 In this video, I want to prove some properties about normal subgroups. The first one you see right here, theorem 1014, this actually gives you a very simple check to see when, well, to identify some normal subgroups, right? It's a sufficient condition to be normal. It's not a necessary condition. Let G be a group with a subgroup H. Let's suppose that the index of H is 2. That is, there's only two cosets for this subgroup. Then we can prove that if there's only two cosets, then GH is equal to HG for all elements G inside of G. That, of course, implies that H is a normal subgroup. And so what's the argument here? Well, one thing you should always notice that in terms of, in terms of like, if you take the identity, right? Well, honestly, if you take anything in H itself, then you're going to take H times H. Well, this is just equal to H itself, which is equal to H times H. Like, if you take a group and you multiply it by an element in that group, whether it's on the left or the right, you're just performing a permutation. You shuffle everything around. But as the set doesn't matter about the order, you get everything back that you had before. This is a consequence of cancellation. So the coset that coincides with the subgroup, remember, one of the cosets is that the subgroup itself, that subgroup is always going to be, it's going to be left and right compatible here. The coset on the left versus the right are the same. That's not a problem. So what about the other option? Since there's only two cosets, if you're not in H, you're in the other coset, the second coset, which let's take such an element for a moment, right? X, take it in the symmetric, the set difference of G and H, right? So take something that's in G but not in H. Because the index is two, there's only two cosets. One of them is H. And since X is not in H, the left coset represented by X there, so XH, that has to, it's from, it's from G, take away H. But since the coset's form of partition, right, G itself is the union of H and XH, okay? Likewise, we're going to see that G is the union of H and HX right here. But the thing is, if you have this disjoint union, right, the intersection of H with XH is equal to the empty set. This actually implies that XH is equal to G, take away H. So when you have only two cosets, the second coset other than the subgroup is just its complement as a set. And this is irrelevant whether you're left or right. If you take H intersect HX, right, that's still the empty set. And since the union of the two sets is equal to HX, that also gives us that G take away H is equal to HX. And that's the crux of the argument right there, that because you have only two cosets and the cosets partitioned the group, that coset, which is not the identity, that coset that's not H, right? Because these two match up, the other two chunks, right, is they're going to have to agree with each other, right? HX is going to have to equal XH. So subgroups of index two have to be normal because their cosets have no choice but to agree with each other. One important example of this is the alternating group, right? The alternating group always has as its order, in factorial over two, the symmetric groups, orders always in factorial. The argument behind this order is actually half of the permutations are even, the other half are odd, the even ones of course belonging to AN. And so this then tells us as a consequence that the index of AN inside of SN is equal to two. So we always have that AN is a normal subgroup of SN. And it turns out that when AN is greater than equal to five, then AN is the only proper non-trivial normal subgroup inside of SN. Turns out where our normal subgroups are actually quite rare in the symmetric group when the degree gets sufficiently large. It's also true that if you're looking at the quaternion group, right? In the quaternion group, you have that the subgroup generated by I is a normal subgroup because it's index two. So is the subgroup generated by J. So is the subgroup generated by K. If you're looking at like the dihedral group, right? You have the dihedral group. It has always sitting inside of it, the subgroup generated by R, right? Because the subgroup generated by our rotation here, this is just isomorphic to ZN, the cyclic group of order N. And so the order of this group is N. On the other hand, the order of the dihedral group is two N. And so that tells us that the index of R inside of DN right here is gonna equal two. So we get that the rotational subgroup, that is the subgroup of rotations inside of the dihedral group is always gonna be normal. That's a pretty nice consequence. So there's a lot of subgroups we can argue are normal because the index is two. Now, a nice little trick I wanna mention to you, and this will go without proof, is that you can slightly generalize this argument right here. If you were to slightly change it to that, the index of H and G is P, where P divides the order of G, that has to happen if it's an index, but also it is the smallest prime, smallest prime. Okay, so if P divides the order of G and it's the smallest prime that does it, turns out that also is gonna, that necessarily gives you a normal subgroup. And so if you were looking at a group of order 21, let's say you have some non-Abelian group of order 21 assuming such a thing even exists, we won't get into that conversation right now, but what I could tell you is that, oh, if you have a subgroup of G and the index of G, let's say that the order of H is equal to seven, right? And then that would tell us that the index of G, the index of H inside of G would have to be three. Three is the smallest prime that divides 21. So that would tell you that H, H would necessarily have to be a normal subgroup of G because it's the smallest prime. Now that proof is a lot harder than what you see right here. And