 In this video, I want to provide a stronger version of the Cayley's theorem we proved in abstract algebra one. And so without a better name, we're going to call it the strong Cayley's theorem. Remember previously that Cayley's theorem says that every group is isomorphic to a subgroup of SN. So essentially we can think of every group as a permutation group. The proof utilized the fact that we can identify group elements with permutations by associating to that group element its left multiplication. That is to say we identify G with the function lambda G which goes from G to G, which is the function lambda G of X is just left multiplication GX. Like so, so we can identify group elements with permutations. I want to now point out that this identification, a group element with its left multiplication, this is just the left regular action of the group onto itself. And so what we're doing here is we're associating a group with its left regular action which then is a permutation action which then connects you to a permutation group. It turns out that in general we can associate a group with basically any of its group actions and get a stronger condition there. All right, so let's state specifically what's the strong Cayley's theorem states. G is a group acting upon some G set, call it X. Associated to X we can take S sub X which is then the symmetric group on the set X. So S sub X is just the collection of all permutations on the set X and that does form in fact a group. Okay, so for any element G inside of the group G we can come up with a map we're gonna call it Pi sub G. This will be a permutation from X to X where we associate, well the function is defined by the rule Pi sub G of X is then just D G dot X. It's the group action. So our first claim is that this map, Pi sub G is in fact a permutation so it belongs to S sub X. And then furthermore the map phi from G to S sub X which S X is a group of course. We identify an element group, the element of the group G with its permutation Pi sub G. This is in fact actually a group homomorphism. So with respect to any group action we have a homomorphism from a group to a symmetric group. Now of course in the case of Cayley's theorem we have the regular action and this homomorphism will be one to one therefore we're isomorphic to a permutation group but we don't necessarily get isomorphisms in general this could be a homomorphism and this could be the fact that two different group elements have the same they act the same way. And this is very, very beginning of what's called representation theory we can represent any group as a permutation action. That's what we're doing here with the strong Cayley's theorem. So the first thing we have to prove is we have to prove that Pi sub G is a bijection. By construction it's a map from X to X. Why is it bijective? Well to show that it's bijective we could show individually that's one to one and onto but an alternative approach will just be to produce its function inverse that is there is a function with compose with Pi G to give you back the identity. And because we're identifying already this function with a group element there's a candidate on who the inverse should be. The inverse of Pi G as a function could be the function associated to the element G inverse. So that's gonna be our claim. I wanna show you that Pi G composed with Pi G inverse is the identity function and therefore Pi G and then this Pi G inverse. Well I should say this Pi G gives you the identity so it's in fact the inverse proving the claim here. So that's how we're gonna get that it's a bijection. And so let's go through this calculation. The way that I wanna show that Pi G composed with Pi G inverse is the identity is just look at it element wise. If X is an arbitrary element of the domain if we can prove that we get back X then this would prove that the composition of the two functions is the identity. So let's start off with exactly that. Pi G composed with Pi G inverse of X. What does it do? Well by definition here Pi G inverse of X is G inverse acting upon X and then Pi G by definition is G acting on whatever that turns out to be. But as we have a group action here compatibility comes into play and we can re-associate this to become G G inverse dot X which G G inverse will be the identity. What does the identity do to X? Well it does nothing you get back X but this is exactly what the identity function should do to X. And as X was chosen arbitrarily we then see that G composed with G excuse me Pi G composed with Pi G inverse is the identity and by similar reasoning we get that Pi G inverse composed with Pi G gives you the identity. So this shows that Pi G is in fact a permutation because it's bijective. So Pi G does belong to the group S X. So we can identify G with its permutation in S X that's what Phi did right? So Phi sends G into S X in such a way that G is identified and little G is identified with Pi G. So we have a function from a group into another group. Is this a homomorphism? We now have to prove the homomorphic property here that multiplication is preserved. If we take two different elements of well I mean they could be the same element I don't know it doesn't matter. We just take two elements of the group G call it G and H we want to then prove that the function Pi sub G H is identical to the function and the composition of functions Pi G and Pi H. We're going to do this element wise. If we again if we take an arbitrary element of the domains and show that these two functions always agree with each other then the two functions are in fact the same. I'm going to start with the permutation associated to G H. So by definition this is G H dot X. By compatibility this becomes G dot H dot X for which then what does H do to X? Well it does the same thing that Pi H does to X. What does G do to this? Well G dot whatever that's just Pi G there. And so we get Pi G of Pi H of X. And this is just function composition here. This is Pi G of Pi H of X. So this then shows us that Pi G H is the same thing as Pi G composed with Pi H. So therefore since Pi G H is the same thing as Pi G composed with Pi H that gives us the homomorphic property because Pi of G H is the image of G H. And then Pi G composed with Pi H that is permutation multiplication. This is Pi G times Pi H. And so that's in the product of the images of G and H like so. This then proves this strong Cayley's theorem that every group action produces a homomorphism from a group into a symmetric group.