An example of a Hausdorff topological group which is not normal

Question

I know the definitions of Hausdorff topological group. Can you give me an example of Hausdorff topological group which is not normal. (I know $\mathbb{Z}^\mathbb{R}$ but I can't prove that:(.)

Answer 1

Let $X$ be any space which is completely regular but not normal, then the free topological group $F(X)$ over $X$ provides an example of a topological group which is Hausdorff, completely regular and not normal.

By definition the free topological group over $X$ is a topological group $F(X)$ together with a continuous map $\sigma\colon X\to F(X)$ with the following universal property: for every continuous function $f\colon X\to G$ into a topological group $G$ there is a continuous group homomorphism $\tilde f\colon F(X)\to G$ such that $\tilde f\circ \sigma=f$.

By standard results when $X$ is completely regular the map $\sigma\colon X\to F(X)$ is a topological embedding, indeed the definition above turns out to be equivalent to letting $F(X)$ be the (algebraic) free group over $X$, equipped with the finest possible group topology in which $\sigma$ is a topological embedding. Moreover the image $\sigma(X)$ is a closed subspace of $F(X)$ (for a reference see Chapter 7.1 of Topological Groups and Related Structures by Arhangel'skii and Tkachenko). This immediately implies that if $X$ is completely regular but not normal, then $F(X)$ is not normal either, indeed if it were then $\sigma(X)$, which is homeomorphic to $X$, would have to be normal too, being a closed subspace of a normal space.