A pair of non group homeomorphic topological groups The following assertion is true:
If $A$ and $B$ are dense group homeomorphic subgroups of complete Hausdorff topological groups $X$ and $Y$ respectively, then $X$ and $Y$ are group homeomorphic.
I wonder if we can find $A$ and $B$ such that $A$ is homeomorphic to $B$, but not group homomorphic and that $A$ and $B$ are dense in topological groups $X$ and $Y$ respectively, but $X$ and $Y$ are not homeomorphic.
Then what about $X$ and $Y$ are homeomorphic but not group homomorphic?
The the topological groups are assumed to be commutative, by group homeomorphic we mean that (topological) homeomorphic that is also (group) homomorphic.
 A: Sure, there's lots of examples.


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*$A$, $B$ homeomorphic, not group homeomorphic, and $X, Y$ not homeomorphic: Take $X=\mathbb{R}$, $Y=\mathbb{R}^2$ with the usual topologies and group structures. Let $A=\mathbb{Q}\subseteq X$ and $B=\mathbb{Q}^2\subseteq Y$. Then $A$ and $B$ are dense in $X$ and $Y$ respectively, are topological subgroups, and are homeomorphic (this might be a bit surprising, but it's true, and a good exercise) - but clearly $X$ and $Y$ are not homeomorphic!

*$A$, $B$ homeomorphic, $X$, $Y$ homeomorphic, but $X$, $Y$ not group homeomorphic: Consider the sets $2^\omega$ and $3^\omega$ of infinite strings from $\{0, 1\}$ and infinite strings from $\{0, 1, 2\}$ respectively. These sets carry natural group structures - the $\omega$-fold products of $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$, respectively. They also have a natural topological structure (the product topology, with the discrete topology on each fact), and form topological groups. As spaces, they are homeomorphic, but obviously they are not group isomorphic; these will be our $A$ and $B$. Now let $X$ and $Y$ be arbitrary countable dense subgroups of $A$ and $B$, respectively. We will have $X$ and $Y$ homeomorphic as topological spaces.
