# If you equip two isomorphic groups with homeomorphic topologies, are they isomorphic as topological groups?

I'm wondering if anyone has any insight regarding the truth of the above statement. Intuitively, if I have two topological groups in which their algebraic group structures are the same up to relabelling, and toplogical spaces that behave the same, it seems as though as topological groups they would have the same structure and behaviours, up to relabelling of course. Or is there an obvious counterexample that I'm missing?

I've seen a similar question posted with a counterexample, however I believe that the counterexample proposed did not accurately satisfy the hypothesis.

• For this to be true you would have to find a general way to functorially turn a map of the underlying topological spaces into a continuous group homomorphism for all topological groups which is an (I'm very strongly suspecting) impossible thing to solve. You also don't require that the homeomorphism of the underlying spaces has any relation to the group structure. Mar 16, 2020 at 23:26
• I'm not sure if this is the "similar question" you're alluding to, but the answer is no. A counterexample is provided here: mathoverflow.net/q/44060/145247 Mar 16, 2020 at 23:32
• @HallaSurvivor: good link: I find the example involving $\Bbb{Q} + \pi\Bbb{Q}$ and $\Bbb{Q} + \sqrt{5}\Bbb{Q}$ given there very appealing. Mar 17, 2020 at 0:45
• Does this answer your question? Diffeomorphic, group-isomorphic Lie groups that are not isomorphic as Lie groups Mar 17, 2020 at 3:02
• @MoisheKohan and other close-voters: I don’t think that question about Lie groups is a duplicate, even though purely technically its answer does answer this one. In particular, the question and answers there both require more background in topology/manifolds to understand; the OP here, and other students finding this question later, may not have that background. Mar 17, 2020 at 8:55

If $$G$$ is a finite topological group and $$N$$ is the connected component of the identity, then $$N$$ is normal and the coset space $$G/N$$ forms a basis for the topology. Conversely, one can create any finite topological group given a choice of finite group $$G$$ and normal subgroup $$N$$ to be the connected component. (See this question for the result.)

Thus, if we can find a finite group $$G$$ with two normal subgroups $$N_1$$ and $$N_2$$ that are not related by any automorphism but are nonetheless the same size, we can have $$(G,\tau_1)$$ and $$(G,\tau_2)$$ homeomorphic, but a continuous isomorphism would have to preserve identity's connected component, i.e. send $$N_1$$ to $$N_2$$, which is impossible, and we would have a counterexample.

For this, we can pick $$N_1$$ and $$N_2$$ to simply be nonisomorphic. For instance, if $$G=\mathbb{Z}_2\times\mathbb{Z}_4$$ then we can use the subgroups $$N_1=\mathbb{Z}_2\times\mathbb{Z}_2$$ and $$N_2=\mathbb{Z}_4$$.

The excellent answer by @runway64 gives a counterexample; I’d like to say a little more about the intuitions you sketch in the question.

Intuitively, if I have two topological groups in which their algebraic group structures are the same up to relabelling, and toplogical spaces that behave the same, it seems as though as topological groups they would have the same structure and behaviours, up to relabelling of course.

With a counterexample $$G_1, G_2$$, we know that their group structure is “the same up to relabelling”, and likewise their topological structure is “the same up to relabelling”. But the relabellings may be different! That is, we know there’s a group isomorphism $$g : G_1 \to G_2$$, and a homoeomorphism $$h : G_1 \to G_2$$. But as functions, $$g$$ and $$h$$ may be different, and there may be no function $$G_1 \to G_2$$ that is both a group isomorphism and a homeomorphism at the same time — as would be needed for them to be isomorphic topological groups.

Similarly, thinking about their behaviours: We know from the group isomorphism that they will have the same purely group-theoretic properties, and from the homeomorphism that they’ll have the same purely topological properties. But they may differ with properties involving the interaction of the group structure and topology: for instance, in @runway44’s counterexample, the property “the connected component of the identity is cyclic” holds in one of the two groups but not the other.

Here's an example in the realm of connected Lie groups.

For $$d\ge 7$$ there exists a 1-parameter family $$G_t$$ (which is explicit) of complex nilpotent Lie groups, which are pairwise non-isomorphic as topological groups (= as real Lie groups), except $$G_{\bar{t}}\simeq G_t$$. But they're all homeomorphic (they're all biholomorphic to $$\mathbf{C}^d$$, hence homeomorphic to $$\mathbf{R}^{2d}$$). And for conjugate $$t,u$$ under $$\mathrm{Aut}_{\mathrm{field}}(\mathbf{C})$$ (e.g., $$t,u$$ both transcendental), $$G_t$$ and $$G_u$$ are isomorphic as groups.