 Right now we're going to talk about something really exciting, subgroups. You know, we've been filling the box, relations and all sorts of things. In the beginning it was boring, we filled it, we don't know where it belongs. We filled the box, we've eventually started looking at groups. Things got interesting, showed you integers, mod n, some other groups. So, you know, we're using stuff out the box, it's getting more exciting, it's getting more exciting and now the excitement just ramps up because we're going to talk about subgroups. So remember g is a group if I have to set a and its binary operation with the four properties. There's closure under that binary operation, associativity holds, associativity holds for any a, b and c element of a. We have a unique identity element called e and then there's a unique inverse inverse for every element, for each element in a. So it contains all its elements and for every element the unique element is also there. One of them is the identity element, there's associativity and there's closure. Now imagine I have a subset of a called a prime, a subset of a called a prime and I have the same binary operation. Now I can create this subgroup of a prime and that same binary operation. If, you know, these properties hold for this subset of a, the elements in a, then h will also be a group and this will be then called a subgroup. That's the definition of a subgroup of g, the definition of the subgroup of g. All these things must hold though for that to be to be so. Remember I can have two special ones though. I can say that a inverse contains only the identity element and I can have that a inverse, you know, is a equal to a and yes, that will make a subgroup of well of h but those are called improper subgroups sometimes and that's not the ones we are after. We are after taking some elements out of a so that it is a proper subset, compion empty set and not one of these two and if I have that same binary operation, they've got to be the same and all four hold then h is a subgroup. That's our definition. Turns out though, I actually don't need all four because there's this following theorem, a non-empty subset h. Now here I just used the a and the subset of a. I'm just going to shift a bit and just call this thing the g and h both a set and group. I'm going to use the term interchangeably for these and please just know that it includes the binary operation as well. So I'm going to say a non-empty subset h of a group g is a subgroup of g if and only if the closure property holds. So if a and b are an element of h then a b and that should be a with its binary operation there. Most of the time we just leave it out as an element of h and every element of h has a unique inverse in h. So if a is an element of h it implies that a inverse is also an h. Turns out if those two things are true then h is a subgroup. I don't need this one and I don't need this one. Let's prove this and as I say that's why this gets exciting because we're going to do proofs and I'm going to show you actually just how easy it is. Because we have an if and only if statement it goes both ways. You've got to prove you've got to take this and prove this you're going to take this and prove that. Let's do it in the what's sometimes called the forward direction. So I'm given that h is a subgroup. It's the easy part so given that h is a subgroup then by the definitions of a subgroup which I've just given it to all four holds meaning those two hold as well. Given that h is a subgroup by the definition of a subgroup by the definition of a subgroup you can write those two properties do hold. So it's proven in that direction now for the interesting one let's prove it in this direction. Given that I do have this closure property so I'm given that there is this closure property. Can I prove can I prove then that given these two what I'm trying basically to do is given these two can I prove that all four of them hold can I prove that all four of them hold. So I need to prove this associativity and I need to prove that the unique element is also an h. Given just these two facts can I do that well let's say for all ab and c elements of h given that it because it is a subset you can write out the words nicely it implies that ab and c are also in g because h is a subset of g and g is a group we were told that is a group so because it's a group abc equals abc because that is a group so association holds but that same abc is also in here so associativity holds for a we were given this one we were given that one I just showed that this one is implied let's look at the other one that the unique I'm told that every has its unique there so for all a element of h it implies that a inverse is also an h because I'm given this does this help us make a group well we've just we've just been told or let's just remember by definition if I have a inverse and a I get the identity element I am given the fact that the closure property does hold that means e must be in there therefore that implies e is an element of h so we also have that so given just these two all four of them are there and that implies that h is a subgroup I've proven it QED I've proven it in both directions so the proof is very very easy so now we know if we have this group g and if we take some of the elements out and we keep the binary operation the same if we look at this new subset h and only these two properties hold we know that that subset and its binary operation does form a subgroup and the proof is very easy I'm going to clean the board because it turns out this is not the