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Let $(G,\cdot,T)$ and $(H,\star,S)$ be topological groups such that
$(G,T)$ is homeomorphic to $(H,S)$ and $(G,\cdot)$ is isomorphic to $(H,\star)$.

Does it follow that $(G,\cdot,T)$ and $(H,\star,S)$ are isomorphic as topological groups?
If no, what if they are both Hausdorff? What if they are both Hausdorff and two-sided complete?

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Like your previous question… this is very awkwardly written. Nobody says "homeomorphically isomorphic", but rather "isomorphic as topological groups". A better way of expressing yourself would be: "Is there an example of non-isomorphic topological groups which are isomorphic as topological spaces and isomorphic as groups?" – KConrad Oct 29 '10 at 6:00
I didn't mean only the phrase "homeomorphically isomorphic" is awkward, but the way of talking about topological groups as ordered triples is awkward. What you wrote is logically correct, but is too pedantic. Rudin has a comment along these lines about writing measure spaces as ordered triples (4-tuples?) in the first chapter of his Real and Complex Analysis. Admittedly your question is precisely about this kind of pedanticness, but it's still better here to use words and talk about the sense in which objects are isomorphic ("as topological spaces", etc.). – KConrad Oct 29 '10 at 6:33
As for different wording in the title: "Does group isom. and homeomorphism imply topological group isom.?" – KConrad Oct 29 '10 at 6:33
@KConrad: Rather "group isomorphic + homeomorphic => topological group isomorphic" because we are talking about objects and not about morphisms. – Martin Brandenburg Oct 29 '10 at 6:51
Anyway "homeomorphically isomorphic" is pretty clear and not ambiguous. Sometimes it's better to use a clear wording than the usual one. Here I wouldn't say that "isomorphic as topological groups" is bad. Also to view a topological group as a triple (set, group law, topology) is just a point of view and I wouldn't call it pedantic. (wow I just realize that the thread is 3 years old... still I post the comment) – YCor Feb 21 '13 at 16:55
up vote 103 down vote accepted

The 2-adic rationals $\mathbb{Q}_2$ and the 3-adic rationals $\mathbb{Q}_3$ are homeomorphic, because each one is a countable disjoint union of Cantor sets. They are also isomorphic as groups if you assume the axiom of choice, because they are both fields of characteristic 0 and therefore vector spaces over $\mathbb{Q}$ (of the same cardinal dimension). However, the 2-adic integers $\mathbb{Z}_2$ are a compact subgroup of $\mathbb{Q}_2$ in which every element is infinitely divisible by 3. On the other hand, in $\mathbb{Q}_3$, any non-trivial sequence $x, x/3, x/9, \ldots$ is unbounded in the complete metric, and is therefore not contained in a compact subgroup.

Keith Conrad asks whether these is an example without the axiom of choice, and Jason De Vito asks whether there is an example using Lie groups. In fact, there is a cheap example using disconnected Lie groups. Let $G$ and $H$ be two connected Lie groups that are homeomorphic but not isomorphic. For instance, abelian $\mathbb{R}^3$, the universal cover $\widetilde{\text{SL}(2,\mathbb{R})}$, and the Heisenberg group of upper unitriangular, real $3 \times 3$ matrices are all homeomorphic, but not isomorphic. If $G'$ and $H'$ are $G$ and $H$ with the discrete topology, then $G' \times H$ and $G \times H'$ are explicitly isomorphic and explicitly homeomorphic. But they are not continuously isomorphic, because the connected component of the identity is $G$ for one of them but $H$ for the other one.