again, I'm not gonna provide that detail to you, but we could tell, but I could tell you here that, yeah, since three is the smallest prime divisor of the group, every group of index three would have to be normal inside that group. Again, the proof of that statement's a lot more advanced and you're not gonna see that in our lecture series here. So another important result about normal subgroups is actually gonna be a list of equivalents, right? So you often see mathematical proofs of the following way. The following are equivalent. You sometimes see the acronym, the following are equivalent. T-F-A-E, of course, I snuck in the word statements, to be clear, the following what? The following kitty cats are all equivalent? No, the following statements. And so the following three statements are actually equivalents to each other. So the first one is, well, so we have a group G and we have a subgroup N inside of G. So the first condition is that N is normal inside of G. So what does that mean, right? So for all little G inside of G, we have that GN is equal to NG. So that's what it means to be normal. Left and right cosets are the same, right? The coset, left and right cosets give the same partition of the group. Left and right cosets are the exact same. So that's the first statement. The second statement is gonna be the following, that for all elements of G inside of the group G, if you take the set N, or G times N times G inverse, that is a subset of N. I don't claim it's a subgroup because I'm not even saying that this thing is a subgroup, although it actually is a subgroup. But again, we're not providing that argument right here. So we have GNG inverse, that's a subset of N. So when you take this element G on the left and G inverse on the right, that gives you a subset. Then the second, the third statement is actually that in this situation, the two sets are equal to each other, right? This is not the same statement right here, although one implication is quite obvious, right? Noticing if I go this one, statement three implies statement two from a tautological point of view. And then, so I clean that all three of these statements are equivalent to each other. So when it comes to proving the following are equivalent, the simplest proof is to do the following. I'm gonna take the first principle and show that it implies the second principle. I'm then gonna take the second principle and show that it implies the third statement. And then I'm gonna show that the third statement implies the first one. Because otherwise this can get a little bit messy, right? Otherwise, we would have to take some, we'd have to first show that one implies two, one implies three, two implies one, two implies three, three implies two, three, you know, three implies one, right? So that's like six proofs right there, that's a lot. Okay, I don't wanna do all those. By, by, by strategically right now, we have to just do one, two, three proofs. So we actually have to do half of them. And that's because if I wanna prove that one implies three, I just take the composite of these two proofs. One implies two and two applies three. So by transitivity, one would imply three, okay? If I wanna prove that two implies one, we'll all just take one imply, or two implies three and then compose that with three implies one. So again, the composite proves it because of the cyclic nature here. And if I wanna prove that three implies two, well then I'll just take three implies one and one implies two because of the cyclic nature of this proof here. I'm gonna rewrite it a little bit, right? So really we wanna think of it in the following way. One implies two, two implies three and then three implies one. So because of the cycle proof, we actually can prove all of the equivalencies by transitivity alone. Which again, for three, that basically cuts the proof in half. But if you have like four equivalent statements, you'd get four factorial proofs that you'd have to do with just 24. But with our cyclic approach, you'd actually still only have four. So the difficulty of the proof only grows linearly on the number of statements as opposed to factorially, which is definitely a good thing, right? So let's then proceed with the proof. So our first one is we're gonna prove that one applies two. So what that means is we're gonna assume number one and then prove that under the assumption of one holding that two is a true statement, okay? So what we have to do is we have to show that NG inverse is a subset of N. How do you show that a subgroup is a subset of another? I'm gonna take an arbitrary element of GNG inverse and show it's contained a set of N. So that's where we begin. So X belongs to GNG inverse. But by assumption, right, by condition one, which we are assuming to be true right now, the set N or GN is equal to NG. We're gonna get to that in a second, right? That's gonna be the kicker going on here. Now, because X belongs to GNG inverse, that means there exists some little N that can side this in the subgroup N so that X equals GNG inverse, okay? But then like I said, little G little N belongs to little G big N, right? And so by assumption, since this set is equal to NG, there's gonna exist some element, we'll call it N prime, so that GN little G little N equals N prime G, okay? Now, I'm not saying N and N prime are the same element, but what I'm saying is if you wanna compute G past some element of N, you could get a different element of N potentially, but you can pass it by by swapping N to N prime, okay? So we're gonna play around with that in the equation we had up here. So if we take this, the GN times, so you have GN times GN inverse, right? That the GN becomes an N prime G times GN inverse, for which though, okay, then those elements are gonna cancel out as we're seeing happening right here, for which then X simplifies