, it's fairly simple. An element will be clearly every element, like if you just take an arbitrary rotation R to the K, it's clearly going to be conjugated to itself because congenitally gives us a equivalence relationship. But in the dihedral group, the only other element that a rotation can be conjugated to would be its inverse. Now, assuming this is true, this gives us a very interesting corollary. When n is an even number, that means R to the K will be conjugate to itself. The only other element it would be conjugate to would be R to the negative K, which it's itself again, if the number is in fact even. Which would tell you in that situation that if you take half of the rotation, like half of the degree there, that element is going to be central because the only thing conjugate to it is itself. That helps one determine the center of the dihedral group. The dihedral group, if you have the dihedral group 2K, then its center is going to be the identity and this half rotation, half spin. On the other hand, when you have an odd dihedral group, then it will be centerless. That is the only thing in there is the identity. That's a happy little consequence of this observation here that rotations are only conjugate to each other. Now, you might be like, why aren't there any reflections in the center? We'll see that in just a second. When it comes to the reflections, there's two possibilities and this depends whether you're even or odd. If the degree of the dihedral group is odd, then it turns out there's only one congenit class that contains all the reflections. All the reflections would be conjugate to each other. On the other hand, if n is even, there'll be two congenit classes based upon parity that you'll have, you'll have because after all, every element of the dihedral group can be written in this normal form, r to the ks or r to the k, but for reflections that look like r to the ks. If the exponent is an even number, then that'll be in the even class. If k is an odd number, then it'll be in the odd class. You have these two congenit classes based upon parity when the degree is even, but when the degree is odd, they all get glued together. Let's prove the first statement. I guess before we prove that, just came back to the comment about the center here. Well, because the center, since these reflections are always glued to something, whether it's glued to all the reflections or half the reflections, there is no reflection in the dihedral group that can be central. So we get the claims we had earlier, if this theorem was true. So now let's prove it. So to prove it, we're gonna first start off with the statement about the rotations, right? Why are rotations only conjugates to their inverses amongst other things, right? So let's take a typical rotation r to the k. Now we know that powers of r will commute with each other. So if I take r to the k and I conjugated by r to the m and it's inverse r to the negative m, well, these are just powers of r, right? So they're just gonna commute with each other. So you can commute to the rm with the rk, in which case you get this, r to the m and r to the negative, m cancels into r to the k, right? So when you conjugate r to the k by a rotation, you'll just get back itself. It's left unmoved. The reflection is what's gonna produce this inversion going on right here. So if you have a reflection, you have a typical reflection will look like r to the ms. You're gonna conjugate r to the k by that reflection and it's inverse r to the ms inverse, right? So some things to remember here about reflections, very cool to their own inverses. But if you use the shoe sock principle, you're gonna switch these things around their order when you take the inverse, in which case then the inverse here is gonna look like s inverse r to the m inverse. So taking the shoe sock principle here and also redoing parentheses, I first have to conjugate r to the k by s first. And so what happens in that situation? We've seen previously that if you want to pass a reflection, pass a rotation inside of the dihedral group, then you can move s past r, but you have to take the inverse of r. And so that's gonna end up giving us r to the n minus k ss inverse. You get its inverse like so. Then the s's will cancel out. So we then get rm r to the n minus k r to the n minus m. Like we said before, rotations can be with each other. So you get r to the n minus k then rm r minus m, they cancel out. So you just get r to the n minus k r. You can just abbreviate that as r to the negative k. And so we see that when you conjugate a rotation by rotation, you get back itself. If you conjugate a rotation by a reflection, you get back its inverse. And as every element of the dihedral group is either rotation, reflection, that exhaust all possibilities. So the constancy class of a rotation will always be itself and its inverse, those only possibilities. Okay, let's take a look at reflections. Reflections, you have to be a little bit more careful, but we can handle it, no problem here. So if you take a typical reflection in the dihedral group, its normal form will look like r to the k times s. Well, we can conjugate this reflection by a typical rotation. What happens there, r to the m and r to the negative m. So again, looking at multiplication here, what I wanna do is I want to switch the order of s and r to the negative m. But like we said a moment ago, to pass an r by an s, you have to take the inverse of the r. So r to the m when it moves to the left will become, excuse me, r to the negative m, when it moves to the left will become r to the m. And then as these are all just powers of r, you can squish them all together. And we see that we end up with a reflection right here, a reflection of r to the k plus two m s. So we get that r to the k s will be conjugate to r to the k plus two m s, right? So notice in this situation, if