 Suppose that we have a group G that's acting upon a set X, so X is a G set. And suppose we have some fixed element G inside of G. Using this element, we can define what's called the stable set, sometimes called the fixed point set associated to G inside of X. This is commonly denoted as X sub G. X sub G is important to point out here. It's a subset of X. It's some collection of the points that we're acting upon. And this is going to be a set of all points which are fixed by little G, all right? So X sub G by definition is a set of all elements, little X inside of X, so that G dot X is equal to X. These are the stable set is all of these elements that G does nothing to, all right? In particular if G is the identity, right? If you take X sub E, this is all of X by construction, the identity doesn't do anything to the set. So the stable set associated to the identity is always the whole set. The kernel of a group action is defined to be all of those elements which stable set is nothing, right? So if you have some group action, it's so-called kernel, this is gonna equal all of the elements G inside of G such that XG is equal to X in that situation. So this stable set's a pretty big deal. It does say a lot about the group actions, what these do individually, that is these individual stable sets, these fixed point sets, what do they do? Now on the flip side of the coin, now let's take an element X inside of X. What does its stable set mean? These are typically called stabilizers. Sometimes they're called the isotropy subgroups. The term subgroup is somewhat getting ahead of ourselves. Why is it a subgroup? Well, we'll get to that in just a moment. So given an element of the G set X, we can define its stabilizer, which is commonly denoted as G sub X. So G is the group, X is the element that's getting fixed right here. And that's what G sub X is. This is a set of all elements of the group such that G dot X just gives you back X, right? So these are the things that stabilize X. And so like I said before, X sub G is a subset of X. These are all the elements which are fixed by G. Now G sub X is a subset of G and this is the collection of all elements G that fix X. So in both situations, the subscript is the element that's doing the fixing, right? So X is fixed in this situation or G is the thing that's fixing. And then the large set, the large set X and G, this is then the subset of X and G respectively. And there's some possible collection there, right? So we're taking, you take X sub G here, all the things inside of X that are fixed by G and then G X here, these are all of the elements of G that fix X. So there's always this relationship of stabilizing something, but the notation tells you which subset we're talking about. But like I said, why are stabilizers sometimes called isotropy subgroups? Well, that's because they're subgroups. That's our proposition that we're gonna prove right now. If you have a group G that acts upon a set X and you pick any element X inside of that, then it turns out that the stabilizer is in fact a subgroup of G and therefore it deserves the name isotropy subgroup. So to prove that G is a subgroup, well, there's basically three things we have to check. We have to show that it's closed under multiplication, closed under identities, well, there's only one identity, and then closed under inverses. So we'll do those bit by bit by bit. So let's take two elements of the stabilizer, call them G and H. So by assumption, G acting on X is X, and likewise, H acting on X gives you X as well. That's what it means to be in the stabilizer. They stabilize the element X. So what happens when we take the product? Well, because this is a group action, G H acting on X is the same thing as H acting on X and then G acting on that. But by assumption, H stabilizes X. So H dot X is just X. But also by assumption, G acting on X is just X because they're both in the stabilizer. So therefore the product G H also stabilizes X. So thus the product G H is inside of the stabilizer. G X is closed under multiplication. The next one's a pretty easy one. What about the identity? The first axiom of a group action is that the identity stabilizes everything. Therefore the identity will belong to G H for any X, well, excuse me, the identity will belong to G X for any X whatsoever. So G X always contains the identity. It's closed under identities. So the last one about inverses. Suppose that G belongs to G X. So that means G stabilizes X, G dot X is equal to X. What does the inverse of it do? Let's look here. G inverse dot X, we want it to be X and we'll see that in just a second. Since G stabilizes X, X is the same thing as G dot X. Using compatibility, we can re-associate and get G inverse G dot X, which G inverse G is the identity, right? The identity stabilizes everything. And so this then implies that G inverse stabilizes X as well. So G inverse belongs to the stabilizer and the stabilizer is closed under inverses. Since the stabilizer is closed under inverses identity and multiplication, the stabilizer is a subgroup again justifying the label of colon and isotropy subgroup. Let's look at some examples of these stable sets here, right? So let's take the set to be one, two, three, four, five, six. We're just gonna take the permutation action here. We're not gonna do all of S6. So I'm gonna take a subgroup that looks like Z4. So we're gonna take the identity. We're gonna take one, two, three, four, and five, six. So that's a two cycle times a four cycle. Then we're gonna have a two, two cycle, three, five, and four, six. And then another two, four cycle, one, two, and three, six, four, five, like so. And you can verify that this is in fact a subgroup and G is gonna act on X by the usual permutation action, okay? So let's look at the very stable sets. So as I go through these, as I go through these, excuse me, if I go through the elements of G, so we have the identity, we have this element, this element, this element, what are their stable sets? So like I mentioned before, the identity is gonna stabilize everything. This is the group identity just so we're clear here. Because the identity by construction fixes everything. So the stable set of the identity is always the entire G set. That's perfectly good. Be aware of that. Next, what do we get here? If we look at the stable set associated to three, five, and four, six, who doesn't get moved by that element? Well, notice that one is left fixed by three, five, and four, six. And so