 In the previous video, we saw a counter example of a subset of a group that could not be a subgroup, not every subset will be a subgroup of a given group. There are some conditions that have to be met, right? The subset itself has to be a group using the operation that's already on the group, but with restriction. This idea of restricting the operation to a subset, if this is done successfully, this is referred to as the closure property. That will say things like, oh, the subset H is closed under multiplication. If whenever you take a product of two things in H, it lands back in H itself. So if G and H are in H, then G times H is in H as well. This is called the closure property. If H is a subgroup, then restriction of the binary operation to H, if that is also an operation in H, that means that H is closed under the operation. That's an important principle here. And so with that in mind, a subset is a subgroup. You know, by definition, that means it's a group inside of a larger group using the same operation. But that can be kind of difficult to show at times. Alternatively, we're going to see in this proposition, there are three alternative conditions we can check to see when a subset is a subgroup or not. So the first one, like I already mentioned, is this closure principle, that the product of any two elements in the subset lands in the subset itself. G or H is closed under the multiplication. To be a subgroup, you also have to contain the identity. So the identity has to be inside of H. And given any element in H, the inverse of that element must also be H. So these things, we say things like this first principle to emphasize, you might say that it's closed under your operation, which we'll just call it multiplication to be general. It's closed under multiplication. You might say that for property B that the set is closed under the identity, which, you know, that just means I did the identities in there says no one usually says it that way, but you could if you wanted to. The last principle, we often say that it's closed under inverses, because you can think of like the inverse map as this unary operation that is just a function with one input. And so our set H has to be closed under that function as well that have taken the element when you restrict the inversion map to the to the set that will also give you that it will restrict properly so that inverses are always contained in there. So I, I claim this is an if and only of statement right that H is a subgroup if and only if it's closed it has the identity and it has inverses. And so as it's an if and only if statement this is a this is a proof for the cost of two right we have to prove both directions. So what we're going to do first is assume that H is a subgroup. And because H is a subgroup that means it is itself a group inside of G but using the same operation as G itself. And so like I was mentioning before we started the proof that would imply that H is closed right because we have this operation G times G, which goes to G this is the group operation. If H is a subgroup then the restriction of the operation will be to H as well so we get the operation h cross H to H. And therefore, if we take elements G and H, which are in H, we see that G times H, this will have, well I should say the pair G times H would have to map to an element GH, which belongs to H. So you get the you get the closure principle pretty quickly from the definition of a subgroup. So that's pretty nice. What about the identity right. Well, H has to contain an identity if it's a group, I mean if it's a subgroup it's a group in its own right, in which case it has an identity, which means that E prime, let's call that identity E prime. So that means E prime H is equal to H E prime which is equal to H for all elements inside of H right there. This E prime potentially could be a different element from the group it's from the identity of the group right. In particular, since E prime is inside of H, it must be true that E prime times E prime is equal to E prime. This is what we refer to as an item potent element that it's square is equal to itself. This is something we see with sets all the time like the intersection of two sets. I should say the same set right the intersection is itself on the union of a set with itself is a again in a group an identity element is item potent. But in fact in a group the only item potent element is the identity itself and this is sort of going to be the problem right here that if you take an item potent element like this E prime it has an inverse right it has an inverse in the group because G is a group so it has an inverse. So what happens when we take. I, or E prime times its inverse one way times E prime times inverse well it'll give you the identity element in G right, but since E prime is item potent I can replace E prime with E prime times E prime, then by associativity you're going to multiply E prime by its inverse that'll give you the identity of G, and then E prime times E will just be E again because you're times it by the identity in G, not sure why there's two, two periods right there. But the thing is what we see here is that E is an equal to E prime. So what we've basically shown here is, this is often what's referred to in the literature as a porism. Basically, a theorem we proved while we were proving something else. So this is something to kind of note here that any, any item potent, any item potent element you know, I'm going to try to write this with my poor handwriting which is not going to do any, any favors right here. We're going to do the follow I'm going to type it. So this is our porism. It's something we proved in the proof of something else. Let G be a group. And let we'll say G be a member of G right here. And so what we've shown is the following, then an element then G is in fact item potent. If and only if G is the identity, a group only has one item potent element, and that is the identity itself the identity is clearly going to be item potent, but we've now shown that if you have an item potent element it is the identity. And so that's helpful here, because