@BalarkaSen from my understanding this is a permutation group for a set $X$. We have a bijection that maps elements to themselves so that's our neutral element. It's a bijection so there exists an inverse function that negates another function, so each element has an inverse. Function composition is closed(I think) and it assures closure. Last, of course we have association as we have function composition as group operation. It's non-abelian as we don't have commutativity.
@UserX That's right. Indeed, Map(X) is the permutation group on X. These are "natural groups" meaning that they come up natural from fundamental set theory.
@Balarka Pedro suggested to look at the homomorphism between $G$ and the permutation group of the set of right-cosets of $H$ (there are 5, so that will be $S_5$)
To be honest, group theory is the only subject apart from topology that hasn't come natural to me. When I first learnt calc or analysis, I could guess what an example to a statement would be or what would be taught next. On group theory, I lost that intuition almost immediately
I wonder if there is a way using the fact that $S_5$ has no subgroup of order 15, because for some reason I was heading towards there - and I no longer recall why. I thought there was a hint saying that but I can't see any, so no idea why I was thinking that. Any idea @Balarka?
@r9m I've done the estimates here, and also found an expression of the Dirichlet series in terms of $\zeta$ (unfortunately, it's an infinite sum, but it's $\zeta$, man).
None. I at first used online sources but then moved to various lecture notes. I haven't found a simple group theory book that doesn't include the rest of abstract algebra yet.
@BalarkaSen if you were talking about topology I started with Munkres but moved to a less-formal "Topology without tears" in which I like the attitude and teaching style of the Author
Oh, I have it @Balarka. If there is such homomorphism, and I know $S_5$ has no subgroup of order 15, I also know $|G|$ divides $5x3$. However, it is not 15 as otherwise there would be subgroup of order 15 in $S_5$ by the homomorphism that injects into $S_5$.
So it must be 5. And that doesn't make any sense, there's a mistake somewhere in the argument
@BalarkaSen Cayley's theorem includes infinite groups too. How can $(\Bbb R^*,\times)$ be a permutation group of a set? What's that permutation group called and what are it's elements?
Oh that makes more sense now. What about the generalization of Cayley's theorem on infinite groups? The question "what's that permutation group" still remains...
clearly $G$ acts on $M$ by acting on the vertices of $\Gamma$ freely and properly discontinuously. an interesting question would be what is $\pi_1(M)$.
Then I guess we can define our group operation as a function f: R to R , find a map that will satisfy the group axioms and we have a permutation group right?
@MikeMiller $\Gamma = \Gamma(G, \mathcal{S})$ be the Cayley graph of $G$ with generating set $\mathcal{S}$. $M$ be the smallest genus surface on which $\Gamma$ can be embedded planarly. Genus of $M$ is equivalent to $1 - \chi(\Gamma)/2$. What can we infer about $M$?
@Balarka I may be wrong, but a bit of googling leads me to think it has to do with the small cancellation theory. You will probably understand it all much better
Oh it's just a tiny bit of knowledge about these surfaces it seems...
I recall the argument went along the lines of "relative to any finite set of generators, the cayley graph of a finite group is hyperbolic as it is bounded"
@Balarka there are two main models, G(n,p) and G(n,m) (there are others). G(n,p) is the undirected graph on $n$ vertices, so that $P(ij\in E(G))=p$ for every two vertices independent of the others.
$G(n,m)$ is the graph on $n$ vertices chosen uniformly from the graphs with $m$ edges (and $n$ vertices)
For example, by studying the properties of G(n,1/2), you are basically studying the properties that apply to almost all graphs
I wonder, I know isomorphism doesn't change the order of an element. Can I conclude from that, that if there is isomorphism $G\to H$, then $o(G)|o(H)$?
@DanielFischer I suspect the guy you just commented to is not asking in good faith. Look at his question about whether $\nabla$ is a vector and look at the comments under the answers.
