 Welcome back to our lecture series Math 4220, abstract algebra one for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. In this video, I want to talk about the idea of a normal subgroup, which honestly is kind of a silly name. I don't know what makes these things so normal. But if you're not normal, then I guess you're abnormal, which is actually a thing in group theory. There's a lot of puns off of the word normal subgroups. I want to see a paranormal subgroup at some point. But the definition is the following, whether it should be called this or not. A subgroup H inside of some group G is called normal if G H is equal to HG for all little G inside of the group. In particular, we say that a subgroup is normal if every left coset is equal to its corresponding right coset. And so it doesn't matter which element you choose in the group, the left coset is equal to the right coset. And this will be denoted using the following symbol. You have H. There's triangle less than or equal to G right here. So normally when we talk about a subgroup, right, we just say H is less than or equal to G, that represents it's a subgroup. If we add this extra line to it, this is what we mean by, that's how we will denote normal subgroup. And for those techers amongst us here, you can use the symbol slash triangle left. And then EQ if you want the line on the bottom. Some people don't put the line. And so if you want, you can just drop that. And that would just give you this triangle looking symbol like so. Okay, and the left is indicating the point. It's pointing to the left when it comes to this normal subgroup. Now normal subgroups are things we've seen before. There's some important examples we should mention that for any group whatsoever, you have the two following normal subgroups. So one important normal subgroup is going to be the trivial subgroup itself. So if you just have the identity, I want you to notice here that with your group, if you multiply the identity by any element G, this is a set is just equal to the element G, which also can be factored the other way around, right? When it comes to the trivial subgroup, the cosets themselves are just singletons. It's just the element G. And so in particular, the left coset is equal to the right coset. So we always get that the trivial subgroup is a normal subgroup of G. And oftentimes the trivial subgroup is just denoted as either E itself or probably more often, you'll see at least in this lecture series, you'll see a one will indicate the trivial subgroup. That is there's really not a huge reason to distinguish a singleton set from the element itself. So you'll see that. Another important subgroup, we see a normal subgroup, I should say, we see when it comes to groups is the group itself, right? It turns out we're gonna see that the group, the improper subgroup is itself a normal subgroup. And to see that, recall that you can take the map of left translation, right? So we called this lambda G in a previous video. This was the map that sends X over to GX. Now, this function which we've established again in a previous video here, this map right here is a bijection. It's a permutation of the set G. So it'll just scramble things around. And so therefore if you take G of G, so little G times G, this will just equal the image lambda G of G, all right? For which as this is a permutation, it's a bijection, the image of lambda under G will just be G itself. So the point is little G is equal to, little G times capital G would just be equal to G right here. Also we introduced previously the map row G, right? Which was the right translation. This goes from G to G. This one's a little bit different, but basically X is gonna map over to XG inverse. We do take inverses on the right-hand side so that the composition works out correctly in this situation. But the point is this function is a bijection as well. And so in contrast, if we take capital G times little G, this is the same thing as row G inverse of G, like so. In which case again as row G inverse is a bijection, its image will be the whole thing. And so we get this is equal to G as well. So the left coset of G is equal to the right coset of G. So when it comes to these cosets of G right, the only coset itself is G, that's the only one there is. When it comes to the trivial subgroup, you're gonna get the identity as a coset, then you're gonna get G as a coset, and then little H and then little K, right? Each individual element, right? So cosets partition the group. The partition associated to the trivial subgroup is just gonna be singletons. The partition associated to G itself though, is you actually just throw everything inside of a single set, like so. And so when it comes to partition to the group, there's these two, these are the two extremes. You put everything together or you keep everything separate. And so when it comes to normal subgroups, you're gonna see that these two normal subgroups are guaranteed, the trivial subgroup and the improper subgroup itself. Now of course, for different types of groups, we wanna know what else is available, right? What are the non-trivial proper normal subgroups? Well, when it comes to abelian groups, this conversation is kind of a mute point here, because if you have an abelian group, that means your operation, let's call it multiplication, is commutative. So for each element, every pair of elements I should say, G times H is equal to HG. So since elements commute on the, since the elements are commuting, right? Multiplication is commutative on the element-wise. This is gonna apply that G capital H is equal to HG. And this is gonna happen for all G inside of G. And that's because as a set, right? As a set, GH looks like you have all the G little Hs as H ranges over capital H right there. But on the other hand, the right coset, this thing is gonna look like HG as H ranges over the subgroup H right there. And so because these element-wise are the same, you see the sets are the same. The reason I make such a distinction here is that the sets could be equal even when the elements don't commute, right? We could have that GH equals HG. This could happen even though element-wise, you have like GH doesn't equal HG. So that's the important thing here. But because we have commutation at the element level, we'll have commutation at the set level, coset level. And so every subgroup of an abelian group is in fact normal. Now you have to be careful though, right? So what I'm saying here is that if you have an abelian group, if you have an abelian group, this implies that all subgroups are normal, right? But the other direction is not true. We're sometimes tempted to go the other direction that if all subgroups are normal, that means the group is in fact abelian. No, that's not the case. There do exist certain groups, not abelian groups where every subgroup is normal. The prime example, sort of the smallest example would be Q8 right here. If you think of the subgroups of Q8, let me just kind of draw the Hase diagram real quick. You get the cyclic subgroup generated by I. The cyclic subgroup generated by, I guess J is the next letter of the alphabet there. Then you get the subgroup generated by K. And then the intersection of each of these groups is going to be the cyclic subgroup generated by negative one. And then this sits above the trivial