only theorem that proves that if I have a subset and the same binary operation not only this one there's another way to look at it as well which also will prove that that is we'll show you that that is also a subgroup and that's a different theorem it's clean the board and we'll do that one so let's have a look at this theorem it also says a non-empty subset h of group g as you can again see I'm using this notation of its subset in a g interchangeably because we're sticking to the same binary operation is a subgroup of g if and only if so I've got to prove both ways for all ab element of h a inverse binary operation b is an element of h so first glance this looks a bit different from the previous theorem but it isn't actually because what we are hiding there is the same it is the same thing that we are hiding so the proof in the forward direction again is very easy because I'm given the fact that this is a subgroup therefore all the properties would hold therefore these things would hold and in reverse what we're actually showing here is that a is an element of h and a inverse is an element of h you know that would imply that by the same way we did the previous proof that e the identity element is also and I needn't stick to ab and c you know ab that could have also been a c so c inverse b would also be so by the same token I have ab and c in h and therefore they are also part of ab and c they're also part of g and since it is a group that means since since it is a group it does mean that this associative property would hold and it's going to equal abc so I'm just showing these other two so it's just a different way of writing but the same implications as the first theorem I just are in there in the statement that you have here as well so the proof other than putting the words out nicely so you sound like a mathematician the concept of these proves are actually quite easy so let's just go on just to look at so and actually I hope you understood those two proofs they are really really simple if you think about it again you are given some stuff we have made definitions those definitions that we make remember they are you know that's what we come up with we decide that as humans that is our definition and the tool set that we have in these definitions we just use them and build new stuff so we put a theorem out there we prove a theorem that's something new that goes in the box and we can from now on just use that we get that that's a subgroup because we've already gone through all the steps of you know using prior definitions prior things things that we've put into our toolbox now this this goes into the toolbox both of these theorems and we can just use them and now we can use them and the next proof because we've already proven them let's look at some examples okay so let's make my group G the A and the binary operation and A is just going to be remember the all the all the fourth roots of one and they were remember they were minus or they were one so that equals minus one one minus one i and minus i those were the four and if I were to have a prime only be one and minus one and my operation that I still have a still multiplication and here it was multiplication for a multiplication is that still is that still now remember I only have to show let's use the second one and so A equals one so A inverse what is its inverse it's also one and B equals minus one so A inverse with B A inverse is one minus one one times minus one that equals one so low and behold this is if I just took those two that is a subgroup okay let's do the following let's take the integers mod four the integers mod four remember those we're just going to be my equivalence class is zero one two three remember those so if I were just to take if so that's my group so my or let's say my group is z four under addition and my A is that and let's make my A prime just going to be zero and two and let's have a look at that let's see let's do a Cayley's table of that we have addition we have zero and two zero and two zero and zero is zero zero and two is two zero and two is two and two and two is four which brings us back to zero which is there so there we go that's my Cayley's table so all I said is I wanted closure and is there closure yes any of the two that I do the solution is also part of the subset so I have closure and the other one that I want everyone's inverse must be there and so something with something must give me the identity element and we know the identity element here is zero if we look at this we just look at this we see the identity element is zero so zero is its own identity element and two is its own identity element because two plus two is zero so I only needed those two all the other two of my four properties hold by the theorem that we've stated and so the fact that I should just call this h or so that is this is g so if h was a prime under addition it means h is a subgroup easy to show that that is a subgroup so in short that those that is subgroups very interesting stuff and it gave us this opportunity to look at proofs and just to show you how easy those proofs are never be afraid of proofs just do so you know as more that you more more that you do the better you'll get at it and it really is easy they don't think you'll forget the concept of those two proofs they really are simple just what does in your toolbox the properties that we have decided as far or the definitions that we've decided that is what we would call a subgroup use that what's in your toolbox and the proof will come and there's two beautiful examples of subgroups