Topological group homeomorphic

Question

Exists a topological group homeomorphic to $(0, 1)$? and to $[0, 1]$? I believe that with includes the extremes is it, to contrary than when they aren't included. I tried to use the definition of topological group, but I'm not sure how to distinguish the two cases because of the extremes.

Answer 1

$[0,1]$ is not homeomorphic to a topological group. To see why, note that the left multiplication action of a topological group on itself is transitive, and it follows that the complements of any two points are homeomorphic. But the complement of each endpoint of $[0,1]$ is connected whereas the complement of each interior point is disconnected.

As said in the comment of @PatrickNicodemus, $(0,1)$ is homeomorphic to the topological group $\mathbb R$.

Answer 2

As mentioned by @Patrick, $(0,1)$ is homeomorphic to ${\mathbb R}$.

However $[0, 1]$ is not homeomorphic to any topological group because every topological group $G$ is homogeneous, that is, for any two points $g$ and $h$ in $G$, there is a homeomorphism of $G$ sending $g$ to $h$, namely left-multiplication by $hg^{-1}$.

However there is no homeomorphism from $[0,1]$ to itself sending $0$ to $1/2$.