 Continuing under discussion of automorphisms, I want to now talk about the idea of inner automorphism. All right, when we defined automorphisms previously, I mentioned it's a very, very important group action that every automorphism group acts on its group via the permutation action. And that this is super, super, super important. But it turns out in some degree, there's a class of automorphisms we've already been studying, and that's conjugation. And that's what an inner automorphism is. If you have a group G, we can define an automorphism from G to G. This is sometimes denoted as I sub G, but this notation is not universal. Defined by the rule that I sub G of X where X is some other element in G there, we define this to be GXG inverse. And as to say the image of X via the map IG is just left conjugation by G, okay? Left conjugation would put the inverse here on the right because the original element here is on the left. You could also, if you wanted to do right conjugation, you get G inverse XG, that's acceptable. And that has to do with between like left actions and right group actions. In this lecture series, we pretty much always stick with left actions, but you can get away with either one. It doesn't make much of a difference, just so you're aware. Because honestly, if you switch it from the left action to the right action, you're just switching from an element to its inverse. And so it makes no big deal whatsoever. The collection of inner automorphisms is often denoted as INN of G, like so. And why are they called inner automorphisms? Well, we'll see in just a second in the theorem down here that these are in fact, automorphisms that conjugation by a fixed element is an automorphism. So it deserves that name. Well, what does inner mean in this situation? Well, the idea is that inner means that the automorphism came from the group itself. You're conjugating by an element in the group. And so the inner automorphisms are those automorphisms of the group, the symmetries of the group that come from the elements of the group itself, as opposed to outer automorphisms that come from something external. That has nothing to do with the elements of the group but some other type of symmetry of the group. All right, well, like I said, as the name suggests, these things are automorphisms. We have to prove that, but we can actually prove something better. Given any element of the group and it's associated inner automorphism, it's gonna be an automorphism, first of all. So we'll prove that, but we can show that the collection of inner automorphisms forms a normal subgroup of the automorphism group. So that's not just it forms a subgroup, it forms a normal subgroup. So this subgroup is closed under conjugation, which when you think of inner automorphisms as those conjugating automorphisms, kind of makes sense when you think of it that way. But let's provide the details here. There's a lot of things to say. So let's first prove that conjugation is an automorphism in the first place. So we have to prove that it's a bijective homomorphism. We'll prove the homomorphic property first. So the automorphism we're gonna consider is I sub G. So we're gonna conjugate by the element G and then take two other elements of the group G. We'll call them X and Y. So the letter G will be what we're conjugating by, X and Y are what's getting conjugated. So we wanna prove this is homomorphic. So take I sub G of X, Y. We wanna prove that this is the same thing as IGX times IGY. Well IG of X, Y, we get G times XYG inverse. For which what we're gonna do is we're gonna insert an identity element in the middle there. So we get G times X, E, Y times G inverse. The identity here is gonna be E. For which then the identity E can be factored as X times G inverse G, Y times G inverse. And redoing parentheses, because we are associative after all, this becomes the product GXG inverse times GYG inverse. For which the first factor, GXG inverse, that's just IGX. And then the second one, GYG inverse, that's just IGY. So in fact, we see that conjugation by fixed element does induce a homomorphism. We see the homomorphic property there. Why is it bijective? Well actually alluded to this already, IG has an inverse. And there's a very natural candidate for this, right? The inverse of the map IG is gonna be conjugation by the inverse element of G. So I want to see that for a moment. If I take IG inverse, so we take the inverse of the element composed with IG of X, well IG of X is going to give you GXG inverse. But then if you conjugate by the inverse of G, you're gonna get G inverse times GXG inverse. Then you're gonna get G inverse inverse, which is just G by the associative property and other axes of group. This will simplify just to be X immediately, which of course is the image of the identity. So this composition gives us the inverse there. So we do get that conjugation is a bijective homomorphism. It's going to be an automorphisms of the group. So IG doesn't belong to the automorphism group. That gives us the very first principle. Now, why is the collection of all the inner automorphisms? Why is it a normal subgroup? Well, let's now prove that it's a subgroup and then we'll prove normality after that. I wanna next prove that the product of 82 inner automorphisms is also an inner automorphism. And there's a natural candidate here. When you multiply together IG and IH, I claim this is gonna be the inner automorphism associated to the product of the two, okay? And so if we take G and H are two elements of the group we're gonna conjugate by, we're gonna look at the image of X. So to prove that IG, IH is equal