How is $\mathbb R^n$ a quotient group of $E(n)$ by $SO(n)$ for any $n$. I am stuck at the following definition which our professor gave during a talk today.
He said that the Euclidean space $\mathbb R^n$ can be viewed as $\mathbb R^n=G/H$ where $G$ is the group of all rigid motions in $\mathbb R^n$ and $H=SO_n$ where $SO_n$ denotes the set of all orthogonal transformations.
I don't understand how did he make this statement.
I know the following definitions:

*

*The set of all rigid motions of $\mathbb R^n$ comprise of translations, rotations and reflections and they form a group which is known as the Euclidean motion group $E(n)$.

*If we consider only the rotations and reflections of the Euclidean space $\mathbb R^n$ then they form a  group which is known as orthogonal transformations denoted by $O(n)$.
However if we consider only the rotations then the group is denoted by $SO(n)$.

My questions are as follows:
If I suppose that my professor denoted $E(n)$ by $G$
How do we know that $\mathbb R^n=G/H$?
I have been scratching my head and searching various articles like this https://www.sciencedirect.com/science/article/pii/S092465090870062X
but I am unable to crack this part.
Can someone please give a step by step explanation of how $\mathbb R^n$ can be viewed as a quotient group of $E(n)$ by $SO(n)$ for any $n$.
 A: I agree with the comment that it should probably be $\operatorname{E}(n)/\operatorname{O}(n)$. $\operatorname{SO}(n)$ is not the stabilizer of any point in $\mathbb R^n$, so we can't find a bijection to $\mathbb R^n$ via group actions. It also isn't a normal subgroup of $\operatorname{E}(n)$, so we also can't treat the problem via factor groups. But the situation looks different for $\operatorname{O}(n)$. There we have two approaches.
First, consider the homomorphism
$$\varphi:\operatorname{E}(n)\to\mathbb R^n,~f\mapsto f(0),$$
where $\mathbb R^n$ is the additive group. The kernel is clearly $\operatorname{O}(n)$, and it is also surjective, so the first isomorphism theorem states $\operatorname{E}(n)/\operatorname{O}(n)\cong\mathbb R^n$ as groups.
Second, $\operatorname{O}(n)$ is the stabilizer of the origin under the standard group action on $\mathbb R^n$, and the orbit of the origin is all of $\mathbb R^n$, so we have a natural bijection $\operatorname{E}(n)/\operatorname{O}(n)\to\mathbb R^n$.
A: Based on what you write I think it may be a problem of notation. You say that $\mathrm{SO}(n)$ denotes the group of all orthogonal transformations. The usual notation for the group of orthogonal transformations is $\mathrm{O}(n)$, whereas the group of orthogonal transformations with positive determinant is denoted by $\mathrm{SO}(n)$ (it is the connected component of the identity in $\mathrm{O}(n)$, its elements are sometimes called proper rotations).
You write that if we consider rotations and reflections of $\mathbb R^n$ they form the group $\mathrm{O}(n)$, and if we consider only the rotations then we get $\mathrm{SO}(n)$. Actually every orthogonal transformation can be expressed as a product of reflections, see the Cartan-Dieudonné theorem.
Now $E(n)$ acts on $\mathbb R^n$ transitively, and the subgroup fixing the origin is the orthogonal group $\mathrm{O}(n)$ (reflections and rotations about the origin, non-trivial translations clearly move the origin somewhere else). Thus $\mathbb R^n$ has a unique smooth structure making it diffeomorphic to $E(n)/\mathrm{O}(n)$, see here.
A: As I said in my comment, I think the accurate theorem is:
$$ \frac{E(n)}{O_n} \cong \mathbb{R}^n$$
The key point is that $O_n$ is the set of all rigid motions which fix the origin. I'll assume this is obvious.
$E(n) / O_n$ is the set of cosets of the group $O_n$ in $E(n)$, namely the elements $eO_n$ for $e \in E(n)$, with the operation $(e O_n)(fO_n) = (ef)O_n$. For this operation to be well defined, we require that if $ef^{-1}\in O_n$, then $eO_n = fO_n$.
So let $e$ be any rigid motion. Consider the transformation $e': x \to e(x) - e(0)$. Clearly, $e'(0) = 0$. Since $e'$ is a composition of rigid motions, it is a rigid motion too. By the "obvious fact", $e'=e$ if and only if $e \in O_n$. As a consequence, $eO_n = e(0)O_n$ for every $e$ (where by $e(0)$ here I mean the translation $x \to x + e(0)$). The group of such $e(0)$ is clearly $(\mathbb{R}^n, +)$.
