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The construction in the following uses a non-continuous additive map R->R whose graph is connected to build the subgroup described in its title. Ryuji Maehara, On a connected dense proper subgroup of R^2 whose complement is dense. Proc. AMS, Vol 97, no. 3, July 1986 For further reading, here's another interesting take. The construction is of a connected ...


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As it was mentioned in the comments, the pseudo-arc and the Hilbert cube have the fixed point property so they cannot be homeomorphic to a topological group. On the other hand it was proved by G.S. Ungar in "On all kinds of homogeneous spaces" (TAMS, 1975), that any homogeneous compact metric space is homeomorphic to a coset space. In particular this is ...


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This is a longish comment on James Schwass's answer, not an answer to the original questions. Have to be a little careful here. We are deliberately informal (p.97), but we are working in a category of "cocellular" spaces and maps on which localizations can be constructed to be product-preserving on the point-set level. Every nilpotent space $X$ ...


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In the paper Arkowitz, Martin Categories equivalent to the category of rational H-spaces, Manuscripta Math. 64 (1989), no. 4, 419–429 it is shown that the rational homotopy equivalence $G_\mathbb{Q}\cong\prod_n K(\pi_n(G_\mathbb{Q}),n)$ is an equivalence in the category of rational $H$-spaces if and only if $G$ is homotopy abelian (see Proposition 3.1). ...


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A partial answer to your first question can be found in the More Concise Course in Algebraic Topology by Ponto and May. There you can see (pg. 97) that if $X$ admits a "good Postnikov tower" (which is the case when $X$ is simply connected, or more generally nilpotent) then the rationalization $(X\times X)_{(0)}$ of $X\times X$ is naturally homotopy ...


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For a negative answer to the first question: In a topological group discrete (and more generally locally compact, or even locally closed) subgroups are closed. So in a compact group the discrete subgroups are exactly the finite subgroups. Taking a compact group with a discontinuous group automorphism (example: the unit circle, using choice to decompose it in ...


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See Lian B. Smythe's paper on arXiv.org (2011).


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It seems the answer is no. Basically, the $p$-adic and Archimedean topologies on $\mathbb{Q}$ are incompatible enough that the maximal compatible topology contained in both of them is indiscrete. Here is an outline of the proof (I'm heading to bed so I haven't filled in the details): $\bullet$ Any subset of $\mathbb{Q}$ which is open for both topologies ...


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Edit: This answer is completely wrong. Please ignore it, and see Kevin's answer. A $T_{0}$-topological group $G$ is Hausdorff ($T_{2}$). Recall that $T_{0}$ means that for all $x,y \in G$, there is an open subset $U$ such that $x \in U, y \notin U$ or $y \in U, x \notin U$. Consequently (by homogeneity), a topological group is Hausdorff if and only if ...



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