if the subgroup, if the subgroup has an identity element it will be item potent, but it's an item potent and the larger group G as well, which then means it has to be the group identity. This tells you that H has the same identity as the ambient group, which is proving this condition to be a subgroup it means to have an identity but this statement right here B is saying that it has that it contains the identity of the overgroup. So the last thing we have to do is consider inverses. But since we now know that the identity of H is the same identity as G, the inverses are going to have to be the same inverses as in G as well. So since H is a subgroup, it's a group in its own right it has inverses. So for each element H let's say that H prime is the inverse of little H in capital H. And so that means that H, H prime will equal H H prime or H prime H which is E and now this is the identity in G right. So H prime is an inverse of H, but like we've learned before inverses are unique. So if you walk like an inverse and quack like an inverse then you must be the inverse. The only element which can multiply by H inside of G to give you the identity has to be H inverse, which would imply that H prime is equal to H inverse. And this tells us that the subgroup is closed under inverses. That is the inverses in H were the same inverses in the ambient group of G. So H will have the same identity as G and it'll have the same inverses as G. Although some of those inverses will be missing because some of the elements of G are likely missing as well. So that proves the first direction. If G is a subgroup, because it's a group inside of a group, it'll be closed. It has the same identity as G and has the same inverses as G. So every subgroup will have those three properties. Now let's go the other way around. Let's suppose that H satisfies those three conditions. It's closed under multiplication. It's closed under the identity and it's closed under inverses. It contains the same identities and inverses as above. Well, since the operation is closed, this gives us a well-defined operation on H. So the closure principle is saying that we can restrict the operation to give you some type of H cross H to H. That the product of any two elements of H is well-defined, yes, but also it'll be inside of H itself. That's the important part there. So we have a well-defined operation. So we can then talk about whether that operation is going to give us a group operation or not. So the next thing to mention here is that, well, by assumption, it has an identity element. The identity of G is an identity for that operation. So we restrict it. Identity will still be an identity. And so this operation will have an identity. And because it's closed under inverses, every element in the subgroup, the potential subgroup, right, will have an inverse. So the inverse axis is satisfied as well. So basically what we have left is to worry about associativity. Why does associativity work? Well, if we take elements G, H, and K, they're inside of H. Since H is a subset of G, this means that G, H, and K are in G. So when you look at statements like G times HK, while this is a product in H, it's also, you know, it's a product in H, it's also a product in G. And in G, this operation is associative, so we can redo parentheses, like so. And so it's associative in G, which also means it'll be associative in H. There's really not much more to it. And in this situation, we often say, we say that associativity is inherited from G, because every element in H belongs to G itself. If things are associative in G, and the operation when it's restricted down to H, it'll still be associative. You don't lose properties when you restrict. And so I should mention that in this proof, it's not just associativity that would be inherited. Associativity is inherited because if the property holds for all elements in G, then certainly it still remains true for the restriction to G. This is also true for commutivity, that is, if every group, every subgroup in an abelian group is abelian necessarily, because just like associativity, commutivity of the subset is inherited from the ambience commutative group. On the other hand, though, not every subgroup in an abelian non abelian group has to be non abelian. And you might ask, why is that? Because non abelian just means that commutivity is not guaranteed. It doesn't mean that commutivity never happens. In every group, there will be some elements that commute with other elements, like the identity commutes with everything, an element will commute with its inverse, it'll commute with its powers. So there is always some commuting. A non abelian group just means that commutation is not universal, not every product will necessarily commute. But what if we restrict to a certain subset? Maybe on a subset, commutivity does in fact happen. So while properties like associativity and commutivity are inherited by any closed subset, issues about containment, like containing identities or inverse elements are not guaranteed. That's what this proof is basically trying to show us. When you check for a subgroup, you do not have to check whether the operation is associative. Because if you're closed, and the restriction of the operation is well defined, it'll be automatically associative. And if it was abelian, then the restriction will still be commutative. But the containment, like what if I take a subset? That subset, if chosen poorly, might not have the identity. It might not have inverses. So considering that a subset is likely a collection of G with some missing elements, we should be highly curious whether the elements omitted from the group G was the identity, was it inverses, or maybe there was, maybe we omitted specific products, like we took H1 and H2 but we forgot to include the product H1, H2 in the subset. So when it comes to checking whether you have a subgroup, we really just have to look for did we forget to include things like products, identities, and inverses, because properties of the operation like associativity will be inherited because our subset is closed under set operation.