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Nice example ! – Martin Brandenburg Oct 29 '10 at 6:49
Also, thumbs up for not ignoring the use of AC. – Ricky Demer Oct 29 '10 at 6:57
I do like this example, but it raises the question of whether one can give an example without relying on AC. – KConrad Oct 29 '10 at 15:13
Greg, thanks for the update in the answer. I was hoping there could be an example cooked up using two non-isom. top. groups whose underlying topological spaces are homeomorphic, such as R*xR and the ax+b group over R. You showed how to do it. – KConrad Oct 31 '10 at 21:10

Sorry for the necromancy, but the following was too cute to resist:


$A$ be $Z_4$ with the discrete topology

$B$ be $Z_4$ with the indiscrete topology

$C$ be $Z_2 \times Z_2$ with the discrete topology

$D$ be $Z_2 \times Z_2$ with the indiscrete topology.

Then $A \times D$ is not isomorphic to $B \times C$ as a topological group, but the the underlying spaces are homeomorphic and the groups are isomorphic.

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Is this counterexample minimal? Are there a pair of topological groups with cardinality less than 16 which are homeomorphic, isomorphic, but not isomorphic as topological groups? I guess this is a small enough problem that I could just do an exhaustive search. – Steven Gubkin May 19 '11 at 14:08

Another example: there are uncountably many different (count-ably based) abelian pro-$p$ groups isomorphic to the product of all the cyclic $p$-groups. Each of these must be homeomorphic to the Cantor set.

[This is in my transfer report, and shortly to be in a pre-print on the arxiv.]

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This is interesting because it would give you compact examples. – Greg Kuperberg Oct 29 '10 at 17:48
@Jonathan: what do the Pontrjagin duals of these groups look like? – Kevin Buzzard Oct 29 '10 at 19:51
@Kevin: Essentially, any countable p-group $G$ such that the quotient of $G$ over its elements of infinite height is isomorphic to the direct sum (i.e. restricted product) of all the cyclic $p$-groups. The isomorphism of groups is not at all obvious: indeed, it depends on the axiom of choice. – Jonathan Kiehlmann Oct 29 '10 at 22:28

The Banach spaces $c_{0}$ and $\ell_{2}$ may be viewed as isomorphic Abelian groups, that are also homeomorphic (due to Kadec). Still, they are clearly not isomorphic as topological groups.

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Kadec proved that all separable infinite dimensional Banach spaces are homeomorphic and all are clearly isomorphic to one another but as topological groups there are a continuum of different ones. – Bill Johnson Oct 30 '10 at 15:49
A nice example. But again in this case, the isomorphism as groups requires AC for proof. – Gerald Edgar Feb 22 '13 at 15:27

Any two $2$-dimensional rational vector spaces in $\mathbb R$ are isomorphic as groups and also order-isomorphic (I believe), hence homeomorphic when given the relative topology from $\mathbb R$. But any isomorphism as topological groups would have to be given by multiplication by a real number. So for example $\mathbb Q+\mathbb Q\pi$ and $\mathbb Q+\mathbb Q\sqrt 5$ are a counterexample.

EDIT: Or the same thing with $\mathbb Z$ instead of $\mathbb Q$.

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Cute. And yes, any two countable dense linear orders without endpoints are order-isomorphic. – Todd Trimble Feb 22 '13 at 12:46
Yes, and another point that could be overlooked: order topology and subspace topology do not always coincide for subsets of the line, but they do when the set is dense in the line. – Tom Goodwillie Feb 22 '13 at 17:14
(When I say "could be overlooked" I mean "might be easy to overlook".) – Tom Goodwillie Feb 22 '13 at 17:15

I see above counter-examples in this thread. On the other hand, in the Abelian case there are positive results, which go way further than the question. In the case of (general) solenoids, they themselves are their own first homology groups with coefficients in $S^1$. It follows that when two of them are homeomorphic (or just homologically equivalent) then they are isomorphic as topological groups--see Karol Borsuk and myself, Fund. Math. 1970. This line was developed much further in later papers by James Keesling.

The following five classifications of (general) solenoids are identical: (i) homological (with coefficients in $S^1$; i.e. cohomological with coefficients in $Z$; for Čech theories), (ii) shape-theoretical, (iii) homotopic, (iv) topological, (v) as topological groups.

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