subgroup itself right there. So these are all the subgroups of the quaternion groups. So just for simplicity, we'll call this one capital I, capital J, capital K. And then for the second one, we'll call this one H, right? If you think about the subgroups, right, we already know that the improper subgroup is normal and the trivial subgroup is normal. Well, this group right here, this group H, this is none other than the center of the group. That is, this is the set of all elements Z inside of your group G right here. In this case, it would be Q8, such that ZG equals GZ for all G inside of the quaternion group. And again, you can replace Q here with any group you want. So this is the center of the group. One thing that we can make the argument is that the center of a group is always, in fact, normal. So this, and we'll talk about that in another video, but you could show that this thing is normal here, but you can also do this element-wise. If you take plus or minus I, and you times it by one and negative one here, what's gonna happen is that when you times plus or minus I by one, you'll just get back plus or minus I. But if you times plus or minus I times negative one, you'll get, you'll switch the sign. So if you start with positive I, you'll switch to negative I, and if you take a negative I, you'll switch to a positive I. So this will just turn out to be, this will just be plus or minus I, and then negative plus I. So you'll switch the sign, right? But it doesn't really matter. And then going the other way around, if you take one and negative one, and you times it by plus or minus I on the right, same thing is gonna happen. When you times it by one, you'll get back what you started with. And then when you times it by negative I, you'll just switch the sign like so. And so these two sets are in fact equal to each other, right? The coset generated by I, and this will just be, this will just be I or negative I, right? That's what these cosets gonna be. Now there's nothing special about I in this situation. If you switched it to a J, you would switch all of these to a J. This thing will just look like J times negative J, right? Same thing for K. So this shows you that the subgroup H, as the one generated by negative one's normal. What if we took I for example, right? If you took I, well this subgroup is equal to, we've listed all the elements, one, negative one, I and negative I. And so if you take the subgroup I, and you multiply it by anything that's in I, you'll get back I. But what happens you take something like outside of I? You take like J, I right here. Well you're gonna end up with J, negative J, K, negative K, like so. Actually I did that backwards, right? J I is gonna give you negative K, but J negative I will give you positive K. But it turns out that little goof I made a moment ago actually doesn't make much of a difference because if you switch things around, when you commute things in the quaternion group, if it's not commutative worst case scenarios, you get a factor of negative one. So when you take one times J, you'll get J. Negative J times J is negative J. But when you switch the things around here, so instead of getting J I, which is negative K, you're gonna get I J, which gives you a positive K, and then you switch the thing around there as well. So if everything works out here, when you commute things, since you only off by a negative sign, you can check this for yourself. We'll also see in a little bit that, in the next video in fact, that also an argument you could take is that each of these subgroups right here has index two and every index two subgroup has, it's normal, right? So with the quaternion group, although it's non-Abelian, every subgroup inside of this group is in fact normal. Let's look at an example where things might not be normal. What's a non-normal subgroup look like? So take the symmetric group S3 for a moment. This is the smallest non-Abelian group there is. Take the subgroup generated by one, two, three, which this is actually just the alternating group A3. Well, what are the left cosets of H? Well, one of the cosets is H itself, which just contains one, one, two, three, and one, three, two. The other coset, because this group right here, I should mention that this subgroup A3, it has index two. So they're only two cosets. So if you take one, two, H, you're gonna get one, two, three, excuse me, one, two, one, three, and two, three, okay? If you look at the right cosets, one of the cosets is itself gonna be H and then the other coset has to be H12, which will be one, two, one, three, and two, three as well. And so the thing is the coset, one of the cosets is itself the subgroup, that's always gonna be itself. H is always gonna equal H. But then the other thing is that the other cosets have to agree with each other. And this is kind of the reason why a subgroup of index two has to be normal is because notice that as the cosets partition the group, you have one coset, which is the subgroup. The second coset has to be everything else. So you actually get G take away H. And this is gonna likewise be G take away H, which as those sets are the same, you're gonna see that you get normality here. So the all-train group is normal inside of the symmetric group. But as an example of a non-normal subgroup, let's take the subgroup K, which is generated by the two-cycle one, two, the transposition there. Well, it'll contain one, two. The K itself is one and one, two. If I take the coset associated to one, two, three on the left, you're gonna get one, two, three, and one, one, three. And then if you take the coset on the left by its inverse, one, three, two, you're gonna get one, three, two, and two, three, okay? But what are the right cosets? The right cosets on the hand K is gonna equal one, two, three. Again, that part is gonna match up. You anticipate these are gonna equal. It's the others that we see at discrepancy, right? If you take the right coset of one, two, three, yeah, you're gonna get one, two, three, that's in common, but you're gonna then get two, three, which actually is in the other coset, one, three, two. And then looking at one, three, two right here, you're gonna have one, three, two. That part's the same, okay? But the other element one, three is actually in the opposite coset. So these things do not equal each other. That's not what I wanted to write. These ones are not equal. Neither are these ones. So there is a disagreement on the cosets here. So if you look at the subgroups of S3, you have A3 over here. There's the identity, like so. But then you have these other subgroups. There's in this case, K, which was generated by one and two. There's the cyclic subgroup generated by one, three. That also has order two. And then finally, there's the subgroup generated by two, three, scooch over just a tiny bit right there. These are all subgroups of order two. So for S3, it turns out the normal subgroups are gonna be the proper subgroup, not the improper subgroup, the trivial subgroup. Those ones are always there. And then A3 had index two. It's this alternating group. Those are gonna be the normal ones. And then these other three are not normal. We showed that K right here is not normal. But by similar reasonings and calculations, you get that the other two, cyclic subgroups of order two are gonna be non-normal in S3.