On the circle $S^1$ there is the usual , i.e. the group isomorphic to $\{e^{i\varphi}\mid \varphi\in[0,2\pi)\}$ with complex multiplication as group operation. This group is a topological group in the sense that $S^1$ is a topological space and the group operation and the inverse are continuous.     Question: Are there other topological groups on $S^1$ essentially different from the circle group, assuming $S^1$ with the standard topology? What about other abelian topological groups?   My question was motivated by  other post. The group described there turned out to be just the usual one.  Two notes:   I am looking for groups involving all of $S^1$, not only a subset. Especially no subgroups of the circle group. I am looking for groups not isomorphic to the circle group.
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Let $S^1$ be the subspace of $R^2$ given the usual topology. How many group structures make $S^1$ a topological group?
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I.e. do there exist non-homeomorphic $X$, $Y$ and covering maps $f:X\rightarrow Y$, $g: Y\rightarrow X$?  I have a basic understanding of covering space theory as its taught in school.  I was inspired by this question   I have done some of the things you would do, looking at what happens to fundamental groups.  And I've tried to find counter examples by drawing  graphs.
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I'm going through Grove's "Algebra." I'm having trouble with this one:  "If $G$ is the union of three proper subgroups, then there is an epimorphism from $G$ to the Klein 4-group."  I think the subgroups should in some sense correspond to the non-identity elements in the Klein 4-group.
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I am interested in the result above, and would like to study a textbook that delves into it deeper.  However the standard algebraic topology textbooks (e.g. Hatcher/Munkres) and standard algebra books like Hungerford do not contain this result.  Is there any suitable book that I can consult? Thanks!
30868
Just out of curiosity: If $G$ is a topological group and $H, K$ are closed subgroups, is $H\cdot K$ a closed subgroup?  Thanks!
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Show that $S^1$ is not homeomorphic to either $\mathbb{R}^1$ or $\mathbb{R}^2$  $\mathbf{My \ solution}$:  So first we will show that $S^1$ is not homeomorphic to $\mathbb{R}^1$. To show that they are not homeomorphic we need to find a property that holds in $S^1$ but does not hold in $\mathbb{R}^1$ or vice-versa. $S^1$ is compact however $\mathbb{R}^1$ is not compact.  The set $\{1\} $ is closed, and the map $$f: \Bbb R^2 \longrightarrow \Bbb R,$$ $$(x, y) \mapsto x^2 + y^2$$ is continuous. Therefore the circle $$\{(x,y) \in \Bbb R^2 : x^2 + y^2 = 1\} = f^{-1}(\{1\})$$ is closed in $\Bbb R^2$.  Set $S^1$ is also bounded, since, for example, it is contained within the ball of radius $2$ centered at 0 of $\Bbb R^2$ (in the standard topology of $\Bbb R^2$).  Hence it is also compact.  However real line $\Bbb R^1$ is not because there is a cover of open intervals that does not have a finite subcover. For example, intervals (n−1, n+1) , where n takes all integer values in $\mathbb{Z}$, cover $\mathbb{R}$ but there is no finite subcover.  Hence $S^1$ can not be isomorphic to $\mathbb{R}^1$.  How to show now that $S^1$ is not homeomorphic to $\mathbb{R}^2$? Can i show it now in the same way? They can not be homeomorphic since $S^1$ is compact however $\mathbb{R}^2$ not. How to show that $\mathbb{R}^2$ is not compact?
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I'm reading John Stillwell's "Naive Lie Theory" and it was mentioned there (without giving a proper definition of what a Lie group is) that the only Lie groups among the unit n-spheres are $S^1$ and $S^3$. Is there a simple or naive explanation of what makes $S^2$ so different from $S^1$ and $S^3$?
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I am a beginning learner of group theory.  Which of the following are isomorphic:$$\mathbb{Z_{24}}, D_{4}\times \mathbb{Z_{3}},A_{4}\times \mathbb{Z_{2}},\mathbb{Z_{2}}\times D_{6}, \mathbb{Z_{12}}\times \mathbb{Z_{2}},D_{12}, S_{4}$$  I know $\mathbb{Z_{24}}$ is cyclic and none of $\mathbb{Z_{12}}\times \mathbb{Z_{2}},D_{12}, S_{4}$ are cyclic since none has an element of order 24. I also know $D_{12}$ and $S_{4}$ are not isomorphic because $S_{4}$ has no element of order 12. Finally, I know $A_{4}$ and $D_{6}$ are also not isomorphic because $D_{6}$, unlike $A_{4}$, contains an element of order 6.  Other than this I don't know how to proceed. Any hints are appreciated. Please allow me to do the work.
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An algebraic group is not necessarily a topological group, since for an algebraic group, the product map $G \times G \to G$ is continuous when $G \times G$ is given the Zariski topology, which is finer than the product topology.   So does this tell you that the additive group $\mathbb{G}_a$ over an infinite field $k$ is not a topological group? The underlying variety of $\mathbb{G}_a \times \mathbb{G}_a$ is $\mathbb{A}_k^1 \times_k \mathbb{A}_k^1 = \mathbb{A}_k^2$, whose Zariski topology is the cofinite topology, where every open set is also open in the product topology (since $\mathbb{A}_k^1$ also has the Zariski topology).
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Let's consider a group $M$ (under multiplication) of all matrices $A$ of size $2 \times 2$ over $\mathbb{R}$ so that $\det(A)=1$. How to show that the group is homeomorphic to the $S^{1} \times \mathbb{R^{2}}$?  Topology on $M$ is induced by the norm $||A||=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}$, where $A = \begin{pmatrix} x_{1} &amp;&amp; x_{2}\\ x_{3} &amp;&amp; x_{4}\\ \end{pmatrix}$ so it's the same as considering matrix as a point in $\mathbb{R^{4}}$. According to $Q = S^{1} \times \mathbb{R}^{2}$,, the topology on $Q$ is induced by the standart one from $\mathbb{R}^{4}$.   The common idea is to start with considering the $x_{1} x_{4} - x_{2} x_{3}=1$, but i can not find some rigorous ways how to conclude that it's precisely homeomorphic to $Q$.  Any help would be much appreciated.
