 Previously we've learned that in order to show that a subset of a group is itself a group that is a subgroup it suffices to show that the subset is closed under multiplication, that it contains the identity, and that it contains all of the inverses for elements in the set. It turns out that we can also rewrite that in yet another way, a potentially simpler way of doing it. In some regard, I kind of feel like this one's a little bit less intuitive, but in terms of checking it's a shorter list, we're going to see that a subset h of g is in fact a subgroup of g if and only if h is not empty, and if the value g times h inverse belongs to h whenever g and h belong to h. So basically what we're trying to say here is a subset will be a subgroup if it's a non empty subset, and it's closed under division, because multiplying by h inverse is multiplying by the reciprocal there. So let's it's an if and only a statement right so there's actually two things we have to do. So let's first argue that if h is a subgroup it has these principles. Well if h is a subgroup it's not empty because it contains the identity, so there's something in there. And then if you take the elements g and h with inside of the subset h, then because it's a subgroup it has inverses. So h inverse will be in there. And so since g and h inverse are in h, then their product will be an h. And as g and h for chosen arbitrarily a subgroup will be closed under division, because it's closed under multiplication and it's closed under inverses. So you saw the three the three actions in play right there right close under inverses close under products closer to the identity that will give you this but it turns out this process is reversible. So suppose that h is not empty. It contains something we don't know that is yet and it's closed under division so where can we go there. Well since it's not empty there's something in there, call it h we don't know what that is necessarily but we know h is in there. Well because h is in there, and it's closed under division right, it'll contain the element h h inverse, but h h inverse is just the identity. Therefore, this set will contain the identity. Oh, that's great. But then what we're going to do is we're going to do the following game. So h is in there e is in there, right. So that means because of this division property, it'll contain the element e h inverse. This will be inside of h as well. But this is just h inverse. All right. And so this set will be closed under inverses as the identity and also be closed under inverses. Oh, two down one to go right we got to show it's closed under multiplication. Well, what we've now done here is we've now shown that it contains e it'll contain h it'll contain h inverse necessarily right. But in fact, to do multiplication like the closure on the multiplication what we have to say is the following if g and h are inside of h so I'm going to stop you right here. It's like the closure principle says if g and h or an h then g times h is an h. Now this is a conditional statement right it's if then right so there's something here and then something here when you prove a conditional statement. One thing that that early proof writers sometimes struggle with is when you prove it when you prove an if and only if statement, you do not have to prove the if part that is we don't have to show that there exists two distinct elements inside the set. For all we know we have the trivial group which only contains the identity. There might not be anything else in there. You don't have to argue that there's multiple elements in there. You don't all you have to do is you have to say if this happens, then this happens. It could be that the if part is completely vacuous. There might be nothing that ever makes the if part true. Don't matter in order to prove a conditional you assume the if, and then you prove that then. So that's what we're going to do as well. So we're going to take two arbitrary elements of h that might not even be any that we only know there's one element they're not there might not be a second element. And after all when you say if G and H are in H, nowhere are we assuming that G and H necessarily are distinct. They could be the same element. Or even if we do assume they're different. What's the big deal right? That's the if part right so take two elements G and H inside of H. Well from what we've already shown, if G and H are in H, then that means H inverse will be an H. And then there's a typo here. Someone's missing their G. Yikes. How embarrassing. Because G and H is an H, H inverse will be an H. And therefore G times H inverse inverse, since you're closed under division, you that'll be an H as well but that's just the element G H itself. So we've now seen that our set here is closed under multiplication and any set that's closed under multiplication contains identities and contains inverses is a subgroup like we've already proven. So this gives you another way of showing that a subgroup that a subset could be a subgroup by showing it's not empty and close under division because basically if you're not empty you have to contain the identity. And when you show something's not empty, you typically just show that the identity is in there. That's the easiest way to do it. And then showing that something's closed under division is basically showing it's closed under multiplication and inverses simultaneously. So I don't want you to think that it's like, oh wow, it's such a short list comparatively. You basically have to show the same things. It's just just kind of squishing two steps into one. And if that seems easy by all means do it. And that's going to bring us to the end of this lecture lecture 10 we'll talk some more about subgroups in the next in the next lecture, of course, take a look at that coming up soon, and I'll see you later. Bye everyone.