to IGH, we just show that if they agree on an arbitrary element of their domains, then they're the same function assignment, all right? So IG, IH of X by composition here, we're gonna first conjugate by H. So we get IG of HX, H inverse. Then you conjugate by G, so you're gonna get G, H, X, H inverse, G inverse, for which then if we rewrite some things, redo some parentheses, we're gonna get GH times X times GH inverse. Remember here that H inverse, G inverse, this is equal to GH inverse by the Shusok principle. So this does in fact lead to that factorization. Then when you look at the last factorization here, this is just the image of IGH on X. So therefore we get that IG, IH is equal to IGH, like so, that the product of two inner automorphisms is in fact, inner automorphism, it's actually the inner automorphism associated to the product of the two conjugating elements. Also notice that if you conjugate by the identity element, you're going to get that that's just the identity. And by our previous observation that the inverse of an inner automorphism is actually the inner automorphism of that group's inverse element, we actually proved that already. So that gives us everything we want. The inner automorphism set is closed under products, identity and inverses so that the collection of inner automorphisms is a subgroup of the automorphism group. Why is it normal? All right, the last thing we wanna show here is that it's normal if we take any automorphism, Sigma does not necessarily have to be an inner automorphism. It could be any automorphism, we conjugate by G and we're gonna hit the element X. I wanna show that Sigma, IG, Sigma inverse is in fact an inner automorphism. It turns out this is gonna be the inner automorphism where you conjugate by the element Sigma G where Sigma G is then the image of G under the automorphism Sigma. All right, this is gonna be a beautiful argument here. So again, to prove that these functions are equal to each other, I'm gonna compare them on an arbitrary element of their domain and show that they're the same. So if we take Sigma, IG, Sigma inverse of X I don't really know what Sigma does other than it's an automorphism but multiplication of automorphism, this is just composition of functions. So this will be Sigma of IG of Sigma inverse of X for which whatever it is Sigma inverse of X that's an element of the group. And so I know what IG is gonna do to that. So IG of Sigma inverse of X this will give us G Sigma inverse of X, G inverse. Okay? And like I said, I don't know what Sigma inverse does because I don't know what Sigma does but I do know that Sigma is an automorphism. So it has the homomorphic property. So if I take Sigma of a product this becomes a product of Sigma's here. So you can get Sigma of G times Sigma Sigma inverse of X times Sigma G inverse. But hey, you have now the composition of the function there Sigma Sigma inverse. Even though I don't know what Sigma is I do know that Sigma Sigma inverse is the identity. So Sigma Sigma inverse of X will just become X right there. And look what we have here. We have Sigma of G times X times Sigma of G inverse. Notice what we did here as well. You have the element inverse of G inside of Sigma because it's a homomorphism. It's in fact an automorphism. This is the same thing as Sigma of G inverse right there. And so this is conjugation by the element Sigma G. And so this is then the image, the inner automorphism of X with respect to the element Sigma G. So we then established that conjugating an inner automorphism gives you an inner automorphism. And so if we have a subgroup that's closed under conjugation that then proves to us that the inner automorphism subgroup is a normal subgroup proving the theorem right here. There's another thing I wanna point out here. We actually proved a stronger result than what we've already listed here. Another result we proved is the following that if you have a group and you have its automorphism group like so, then in fact, the inner automorphisms we can actually view as a homomorphism where you can take an element G and you map it over to IG. So this relationship where you identify a group element with its inner automorphism is actually a homomorphism from the group into the automorphism group. So that is to say every automorphism group contains inside of it a homomorphic image of the group that we are taking the symmetries of which is a very interesting observation. Now be aware that this map is not typically homomorphic and therefore there's not necessarily an isomorphic copy of G inside of its automorphism group but there is a homomorphic one. It turns out with this homomorphism here the kernel would then be the center of the group. The center of the group, remember these are the elements that commute with everything. And so if you commute with everything ZXZ inverse would just become ZZ inverse X which just gives you back X. So this is the identity relation. So this, so we can take a homomorphic image of G inside of the automorphism group. Its kernel would be the center of the group. So what we've really proven with a little bit extra detail here is that the inner automorphism group is actually isomorphic to G mod out its center. And this will always be contained inside of the automorphism group. Let's look at an example of these observations here. Let's consider the group D4 the dihedral group of eight elements. That is, I should say that the dihedral group of order eight it's the symmetry group of a square, okay? We should probably take something non-Avelian so that the inner automorphisms are non-trivial but this is what's gonna