to be the element N prime, which by assumption, it's inside of N, right? And so since X was chosen arbitrarily, this shows that GNG inverse is a subset of N, and that's exactly statement number two. So we assumed one, we proved two, and that's the first implication, one implies two. So now let's do two implies three. So we have to show, so we're gonna assume that N, or excuse me, GNG inverse is a subset of N. We then have to prove that GNG inverse is equal to N. Now, how do you prove that two sets are equal to each other? You show that they're subsets of each other. Now, one of those subsets is already done for us. We have to prove the other way around. So under that assumption, so we have two implies three, we need to show, so our goal here, our goal is to show that N is a subset of GNG inverse, because the other directions are taking care of us for that, in that direction, right? So then let's take an arbitrary element of N, right? That's how we showed that one sets a subset of another. We just did that. So N is a subset of capital N. So by assumption, okay? This is the nifty part, right? By the assumption, let's actually look at it for a moment. By the assumption, GNG inverse is a subset of N for all N. In particular, it's true for GNverse, right? In which case, we get an inverse here, and you get an inverse inverse, which actually cancels it out. So if we apply principle two, statement two to GNverse, we get that GNverse times N times GNverse inverse is a subset of N, which if you're dying to simplify, we can do that, GNverse NG, okay? This is a subset of N, that's the assumption we have here. So the element N we started with, we could factor it somehow, there exists some element N prime inside of N, so it's that GNverse times N times G is equal to N prime right here. So if I take my element N, which I started with, and I multiply it on the left by GNverse NG, then this of course will belong to N, right? By this assumption. So we'll just call that element N prime for a moment. Now let's manipulate this equation. Let's times the left-hand side by G. Let's times the right-hand side by GNverse so things cancel out. We see that N is gonna equal to GN prime GNverse. Which that belongs to GNG inverse, right? And so as N was an arbitrary element of capital N, this then shows that N is a subset of GNG inverse, okay? And then by assumption number two itself gives us the other direction so that then proves the two sets are equal to each other, great. So then we have one more direction to go. We're gonna assume that three implies one. So condition three, remember that GNG inverse equals N for all G inside the group. So that's the assumption we have. We need to show that GN equals NG. That's what our goal is right now. So let's take this equation. So if we take GNG inverse equals N, then if I take, so I have to, so again, we wanna show that GN equals NG. So we need to take an arbitrary element of, because we have to show their subsets of each other, so we're gonna take something arbitrary in GN and show it belongs to NG. That's what our goal is right now. And so we'll take X inside of GN. Well, that means there's some element little N so that X equals GN, that's great. Next, if I were to, so okay, so take that for a moment, X equals GN. Well, of course, by the three, which we're assuming right now, there exists some element N prime inside of N so that XG inverse is gonna equal GNG inverse, which is equal to N. So what are we doing right here, right? Let's be careful. Well, X, like we saw earlier, it's just GN. So if you times X by GNverse, you're gonna get GNG inverse, but that's the situation we're in right here, right? You have something that belongs to GNG inverse, but that's equal to N. So we know this product GNG inverse is something in N, so let's call it N prime for the conversation that we're having right now. And so from here, if we take this equation, right? If we take XG inverse and times both sides by G, you're gonna get that X equals N prime G, which we see then right here, for which, oh, N prime was something inside of N and then G, so okay, X belongs to NG. So we've then just shown that GN is a subset of NG. But by a similar argument, you're gonna get the other direction as well. And that proves condition one, which establishes the equality we're trying to go. So one other comment I wanna do is come back up here, is that when you take this set, GNG inverse equals N. We played around with elements, element-wise to show that the subsets were equal to each other. What I wanna show you here is a much slicker trick here. If I just times both sides on the right by G, GN inverse times G was go away, in which case you're left with then GN is equal to NG. The cool thing about this is, although we went through this long proof element-wise to make sure this is legit, what we see as a consequence is that, hey, you can simplify, you can simplify this equation right here. And that is, you can just multiply on the right by elements here. And you can do algebra with these sets. And you're gonna get the exact same thing, which honestly gives you a much simpler argument in the end. And so these gives us three conditions about when a set is gonna be, when a subgroup's gonna be normal. Some textbooks actually define normal subgroups as this right here. Now this set right here, GNG inverse, this is known as a conjugate subgroup. We'll talk about conjugation a little bit more in our next video. But one could define a subgroup being normal as, oh, all the conjugates of the subgroup are the same. But we're gonna stick with Judson's definition of normal, that left cosets always equal right cosets. But we see that this definition, this equivalent statement about conjugation, well, I guess I kinda said that there, the statement is equivalent, the conjugate condition with cosets being the same.