 So yesterday we just looked at noticing the fact that if we look down at Kaylee's tables row or column that we see that all the values in the set that make up our group are there and they're all they're unique so for a finite set that's easy enough I mean if everyone's unique of course they'll all be there but what about an infinite set the other thing is those are small little small little examples you know it doesn't hold for all so we really have to prove the fact that if we have a set G I should say group G that contains a set of elements and some binary operation on that do we have the fact that the rows and the columns represent unique values and all the values can we prove that first of all let's create a Kaylee's table here there we go we have our binary operation we represent our whole group G here all those and let's select let's select an arbitrary element of our group and we'll call that arbitrary element A and I'm going to look down this row because it's arbitrary it reflects all of those rows every single one of them and what I want to know is you know is each of these entries unique that's the first thing I might want to show so if I take any element G here and I look at A and G the binary operation with each other is that unique first of all let's suggest that they are two here I have also a G prime so that I have this and this and I'm going to suggest to the contrary that it is so that these two are equal to each other so I'm suggesting that there's another one here and they actually the same meaning the entries down the row is not unique well I'm told that this is a group so if A is an element of my set in my group then A inverse better be there and if I do this I've got to do it on this side as well and that's the identity that's the identity so G equals G prime in other words this is actually not two different things because remember we don't repeat it when we write the Cayley stables this one and this one's actually at the exact same thing so it sits on top of that so every element there really is unique there really is unique let's do the column let's look at the columns and see there's my binary operation my whole group G is represented here I again take some arbitrary element and I look down its column it is an arbitrary element so if it's arbitrary that means all of them that are there and again I'm going to look at G here so this is G binary operation with A and once again I'm going to suggest that there's a G prime of a different different from that remember this way around and I'm going to say it is not unique in other words that I have this G and A is going to equal G prime with a I'm going to write multiply write not multiply the binary operation with a inverse and again I get this this fact so it's actually the same thing and these elements that I get here are absolutely unique now I'm going to clear the board because now we need to show that all of them are represented there as I said with a finite set if they are all unique then of course they will all be represented there but what if it's infinite let me clean the board okay so we've proved uniqueness now we have to prove that they're all there so I'm still here with A and I'm still looking at its group I'm saying let me look is G there now G is an element of the group and I'm saying it is arbitrary so it's any one of them and if I take any one of them and prove that it's true it actually it is obviously true for everyone so I'm saying can I can I get any element here can I find G here and since G if I can find G there and G is anyone then that means they're all there and let me suggest that there is a G prime it's not the G prime of the previous bit so I'm saying that a binary operation G prime equals G and the question now is just can I you know which one is this G you know how do I calculate this G prime well it's very easy if I just put the inverses on this side if I put the inverses on this side it just means G prime equals A inverse A inverse with G so I you know this there is closure so this must exist that's an element A is an element in this must be there there's closure I'm told that these all elements of a group that's a given so this one must be there and it's this the result of this must be there and there it is and if I find it there and I find A inverse I actually get G so irrespective I took arbitrary ones I will find it there so even if there's an infinite number of these I will find it in the row so we shown both as unique and all of them are there and the same would go for this can I find G in this column well let's suggest that this is G prime here and again we're going to say that G prime binary operation with A equals G so I can very easily find G prime that it's just going to be G binary operation A inverse that's in the group that's in a group there's a group there's closure so I can do this its result its result is G prime G prime must be in the group so G prime must be somewhere that is I would the binary operation with A inverse will give me the G prime so I know where it is if I do that with A I'm going to find G if I if I do A here with G prime I'm going to find G G was arbitrary I can find it no matter no matter of these entries are infinite I'm going to find it so both in the rows and the columns we prove that all the values are there and each one of them is unique so it wasn't just some special case for the little examples that we've seen before it holds for all of these and as far as a group is concerned and that's a very deep insight we're going to use it later when we look at proper proof for probably look at Kayleigh's tables and what that implies but this is a very deep this is a very deep insight that we've gotten into groups here and that if we look at these rows and look at these columns there is this wonderful property for all of them