happen here. Using the usual symbols here we can describe any element in the dihedral group as R to the K or R to the KS where K will range between zero and three honestly. We're working mod four with the Rs. If you take the inner automorphisms of the identity as we already mentioned that'll just give you the identity map. But also if you take the inner automorphisms with respect to R squared that also will give the identity map and that observation comes from the fact that the center of the dihedral group here is the identity and R squared. Those things commute with everything. Therefore, their inner automorphisms are trivial. Now, if you conjugate by the element R what's gonna happen there? Well, R commutes with every other rotation. So if you conjugate one R, R squared, R cubed by R you're just gonna get back the original element. But the rotation does not commute with the reflections. And so what we see happening here if you take RSR inverse when you commute an R past an S what that does is it inverts the R. So this is the same thing as RRS which is R squared S. So we see that if you conjugate S by R you're gonna get R squared S. Similarly, if you conjugate R squared S by R you're gonna get back an S. And then likewise I'm not gonna go through the details of this one but if you conjugate RS by R you're gonna get R cubed S and if you conjugate R cubed S by R you're gonna get back an RS. So I'm writing these inner automorphisms since they're automorphisms, they're permutations I'm writing these inner automorphisms using cycle notation. So if we think of the eight elements of D4 here then S goes to R squared, R squared excuse me, S goes to R squared S R squared S goes back to S RS goes to R cubed S and then sends it back. Now I should mention that the inner automorphisms that is to conjugate by R cubed does the exact same thing. It sends S to R squared S. It sends RS to R cubed S and vice versa. And why is that? Well, that's because R is congruent to R cubed mod the center of the group here D4 because the center is one in R squared. If you take R cubed you can factor it as R times R squared for which if you forget the R squared a central element R and R3 do the same thing. So with respect to conjugation because R and R cubed differ only by a factor of a central element then their inner automorphisms actually do the exact same thing. You're gonna see the same thing here conjugation by S and conjugation by R squared S do the same thing because they differ by a central element, R squared. And so if you conjugate R by S you get R cubed and conversely R cubed will go to R we saw that one a little bit ago. And if you conjugate RS by S you're gonna get R cubed S and back again. R squared S does the exact same thing. And likewise, if you take RS and conjugate any element you're gonna get the following permutation here. R will go to R cubed, R cubed goes back to R and then S will go to R squared S and back again. Everything else that's omitted here will be fixed by these elements, all right? So what I want you to consider right here is that clean up the screen a little bit when you put all of these things together this is the inner automorphisms of D four, okay? This is actually isomorphic to the Klein four group. You'll notice that each of these inner automorphisms it's a two-two cycle. And so squaring it gives you the identity of course the identity itself is the identity it's not a two-two cycle but it also squares to itself. And this is isomorphic to D four mod out its center which of course its center is one in R squared. Like so this is the observation we saw before that the inner automorphism group is a homomorphic image of D four. In this case it turns out to be the Klein four group. Now this is not, these are not the only automorphisms of D four. There's another automorphism that actually sends all of the, it sends all of the reflections to each other. So there's an automorphism of order four that sends S to RS, it sends RS to R squared S, it sends R squared S to R cubed S and then it sends R cubed S back to S. And it leaves all the rotations fixed. We'll call this, we'll call this automorphism row. It's an outer automorphism as in there's no conjugation that induces this automorphism. It does not belong to the inner automorphism group of G, D four in this situation. And in fact if you take all of, if you put these things together the automorphism group of D four can be created by taking row and IS like so. And this is actually why I called it row because IS is conjugation by S. It is a element of order two. Row here is a four cycle when we view it as a permutation has order four. So you have this element of order four, you have this element of order two and the way they interact with each other is that row IS is actually equal to IS row inverse. That kind of sounds like how R and S behave to each other in the dihedral group, right? RS is equal to SR inverse. And in fact the relations between row and IS are identical to the relationship between R and S. And so in fact the automorphism group of D four is isomorphic to D four itself. It's kind of a fun little observation there that a group is isomorphic to its own automorphism group. That's actually not such an uncommon feature but that's where we're gonna end this discussion here for lecture three. We've introduced the idea of automorphisms in this last video, we talked about the idea of an inner automorphism. So thanks for watching. If you have any questions please post them in the comments below on any of my videos and be glad to answer them as soon as I can. 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