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?



Does homeomorphic and isomorphic always imply homeomorphically isomorphic?
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.




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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.


