TITLE: If you equip two isomorphic groups with homeomorphic topologies, are they isomorphic as topological groups?
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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.