A topological group is a set G with a group structure and a topology such that the binary law

$G \times G \rightarrow G ,\;\; (x,y) \rightarrow xy,$
and the map of the inverse
$G \rightarrow G ,\;\; x \rightarrow x^{-1},$
are continuous. In other words both structures, the algebraic and the topological one, are linked in a natural way. So the following question is of interest.

Question

Are there topological groups which are isomorphic as groups and homeomorphic as topological spaces but not isomorphic as topological groups?

Answer

Yes. The proof goes in three stages:

1. Let G and H be dense subsets of $\mathbb{R}$ and $f :\, G \rightarrow H$ a monotonic bijection. Then f is a homeomorphism.

Without restriction of generality we can suppose that f is increasing. We prove that it is continuous. Let $x_0\in G$ and $\epsilon>0$. Since H is dense in $\mathbb{R}$ we have $H\cap\,]f(x_0)-\epsilon,f(x_0)[\,\neq\emptyset$. Thus there exists $y_1\in H\cap\,]f(x_0)-\epsilon,f(x_0)[\,$. Similarly there exists $y_2\in H\cap\,]f(x_0),f(x_0)+\epsilon[\,$. Since f is onto we can write $y_k=f(x_k)$ with $x_k\in G, k=1,2$. Let $\delta=\min(x_0-x_1,x_2-x_0)$. Then for all x in G

\begin{align*}x_0-\delta

which proves that f is continuous in $x_0$. The proof of the continuity of the inverse function $f^{-1}$ is the same.

2. Let G and H be countable and dense subsets of $\mathbb{R}$. Then they are homeomorphic.

First we write

\begin{align*} G&=\{x_0,x_1,x_2,\ldots\}\;\;\;\;\;(*)\,,&H&=\{y_0,y_1,y_2,\ldots\}\;\;\;\;(**)\,. \end{align*}

We will now list G and H in a new way, $G=\{x'_0,x'_1,x'_2,\ldots\}$ and $H=\{y'_0,y'_1,y'_2,\ldots\}.$ The aim is to make $G \to H, x'_k \mapsto y'_k,$ a monotonic bijection (then it is automatically a homeomorphism). For this we proceed as follows.

• k=0. We take $x'_0=x_0,\;y'_0=y_0$

• k=1. We take $x'_1=x_1$. To choose $y'_1$ we look at the order of $x'_0$ and $x_1'$.
If $x'_1 then we take as $y'_1$ any element in H which is smaller than $y'_0$.
If $x'_1>x'_0$ then we take as $y'_1$ any element in H which is greater than $y'_0$.

• k=2. We take as $y'_2$ the first element of $H\setminus\{y'_0,y'_1\}$ listed in (**). To choose $x'_2$ we look at the order of $y'_0,y'_1,y'_2$.
If $y'_2$ is smaller than $y'_0$ and $y'_1$ we take as $x'_2$ any element in G which is smaller than $x'_0$ and $x'_1$.
If $y'_2$ is greater than $y'_0$ and $y'_1$ we take as $x'_2$ any element in G which is greater than $x'_0$ and $x'_1$.
If $y'_2$ is between $y'_0$ and $y'_1$ we take as $x'_2$ any element in G which is between $x'_0$ and $x'_1$.

• k=3. We take as $x'_3$ the first element of $G\setminus\{x'_0,x'_1,x'_2\}$ listed in (*). To choose $y'_3$ we look at the order of $x'_0,x'_1,x'_2,x'_3$. There are 24 possible orders for these four numbers.
If $x'_3 we take as $y'_3$ any element in H which is smaller than $y'_0,y'_1,y'_2$.
If $x'_2 we take as $y'_3$ any element in H which is bewteen $y'_2$ and $y'_0$.
And so forth.

3. The topological groups $G=\mathbb{Q}+\mathbb{Q}\sqrt2$ and $H=\mathbb{Q}+\mathbb{Q}\sqrt3$ fulfill our wishes.

After what has been said before, G and H are homeomorphic as topological spaces. Clearly they are isomorphic groups. But they are not isomorphic topological groups. In fact, suppose that there exists an isomorphism of topological groups $f :\, G \to H$. One proves easily by induction that f(n)=nf(1) for every integer n, and then that f(r)=rf(1) for every rational r. Therefore by continuity

$f(\sqrt2)=\sqrt2f(1),\;\;\;\;\;\lightning$

a contradiction in $H=\mathbb{Q}+\mathbb{Q}\sqrt3$.