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I wish to thank Professor Claudio Gorodski for his very helpful answers to my question on the webcite: If compact connected Lie groups are homeomorphic as topological space, are they isomorphic as Lie groups?

He said: Let $G_{1}$ and $G_{2}$ be two compact, connected Lie groups with isomorphic homotopy groups in each dimension. Then their Lie algebras are isomorphic.

Now my question is: If $G_{1}$ and $G_{2}$ are two compact, connected topological groups which are homeomorphic as topological space, are there any isomorphism theorems?

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Haven't we had this one before? I think the $p$-adic integers (for different $p$) was the example given -- all of these spaces are homeomorphic IIRC but for different $p$ they're not at all isomorphic as topological groups. I'm going from memory here so apologies if I'm off base. – Kevin Buzzard Jun 7 '11 at 6:59
.. or even as abstract groups. – algori Jun 7 '11 at 7:01
What is IIRC? Is your example a connected topological space? – sife Jun 7 '11 at 7:42
sife -- iirc means "if I remember correctly". iirc, that is.. – algori Jun 7 '11 at 9:23
@Kevin - This question?… – Steven Gubkin Jun 7 '11 at 14:05

If you replace homeomorphic with homotopy equivalent the answer is no. There are infinitely-many non-isomorphic topological groups which are homotopy equivalent (just as spaces) to $S^3$. Actually, these topological groups cannot be connected by a zig-zag of group homomorphisms which are homotopy equivalences. But of course all of them have the same homotopy groups.

Rector, David L. Loop structures on the homotopy type of S3. Symposium on Algebraic Topology (Battelle Seattle Res. Center, Seattle, Wash., 1971), pp. 99–105. Lecture Notes in Math., Vol. 249, Springer, Berlin, 1971.

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Fernando Muro--Are the examples mentioned above compact, connected topological groups? If they are compact, connected topological space, then they must be homeomorphic to $S^{3}$ by Poincaré conjecture . Therefore we need not replace homeomorphic with homotopy equivalent. – sife Jun 7 '11 at 16:00
They're not compact, but even if they were, Poincaré's conjecture would not apply since it is for manifolds. – Fernando Muro Jun 7 '11 at 18:15

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