@MikeMiller The fundamental group is definitely not interesting enough to ask at this point. Here's an example of what I have in mind : PSL(2, 5) is just a dodecahedra which embeds in a sphere. the dodecahedra thing makes me suspsect that Cay(PSL(2, 7)) embeds in the klein quartic
an evidence is that by the 84(g-1) theorem, the klein quartic has 84(3 - 1) = 168 size automorphism group, and order of PSL(2, 7) is precisely 168
@MikeMiller well the smallest genus surface on which some Cay(G) is embedable is homeomorphic to genus 1 - \chi(Cay(G))/2 genus surface so the fundamental group is not at all interesting w.r.t G
it's just the same as asking the fundamental group of an n-torus.
n-torus means the product of $n$ copies of $S^1$. You mean "the aueface of genus $n$". And I know it's not; I even gave you the fundamental group up above :P
In mathematics, the uniformization theorem says that every simply connected Riemann surface is conformally equivalent to one of the three domains: the open unit disk, the complex plane, or the Riemann sphere. In particular it admits a Riemannian metric of constant curvature. This classifies Riemannian surfaces as elliptic (positively curved – rather, admitting a constant positively curved metric), parabolic (flat), and hyperbolic (negatively curved) according to their universal cover.
The uniformization theorem is a generalization of the Riemann mapping theorem from proper simply connected open...
How about the Schwarzian derivative, nice derivation of that from nothing?
@Mike Just an unrelated question : is there a way to homotope two maps by geodesics on some space? i mean, there you have the straightline homotopy which is like homotoping two maps through the geodesics on $\Bbb E^n$, so there must be some kind of generalization or something.
Balark it sounds like you want to homotopically deform the metrics on a space so that the geodesics in each metric deform into each other homotopically
You want to take a map connecting two points, call it the geodesic for some metric, then instead of deforming the maps deform the metrics so that their geodesics are the curves you would have deformed the original curves into
I am just guessing, it sounds right to me, I have never done it and have no idea how to do it in practice
So lets say you want to homotopically deform curve A into curve B. Instead of deforming A into a bunch of random curve in between, you could use this idea select a specific set of maps in between, namely those maps which are geodesics of a specific family of metrics
In other words I guess you could use this idea to choose specific homotopically equivalent maps
Furthermore I know $S_5$ has no subgroup of order 15, and thus $O(G)$ can't be 15. Thus $O(G)$ is 5 or 3. It must be 5 as otherwise $|G:H|=5$ would lead to contradiction.
I don't understand, you said you wanted to homotope two maps by geodesics in some space, if you picture the top half and bottom half of the circle in the R^2 plane, connecting the points (1,2) to (3,2), and you want to deform the top map into the bottom map by saying each map in between is a geodesic then you have to say which metric it's geodesic to! This means you have to invent a metric for which it's a geodesic for, no?
@Balarka Maybe it would help if you could draw a picture. The point of the straight-line homotopy is that it shows that $\Bbb R^n$ is contractible. Manifolds in general won't be...
@bobby I don't think he's interested in changing the structure of geodesics. He's interested in exploiting the existing structure...
Yes, but if you are in the plane and the top half of the circle and the bottom half of your circle are the two maps you are homtopically deforming between, you cannot use the word geodesic because the only geodesic is the straight line splitting the circle in half. If you want "to homotope two maps by geodesics on some space" then you will have to call every map in between a geodesic... The only way to do this is to invent ad-hoc metrics and use embedding and all this stuff
I know that if $G$ is a group, $H$ a subgroup of $G$ and $S$ the set of right-cosests of $H$, than there is homomorphism from $G$ into the set of permutations of $S$, and the Kernel of that permutation is a normal subgroup of $G$, and it is also the maximal one that is subset of $H$.
So I have $G$ as a group, and $H$ as a subgroup. $|G:H|=5$, so I know there is homomorphism into $S_5$, by that theorem - right?