 So in the previous video I gave you some intuition about this idea of group action which I said can be slightly difficult. So that was the intuition and I glossed over a few proofs but I think there's one we really should just spend a little bit more time before I get to the promised non-faithful group actions and some examples there. Remember in the previous lecture and if you haven't watched it this is not going to make sense so I'll put a link up there. So I had my set with three elements and I had the G's. I'm going to write G set that's the set that is part of the group G and that was this e tau 1 2 3 or tau 1 2 1 3 2 3 sigma and sigma squared and the idea is how do we the idea here behind group actions is that these are represented by permutations of that set. So group actions is to have the elements of a set that makes up a group onto another set and those are permutations and what we mentioned quickly was this sigma G function that maps A to A. Now think about what happens when I map a set to itself. So I have a set of objects and I map it to itself that is the definition of a permutation. I'm just going to scramble the order of those of those elements that is a permutation and we define sigma G as mapping A to the action of an element in G onto A. So remember this symbol that doesn't mean this A and this A is the same thing it just means I'm running through all the A's for every G. So if we if we think about what sigma was what the mapping of the sigma was it mapped sigma there it mapped two there's the two and that is the one there and then the three and the two so see these two A's it doesn't mean they the same thing when I use the symbol it's not equal to each other it's something different. So two was equal to then sigma dot one and three was equal to sigma dot two and one was equal to sigma dot three. So it's an action of the group set one of the elements in the group on one of the elements in the set the action of that same element G so this G is the same G G G and one of the G's I picked here was sigma so it's this action of sigma on one the action of sigma on two the action of sigma on three and that's what it did to this side but I'm showing this sigma the other way around now and what we could see here is that really this was a bijection this was a bijection because I have all the elements there and I have all the elements there they are not repeated this is a one-to-one and onto mapping and really so it for our example here at work but it would be much more powerful if I can show in general that this is a bijection because if it is a bijection I'm really dealing here with a permutation because every time I take a different one of these one of these and it acts on there you know I'm going to get this or this is equal to this action on this side so really if this is if this is a bijection this is a biject function then I am really you know representing each of these with a permutation and that is the power of group actions here and I mentioned to you very quickly we use proofs from from set theory we're going to just show that this is actually a two-sided inverse those functions are two-sided inverse and I just want to give you some intuition behind those proofs as well so to show that it's a if it is a two-sided inverse it is a bijection so I'm going to use that form of a proof to show that this is a bijection and what you do is you say that if I have sigma of the inverse of one of the elements so whatever the inverse of that was with that on a that I get back to a if I have this composition of two of these and such that one time I use the the inverse in g and an element in g and it's inverse that I get back a so what I'm saying is I'm mapping the whole set a finally through the whole composite of two actions back to the whole a so I must get all of the elements a here take it through a composition of these actions and I must land up with all those elements once again and the other way around as well so there must be an inverse there so if I have this then the inverse I land up up back with with a again equal sign so I land up with a as as well now I can show you that this one is an injective mapping so imagine here I start with all the a's I start with all the a's and then so my first map is this one that's the sigma of g and then finally I have this one this one is the sigma of g inverse so there's my first map I do this one first and then the outside bit there and then this is the composite now the composite must be an injective must be injective because that's what it's saying there if it's can I prove that this is injective well let's assume to the contrary that it is not now remember what injective is injective means if I have lots of elements on this side each one of these on this side maps to something there's nothing left out here and they they will just map to one to a single element maybe I shouldn't put that in it let's put it in here so that would mean by to the contrary that I would have two elements mapping to the exactly to the same element and even if this part is then one to one so each one of these would map to a unique one there in so that's assuming to the contrary that this composition is not injective that means I cannot have by the composite all of the elements in here being represented in here because certainly it's only one of the two of these that is going to be represented there not both of them so there's going to be a is in here remember each of these represents a there's going to be some a is in here so you know to the contrary then I'm going to have these only that one is mapped and that one is mapped this one's not mapped at all so these two can definitely not equal each other this says that that it should be equal the composite means that all the elements a should eventually appear here but if I let this first step not be to the contrary not be injective so two would go to the same one all of these will not eventually make it to there so by assumption to the contrary proves that that this is injective that doesn't make it so objective specifically if you think about if my set here is infinite in size now let's do it this way so again I have the whole a the whole a the whole a and my first mapping here is now the g sigma g inverse my second mapping here is sigma of g and I'm suggesting that this the whole thing must be that the all of a is all of there but let's assume to the contrary that it is so this way around you look at to show that it is surjective that this is surjective now if I if I say to the contrary let's not make it surjective now it not being surjective means that there's elements here on the second side which don't get mapped to so there's a whole bunch here so there's a whole bunch here that I can just take away they are not being mapped to so to the contrary I assume that this bit is not surjective so there's some that are not mapped to well then you can think of the fact that if there's this one mapped to this one mapped to there through the composite here going through first through this one then through that one there are some here that are not going to land up over there in other words not all of these not all of these make it into there so it's not the whole a that I'm talking about so if I assume that it's not surjective then the composite doesn't work because there are going to be some in here which you know don't end up there so it has to be surjective so this thing a composition of this thing must be injective it must be surjective therefore it is a bijection so if we have a two-sided inverse and that's a two-sided inverse this must be and we saw a beautiful example here it of course it is a bijection because it is a bijection and that links that is the real link between each one of these being represented by a permutation of the set on which it is acting it's a permutation of the set you can see the beautiful example there you can get the intuition to the proof this two-sided inverse effect why it makes this thing a bijection so there's a one-to-one and onto mapping between these two so all of them are there all of them are there in that set and that just means it's just a scrambling of them all the time this just acts to scramble these up each one of these that one doesn't scramble it up that one scrambles those two and that one scrambles those two and that one scrambles these two et cetera they are just a permutation they represent permutations because this is a bijection so you can understand what we mean by the action of one of these on it is a it just scrambles them up that is the action of the elements in a group that make up a set on another set group actions permutations it's as simple as that