 So now that we've looked at conjugacy in the previous video let's just look at what we call conjugacy classes conjugacy classes and I'm going to claim that they are nothing other than orbits and we've looked at the orbits of a set so the orbit of a set now remember here the set is actually the set that makes up the group so this makes it a little bit different and so we write it as the conjugacy class of a instead of the orbit of a and this is just going to be the set where I take all the elements and I act on a so I'm going to choose a specific a specific element this time it's of my group such that g is an element of my group so what am I saying here so I have the group I choose one of the elements then I act on it by all the elements in the group and that gives me this orbit of a or the conjugacy class of a now a very interesting thing is going to happen here if I have my whole again if I have my whole group here I'm going to be able to partition the group by conjugacy classes so let's choose an a that's in there remember we are talking of here about we are talking here about the action of a group on on itself that set is itself so this would be the same as saying g a g inverse running through all the g's of my group this is the exact same thing that's how we defined it in the previous video so this is what I am left with and I'm claiming that this is going to be this is going to partition it and I'm going to have another partition and that is going to be from an element b and I'm going to have to show that if this b is not one of the elements in there then these two sets that I make here has got to be disjoint and we'll have to get there let's start with the first one let's say then that no matter I've got to show that no matter which element I start in here with I'm going to end up you know with another element that's in here so let's choose another element and remember I'm saying that I choose an a from my group and I conjugated like this and I end up with an element so this is actually an element so I've got this this conjugacy class with this new element g a g inverse this new element and my new element here is is this one so a now becomes this thing for one specific g so I'm going to choose a very specific g because remember this is a lot of elements because I'm running through all of the g's here but now for one specific g this is one element so how would I you know how would that now be so that would be the set of all this g dot this is the group action on this thing g a g inverse and that for all this g's but now I've got to make it a different g so I'm going to say g star just to show that I'm still running through all my g's so all the elements in g so I'm running through all of them acting on this very one specific one and I can rewrite that's that I want to rewrite it as that so that is just going to be g star g star composed with g a g inverse composed with g stars inverse okay and we know from before if I have an inverse and an inverse I can rewrite that remember I'm still running through all the all of g and I can rewrite that one the other way around so I'm set here with g star g I'm still with a and now I'm with a g star inverse g g stars g it's inverse I'm still running through all of g still running through all of g now look at this that's the same and that's the same and remember from Cayley's theorem if I took one of the element in a group and I do the binary operation with all the others remember I'm going to end up with all the elements again remember the rows and the columns they are unique and they are exhaustive and otherwise all the elements are there from Cayley's theorem so this is just you know running through this if I take a specific element and I compose it with all of them I just end up with that same group again so let's call this then g double star double star and I have a and I have g double star and it's for all g double star all the elements of g well that and that is exactly the same thing because I'm saying a conjugated with the whole group take a conjugated with the whole group so lo and behold these two are the same c a equals c a g a g inverse so no matter what element other element I take in here I started with another element I'm going to end up with exactly the same set as if I just started with a and that is that is well defined because no matter which one I start with in that set I'm going to end up with exactly the same elements in that set now let's just see that if I take one that is not in there so let's take b and I'm going to define b as let's let's have b as this g star it's not that g star anymore so g star b g star inverse that's going to be my element because I run through all of them and b if these two are not this joint so if I have c a and I have the intersection with c b c b I'm saying that they are disjoint if they are not this joint that means this element must be you know one of the elements in there in other words it's got to be equal to one of the elements in there because it is in there so I'm going to have the fact that g a g has got to equal that and remember I'm saying here it's all the g and g star that's all of g so I take this run through all inverse I run through all of them and take b and I run through all of them and if this was not disjoint that means you know this was one group and this thing is actually inside of there so it is one of the ones that is in there so they are equal now what can I do just to express what b is going to be well I can compose both sides by g stars inverse so this time I'm going to have g stars inverse and then I have g a g inverse and if I have g star inverse that means it's the identity elements I'm left with b g stars inverse and if I write compose with g star I'm going to get rid of that so I have g stars inverse g I have a associativity I have g inverse and I have g star and I have g star because if I put g star there I just have b so now I have this curious thing that b is this well look at it carefully I can rewrite this thing just as we did there so I have your g star it's inverse with g I have a and on this side I'm going to have g stars inverse and g it's inverse because its inverse is exactly that because the inverse of this is g inverse and g star inverse is inverse which is just g star so that's all of that equals b but hang on a minute hang on a minute what's happening here is I am running through all of my g's so by Cayley's theorem again I might as well rewrite this as g star star a g star star and for all g star star elements of g so I'm running through all of them again and what do I have left with here I'm left with c a so if in fact that b was one of a's you know it must just be in a it cannot be in its own so this is you know for that not to be disjoint for these not to be disjoint c that b must actually be in a in other words if not so then these two things are completely disjoint and so I could go for c and d and everything in other words I have partitioned up partitioned up group by these conjugacy classes by these conjugacy classes I think there's one interesting thing I left out and that is if if I choose a very specifically let's choose a to be an element of the center of g now if a is a remember center of g remember what happens then I would have remember what how we define the center of g is that I have g composed with s equals s composed with g for all s element in g in other words there is commutivity and if I defined this as such and this commutivity meaning I can swap those two around so I'll have g g inverse a and that's just a so no matter if I run through the whole of g I'll always just end up with a a is going to be the only thing left it's just going to be a a a no matter what we do because if a was in the center it means whatever I'm composing one in the binary operation I do I can swap those two around or those two and I'm just left with a so a would be the only one if a is from the center if I chose a in the center then a is going to be the only one in that conjugacy class but that's not very interesting I'm I'm choosing my a here such that it's not in the center of g and then I actually get then I actually do get all of these so just watch out for choosing as a center but we can see that no matter what element I choose I'm going to end up in that conjugacy class and if I choose something that's not in that conjugacy class I'm going to end up with two conjugacy classes that are that are this that are absolutely disjoint so that in short is a conjugacy class