91995
How can we prove that the space of homeomorphisms Homeo$(S^1)$ of $S^1$ (strong) deformation retracts onto the orthogonal group $O(2)$?  I know that this result is proved by  in his paper   I want to learn somewhat more elementary proof if possible, or at least the idea of the proof.  Added Later (The topology on the space of homeomorphisms Homeo$(S^1)$ of $S^1$). This is footnote 7 from Hutchings' lecture notes, :      We topologize the space Maps$(X, Y)$ of continuous maps from $X$ to   $Y$ using the compact-open topology. For details on this see e.g.   Hatcher. A key property of this topology is that if $Y$ is locally   compact, then a map $X \rightarrow$ Maps$(Y,Z)$ is continuous iff the   corresponding map $X \times Y \rightarrow Z$ is continuous.
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I know $\pi_1(\mathbb{R}^2\setminus\{x,y\}) = \mathbb{Z}\ast\mathbb{Z} = \langle a,b\rangle$, but its non-abelian-ness isn't obvious to me.   Specifically, I draw a box and two points to represent $x$ and $y$. I call the clockwise loop around $x$ $a$, and the clockwise loop around $y$, $b$.   If I draw the loops $ab$ and $ba$ they look `obviously homotopic' to me (they are both a single clockwise loop containing both $x$ and $y$), so should be the same element in $\pi_1$, which is clearly not the case if we take the free product of $\mathbb{Z}$ with itself.   It seems that I must be making a big mistake somewhere. What is the obstruction to $ab \simeq ba$? I would prefer something concrete, I know this fundamental group follows from standard theorems (van Kampen's theorem), but I want to see the homotopy fail, because I think they are pretty clearly homotopic - so there is something wrong with my intuition which I'd like to fix.
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I'm being asked to find 2 subgroups of $S_3$, one of which is normal and one that isn't normal. I guess, to find the non normal subgroup is easier. I would do this by trial and error, but since the group is $S_3$ and easily visualizable, I guess that there should be a geometrical property that makes it easy to find a normal subgroup. Any of you guys know one? Or should I try this by brute force?
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$(\Bbb R,+)$ is a topological group. Is there any other group structure on $\Bbb R$ such that it is still a topological group and this group is not isomorphic to $(\Bbb R,+)$ ?  Refer to  for the general problem. In that problem I have conjectured that there is no such group however I couldn't prove it.
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I'm having trouble in proving the following:  If $U$ is a neighborhood of $e$, there exists $V$ neighborhood of $e$ such that $V^{-1}V \subseteq U$  Can someone please verify if what I wrote makes sense? And if possible show the way to go further, because I'm stucked.  My attempt:   Let $G$ a topological group.  (Definition: A topological group $G$ is a topological space with a group structure such that the applications $m: G \times G \to G$ and $r:G \to G$, defined by $m(x,y) = x.y$ and $r(x) = x^{-1}$ are continuous.)  $V^{-1}V = \{ x^{-1}y ; x,y \in V\}$   $U$ is a neighborhood of $e$, then is $V^{-1}V$ a neighborhood of $e$?  Since $r$ is homeomorphism and the right-translation $\delta_y:G \to G, \delta_y(x)=x.y$ is homeomorphism, the application $\delta_y \circ r :G \to G$ is homeomorphism. Let $x \in U$. Then $\delta_y \circ r(x) = x^{-1}y$ is in an open set, because of continuity. Why is it a neighborhood of $e$?
18235
I'm trying to show that they're equivalent statements:  1) $1_{S^1}$ is not homotopic to a constant map.  2) $S^1$ is not a retract of $D^2$ ($D^2$ is the closed unit ball).  3) Every continuous map $f:D^2\to D^2$ has a fixed point.  1)$\to$2) I suppose the opposite, then there exists $r:D^2\to S^1$ continuous such that $r(x)=x$ for every $x\in S^1$. How can we show that $1_{S^1}$ is homotopic to a constant?  2)$\to$3) I suppose that there exists $f:D^2\to D^2$ such that $f(x)\neq x$ for every $x\in D^2$, but I don't know how to prove that $S^1$ would be a retract of $D$.  3)$\to$1) Of course I tried to suppose that $1_{S^1}$ is homotopic to a constant, and show a map without fixed points.  Any hint? Thanks.
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I want to show that there are no simple groups of order $p^{k}(p+1)$ where $k&gt;0$ and $p$ is a prime number.  So suppose there is such a group. Then if we let $n_{p}$ denote the number of $p$-Sylow subgroups of $G$ we have that $n_{p}=p+1$. Now by letting $G$ act on $Sylow_{P}(G)$ by conjugation we obtain a group homomorphism $G \rightarrow S_{p+1}$. Since $G$ is simple then either $ker(f)$ is trivial or all $G$. Now here's my question: assume $ker(f)=G$ this would imply then that $G$ has a unique $p$-Sylow subgroup no? but then such subgroup is normal which contradicts the fact that $G$ is simple. So the map in fact is injective but then $|G|$ divides $(p+1)!$ which cannot be.  Basically my question is if my argument is correct, namely thta if $ker(f)=g$ implies the existence of a unique $p$-Sylow subgroup which implies such subgroup is normal in $G$ which cannot be. In case this is wrong, how do you argue that $ker(f)$ cannot be all $G$?  Thanks