Non-isomorphic Group Structures on a Topological Group
I will restrict myself only to Lie groups, since the world of groups outside of this class is way too large and, I do not think, there is a good answer in this context. I will also restrict to simply-connected Lie groups so that the answer is reasonably neat. (This is not a very good reason, but this will keep my answer reasonably brief, I did not think through the details in the non-simply connected case which will have to require some hard case-by-case analysis: If you really want to find out the answer in this case, you would have to do it yourself or ask somebody else who has more time to spare.)
I will say that a Lie group $G$ is rigid if every Lie group homeomorphic to $G$ is in fact isomorphic to $G$.
Now, let's apply some structural results for Lie groups in this context. Every simply-connected Lie group will have the Levi-Malcev decomposition $$ G= S \rtimes H, $$ where $S$ is solvable and simply connected and $H$ is connected and has semisimple Lie algebra. Then $S$ is homeomorpbic to some ${\mathbb R}^n$ and, hence, if $n\ge 2$ then $S$ has non-unique Lie group structure (one can use either commutative or solvable noncommutative one). Since we can always use trivial action of $H$ on $S$ without changing topology of semidirect product, it follows that $G$ cannot be rigid if $n\ge 2$. If $n=1$ then $H$ has to act trivially on $S$ (as $H$ is connected, semisimple and, hence, has only trivial characters) and, thus, $G\cong {\mathbb R}\times H$. Therefore, in both cases $n=0, n=1$ the problem reduces to rigidity of the groups $H$.
Case 1. $H$ is compact. I then found by browsing this mathoverflow post that $H$ has to be rigid. By compactness of $H$ it is also clear that $G$ is rigid as well in this case.
Case 2. $H$ is noncompact. If $H$ happens to have finite center then it has the Iwasawa decomposition $H= U\cdot K$, where $K$ is maximal compact subgroup in $H$ and $U$ is solvable and homeomorphic to $R^m$ for some $m\ge 2$; in particular, $H$ is homeomorphic to ${\mathbb R}^m\times K$. (Algebraically, this is never a product!) Clearly, we can also endow ${\mathbb R}^m\times K$ with the algebraic structure of the direct product, hence, $H$ cannot be rigid; thus, $G$ is not rigid either. This leaves us with the case of infinite discrete center. (The standard example to think about is the universal cover of $SL(2, {\mathbb R})$.) However, every such $H$ can be divided by a discrete central subgroup $Z<H$, such that $Q=H/Z$ has trivial center. Then we can apply the same reasoning as above to conclude that $Q$ is non-rigid and then lift the modified Lie group structure to the covering space $H$ (of $Q$) to conclude that $H$ is again non-rigid.
Thus, the conclusion is that a simply-connected Lie group $G$ is rigid if and only if it is either compact or a product of a compact group and ${\mathbb R}$.
This isn't really an answer, but it's too long for a comment. In any case, I think the answer could depend on your formulation. I can see two ways to interpret your question. Suppose $X$ and $Y$ are topological groups, homeomorphic as topological spaces. Are they necessarily isomorphic as
- abstract groups?
- topological groups?
Obviously 2 implies 1. I think the second question is the more interesting one. I don't know if this fact will lead to an answer, but all separable (infinite dimensional) Banach spaces are homeomorphic as topological spaces. They are isomorphic as real vector spaces for cardinality reasons I believe, thus as abstact groups, but I don't think they are isomorphic as topological groups (this is just my guess.)
One further remark is that your remark on discrete topological groups carries over to discrete topological groups of any cardinality. So $\Bbb Z$ and $\Bbb Q$, both endowed with the discrete topology, are isomorphic as topological spaces, but aren't isomorphic as groups. Maybe this is an indication that the good context for your question isn't abstract topological groups but connected topological groups, or even connected Lie groups.