 In the previous video, we learned an alternative way of checking whether a subset is a subgroup or not. This is actually the preferred way. That we don't have to check things like associativity necessarily because a lot of the properties of the overgroup will, I should say, many properties that the overgroup have will be inherited when you look at the subset like associativity, commutativity, those type of properties of the operation itself. But the inclusion of specific elements like products, inverses, identities, that's not guaranteed. So we have to check basically that the subset is closed under multiplication, identities and inverses. And if you have those closure principles, then you'll have a subgroup necessarily. So consider the group Z with respect to regular integer multiplication, or addition, excuse me. We're gonna define the set NZ to be the set of all multiples of N, so K can range over anything, but we're gonna take multiples of N. So like for example, if we're talking about two Z, we're taking all of the even numbers. If we take three Z, we're taking multiples of three. So zero, three, six, nine, negative three, negative 12, you get the idea. And so we're gonna take NZ right here. I claim that NZ is a subgroup of Z for any integer N. And so to prove this, we're gonna use the proposition we proved previously. We're gonna check these closure principles right here. So let's take first of all two arbitrary elements of NZ. So if they're elements of NZ, this means that there's some type of factorization. So for example, A is gonna equal KN and B is gonna equal LN for some integers, for some integers K and L as is appropriate. What happens when we add these things together, A plus B? Well, when we add these together, we wanna show that since AB isn't NZ, we have to show that A plus B is inside of NZ. So since A is an NZ, we're gonna get that A is KN and B is LN for the same reason, right? Now as these are integers, we can factor set integers. I'll do it right down here. We can factor those integers and we're gonna get K plus L times N. And so this is a multiple of N again. This belongs to NZ since K plus L is itself an integer, right? Since the integers are closed with addition, the sum of any two integers is an integer, right? K plus L is an integer. So we have an integer times Z. So we're gonna get closure right there. This shows us the closure principle that we had listed on the previous slide. This group or this subset is closed under the operation of addition. The next thing to ask is, does it have the identity? Is the identity inside of this? So that is the identity of the group itself of the integers, right? That would be zero. So what I'm basically trying to determine here is zero a member of the set NZ. So notice of course that zero is equal to zero times N for any choice of N. And therefore zero would belong to NZ for any integer N. And so this does in fact answer our question in the affirmative that we do have the identity right here. So the last thing to do is to check inverses. Does this subset contain inverses? So to check inverses, what we're gonna do is we're gonna start off with an element. We're gonna say something like let A be an element of NZ, which like it means before, that means there exists some element K, which is itself an integer, such that A is equal to KN. Now what would the inverse of A be inside of the integer set? That's gonna be negative A, which negative A is of course just negative one times A. That's what we mean there. But A, like we saw before, is equal to K times N. And then by properties of arithmetic on integers, we can write this as negative K times N. This will belong to NZ, I don't know if I can fit it in there, NZ, right? Because negative K is itself an integer. And so this will show that the set NZ is in fact closed under inverses. And so we have all of them, right? We see that the sum of two integers in NZ belongs to NZ. We see that zero belongs to NZ. And given any integer that's in NZ, we get that negative A, well, you know, call that integer A, negative A will also be in there. So by the previous proposition, this shows that NZ is in fact a subgroup. Let's look at another example. So we recall the set GLN of R, this is the set of all N by N real-valued non-singular matrices. Those matrices would have a multiplicative inverse. We are gonna define the set SL in R to be the set of what's called, this is called the special linear group. GLNR was the general linear group. And how we define the SL in R, we're gonna take all non-singular matrices whose determinant is one. So they're still gonna be in my matrices with real coefficients, but we add this special condition that the determinant is equal to one. And so I claim that the special linear group is a subgroup, which it ought to be, because I shouldn't be calling it the special linear group if we don't know it's a group, right? Clearly we're putting the cart in front of the horse here, but that's because of this proof that we're gonna see right here. So we have to check three things. We have to check, is the special linear group closed? Does it have the closure principle? That means if we take two elements that belong to SL in R, so we're gonna take two matrices A and B and say they belong to SL in R, well, what does that mean? That means that the determinant of A is gonna equal the determinant of B, which is equal to one. So by being a member of SL in R, we know these are in my matrices whose determinant is one. Well, what happens when we calculate the determinant of their product, A times B? Well, by properties of the determinant, the determinant actually factors as the determinant of A times the determinant of B, which you get one times one, which is one. And therefore this tells us that the product A times B belongs to the subset SL or N. And therefore the special linear group is closed under multiplication. So we have now the closure principle is taken care of. Well, what about the identity? The identity here. Well, the identity of GL in R, this would be the identity matrix I sub N, which this is the matrix, which will have ones along the diagonal and then zeros everywhere else. Well, as this is a diagonal matrix, the determinant is fairly easy to compute. The determinant will just be the product of the diagonal entries right here. And as you're gonna get one to the N as the determinant, that just means the determinant is equal to one. So the determinant of IN is equal to one. Therefore IN belongs to SL in R. And so therefore it contains the identity. Great. The next thing then to ask is about inverses. Does SL in R contain inverses for any such element? Well, let's take an element in SL in R. And remember that means its determinant is gonna equal one by assumption. Well, determinants also have the following property. Because of that factorization principle we saw earlier, right? Notice that determinant of A inverse is gonna be, well, one time determinant of A inverse. So where are you going with this? Oh, wait, one is equal to determinant of A and the determinant of A times determinant of A inverse. If you factor that thing, or I guess this time you put them back together, the determinant will be the determinant of A times A inverse. But A times A inverse since these are inverse matrices is the identity. And like we already mentioned, that's equal to one. Wow, that's great. I mean, what you see in general is that the determinant has the property that the inverse, the determinant of an inverse will always just be the reciprocal of the determinant. Because after all, A matrix is non-single if and only if its determinant is non-zero. And so if you get like one over one this is just equal to one, right? So the determinant of the inverse will likewise be one which means the inverse is contained inside of there. Therefore, we have in fact proven that the special linear group is a subgroup of the general linear group. And proving that a subset is a subgroup is typically done in this manner. We show it's closed under multiplication. We show that it's closed under the identity, it contains the identity. It's closed under inverses. That is any inverse in every element of the set will also contain its inverse as well. So we can always check those three properties when we want to verify whether a subset is a subgroup or not.