is two. Three doesn't get fixed because this element sends three to five, three to five here. Neither does five gets fixed because it sends five back to three. And then notice that four is not fixed because it sends, this element sends four to six and likewise six will go back to four. So the nice thing about permutation action is that when you look at the stabilizer, the elements that are present in the cycle composition are the things that get moved. Thus those elements, those letters which are omitted, that is going to be the stable set associated to that permutation. Then when you look at the other ones, the two, four cycle one, two, and three, four, five, six, nothing is left fixed. It moves everything. And therefore the stable set will be empty, right? So it's completely empty in that situation. Likewise, if you take the other element, the stable set is empty. So with regard to the permutation action, it's very easy to identify the stable sets because you just look at the cycle decomposition and then the stable set will be all those elements which are not listed in the cycle decomposition because we typically omit elements which are fixed. Now let's go the other way around. What are the stabilizers, those isotropy subgroups? So what is the subgroup that fixes the letter one? Well the identity fixes everything so it'll belong to that. But like we observed earlier, three, four, five, three, five, and four, six fixes one. So the stabilizer of the letter one is then the cyclic subgroup generated by three, five, and four, six. So this is a subgroup isomorphic to Z two. It turns out that the stabilizer associated to two does this exact same thing, right? Because one stabilizes two, three, five, and four, six stabilizes two. The other elements do not. They don't stabilize it. Then if we look at the other elements, the other letters I should say three, four, five, six, the only thing that stabilizes them is the identity permutation. No one else stabilizes them. So in fact, these six stabilizers are subgroups of the group there G. So you just get the trivial subgroup and Z two. All right, so before we end this video, there's one very important property about isotropy subgroups that I want to point out here. And it has to do with taking conjugates of isotropy subgroups. So we have a group action G acting on the set X take a fixed element G of the group, little G and a fixed element, a fixed letter X of the G set X there and we want to prove the following identity. If you take the conjugate subgroup of the stabilizer associated to X with respect to G here. So that's where conjugating by G, the stabilizer of X. This is equal to the stabilizer of G acting on X. Now these are both subgroups, right? Stabilizers are always subgroups and the conjugate of a subgroup is also a subgroup. So what this is basically telling us is that the conjugate of a stabilizer is also a stabilizer and they're connected by the element you conjugated by. But as these are subgroups of the group but as they're sets to prove equality, we only have to show that they're subsets of each other. We don't have to prove that they're groups or anything that's already established. So let's prove their subsets of each other and those are the two directions we have to go. So let's first start this direction. Let's show that the conjugate of GX is a subset of the stabilizer of GX. So let's start off the typical element of G, GX, G inverse, call the element H. Well, since H belongs to this conjugate here that means there exists some element H prime that belongs to the set GX such that H is equal to G, H prime, G inverse. So just the definition of what it means to be inside of that set right there. So what we're gonna do is then we have to prove that H belongs to the set G sub GX. This is the stabilizer of GX. So to be inside this set, H needs to stabilize GX. So if I take H dot GX, I need to get GX. What are the details of that? Well, H dot G dot X by compatibility is the same thing as HG dot X. And now I'm gonna use my formula for H above here, plug it in for H, excuse me. H then becomes GH prime G inverse. You type in that by G. So this is multiplication in the group. So the G inverse G will cancel each other out. We're then gonna get GH prime dot H, dot X, excuse me. Reassociate again, then you're gonna get H prime dot X and then you have a dot G over here. By assumption, H prime belongs to GX, it stabilizes X. So H prime dot X is just X right here. So when you take H dot G dot X, you get back to G dot X. So H does in fact stabilize GX. So H belongs to G sub G dot X. So that gives us the first containment that the conjugate of GX is a subset of the stabilizer of GX. Did I say that right? Let me say it again. The conjugate of GX is the subset of G sub G dot X. Now we wanna go the other direction. So let's take an arbitrary element of G sub G dot X. So something like this, right? So what we know is that H stabilizes GH. So this time if you take H and you act on GX, you get back GX, all right? Well by compatibility, the left hand side can be rewritten as HG, the product dot X. That's going to give you G dot X like so. And then what we're gonna do is we're gonna take this and we're going to dot both sides of the equation by G inverse. You know, we often say like you multiply both sides by G inverse where we're gonna act on both sides on the left by G inverse. Now on the right hand side by compatibility, you're gonna get G inverse times G dot X. This then becomes E dot X for which then simplifies just to be X. Compatibility on the left hand side is much simpler. G inverse dot HG dot X, this just becomes G inverse HG dot X. Again, my compatibility right there. So this shows us that G inverse HG dot X is X. It stabilizes X. So G inverse HG belongs to G sub X right here. Now, if this happens, then our original element H can be factored in the following way. It looks like G G inverse H times G G inverse. You know, for getting some of the parentheses, you have some products of G and G inverses or identities. So this is equal to H, but then look at it. We just showed that G inverse HG belongs to the stabilizer of X. Then if you throw a G in front and a G inverse at the end, this then shows that H belongs to G, the stabilizer of X G inverse, thus giving us the other containment, thus proving equality. And so then we have the statement we were looking for that the conjugate of a stabilizer is equal to a stabilizer and the equality comes from the element you conjugated by.