we have r to the k s, this is, it's conjugate to a reflection, but comparing the powers of the rotation here, you have k and you have k plus two m, right? So what this tells you is that, if you had some typical element r k s, it's conjugate to a reflection r l s, right? What this observation is telling you above is that one possibility here is if l is congruent to k mod two, right? Because when you look at k versus k plus two m, those two numbers have the same parity. They're either both even numbers or they're both odd numbers because you're adding some potential arbitrary, because after all m was arbitrary here, some arbitrary multiple of two to them. So if k and l have the same parity, then they will be conjugate to each other, all right? And so that kind of leads to, that kind of leads to the above case, right? Didn't we mention this earlier that if the degree is even, then the consistency classes will be biased upon parity. The evens are conjugate to each other and the odds are conjugate to each other. We've proven that. When you're odd, it's a single consistency class, which means we take these two sets and we glue them together, which is true in the case we're in. We just have to glue them together somehow. So what we wanna then argue, because we now know if the power of the rotation has the same parity, the reflections will be conjugate to each other. What we have to now prove is that when you're odd, when your degree is odd, you can glue these things together, but when you're even, these are distinguishable, okay? So before we can do that, let's make one other comment here. So this is what happens if we rotated, excuse me, if we conjugate a reflection by rotation. What happens if you conjugate a reflection by a reflection? Well, the calculation's gonna be similar, all right? But if you draw this thing out, you're gonna get RMSRKS. You're gonna get then S inverse R to the negative M, right? The S's cancel out. We're going to then get RMSK, excuse me, R to the K minus M. Commute the S with R, which gives you the inverse, RMRK minus M. We're gonna get S like that, up I forgot to take the inverse there, so take a negative sign. And then when you combine those together, you're gonna get R to the 2M minus K times S, right? Which is what we have right here, which admittedly above I wrote it as N minus K plus 2S, same basic idea. So then this summarizes what we've observed so far. Let me get this out of the way. So notice what we observed so far. So the first condition, this conjugation relationship tells us that, oh, K and L could have the same parity. That would force them to be conjugate. The other option is that N minus K and L must have the same parity, okay? So either, so in order for this to work, either K and L have the same parity or L has the same parity as the inverse, the additive inverse of K in that situation, mod N. All right, so that's the situation we're in. So let's consider the two possibilities. Now, in the case that N is even, N is even, I want you to be aware that these two observations tell us the same thing, all right? Because this equation right here, let me kind of rewrite it here. We're gonna rewrite this equation the following way. You could write this as K plus L is congruent to N mod two, okay? So this first statement says they have the same parity. This one says that when you add them together, it'll have the same parity as N. Well, if N is an even number, right? That means it's congruent to two. I should say it's a multiple two, therefore it's congruent to zero mod two. Now, if K plus L, if their sum is zero mod two, that means they have the same parity because evens plus evens give you an even. Odd plus odd gives you an even. And if it's even on mismatch, you'll get back an odd. So K plus L can be congruent to zero so just they have the same parity. So that means that this statement and this statement are identical, right? They don't give you two different statements, give you the one statement. So the only way that two reflections are conjugate to each other is if their powers of the rotations have the same parity. And that then gives us the two classes we had before. For even degree, the two classes are based upon parity. There's no way to mix them together. Now, when you're odd, right? Let's look at the statement K plus L. Well, if you're odd, okay, then n being odd means that it's gonna be congruent to one mod, mod, what? Mod two, clearly that was the typo there, sorry. So K plus L being congruent to one mod two means they have different parities. So wait, okay. If they have the same parity, they're conjugate. But if they have different parities, they're conjugate. Doesn't leave a lot of wiggle room, right? I mean, if your two numbers either have the same parity or they have different parities. And so in either situation, boom, they're conjugate. And so this then explains that all of the reflections are conjugates to each other in an odd dihedral group. Okay? And so another takeaway I wanna mention about this theorem here, because we've now finished the proof, is look at the rotation subgroup. That is the cyclic subgroup generated by a single rotation. Well, we mentioned earlier that the congisly classes of rotations are just the rotation in its inverse. This tells us that the rotation subgroup is a union of congisly classes, which implies that it's actually a normal subgroup. So we actually always get that the rotation, the rotational subgroup, the dihedral group is a normal subgroup. It's a very important normal subgroup. Another argument you can make is that the rotation subgroup is actually a subgroup of index two, which also is another argument that it is a normal. But in this regard, I wanted to make the mention that since it's a union of congisly classes, that implies it's normal because normal subgroups are exactly those subgroups which are closed under conjugation.