Let G be a subgroup of $E(n) = \mathbb{R}^n \rtimes O(n)$(the rigid motions of $\mathbb{R}^n$ ) with orbit space as a point.


Example: the group of all translations of $\mathbb{R}^n$ and of course any group containing it.


Q:
1.Are there other examples of G not containing all the translations? If such G exist , How many different types are them?


2 .What about the general propositions of these groups?



Subgroups of $E(n) = \mathbb{R}^n \rtimes O(n)$ with trivial orbit space
In the simply-transitive case, there is the following result. 


**Theorem**. Let $G$ be a simply-transitive group of affine rigid motions acting on a finite dimensional Euclidean space $V$. Then $G$ is a connected solvable group, and there exists an orthogonal decomposition $V=U\oplus W$, where $U\neq0$, and a homomorphism $\varphi:V\to \mathfrak{so}(V)$ such that $\ker\varphi=U$, 
$\mathrm{im}\,\varphi$ is an Abelian subalgebra of $\mathfrak{so}(V)$
annihilating $W$ and preserving $U$, and 
\begin{equation}
 \mathfrak g=\{\tau\_v+\varphi(v):v\in V\}, 
\end{equation}
where $\tau\_v:V\to V$ denotes translation by $v$. 


*Proof.* $G$ is diffeomorphic to $V$, thus connected and contractible. 
By The Levi-Malcev theorem, it is solvable. 


Consider the homomorphism
$\pi:G\to SO(V)$ given by $\pi(g)=g\_\*$. 
Then $\mathfrak n=\ker d\pi$ is an 
ideal of $\mathfrak g$ and $\pi(G)$ is a solvable subgroup of $SO(V)$. 
The closure $\overline{\pi(G)}$ in $SO(V)$ 
is a compact solvable group, hence, Abelian. It follows
that $\pi(G)$ is Abelian. 


The map $\theta:\mathfrak g\to V$, $\theta(X)=X\cdot 0$, is an
isomorphism of vector spaces. Let $U=\theta(\mathfrak n)$ and $W=U^\perp$.
If $U=0$, then $\mathfrak n=0$ and $d\pi[\mathfrak g]$ is an Abelian subalgebra
of $\mathfrak{so}(V)$ of dimension $\dim\mathfrak g=\dim V$. However, 
the rank of $\mathfrak{so}(V)$ is at most $\frac12\dim V$, so this 
cannot be. It follows that $U\neq0$.


Set $\varphi=d\pi\circ\theta^{-1}$. Note that $\varphi$ is a 
homomorphism between the Abelian Lie algebras $V$ and $d\pi[\mathfrak g]$.
It is also clear that $\mathfrak g$ has the required form. 
Of course, $\ker\varphi=\theta(\ker d\pi)=\theta(\mathfrak n)=U$. 


Let $u\in U$, $w\in W$. Then
$$[\tau\_w+\varphi(w),\tau\_u]=\tau\_{\varphi(w)u} \in\mathfrak g, $$
so $\varphi(w)u\in \ker\varphi=U$. This shows that $\mathrm{im}\,\varphi$ preserves
$U$ and, hence, the orthogonal decomposition $ V=U\oplus W$.
Finally, if $w$, $w'\in W$, then $[\varphi(w),\varphi(w')]=0$, and so
$$ [\tau\_w+\varphi(w),\tau\_{w'}+\varphi(w')]=\tau\_{\varphi(w)w'}
-\tau\_{\varphi(w')w}=\tau\_{\varphi(w)w'-\varphi(w')w} \in\mathfrak g. $$
This shows that $\varphi(w)w'-\varphi(w')w\in W\cap U=0$. Now 
the trilinear form on $W$
$$ (w,w',w'')\mapsto \langle\varphi(w)w',w''\rangle $$
is symmetric in the first two variables and skew-symmetric in the last two 
variables, hence, identically zero. This proves that $\varphi(W)W=0$
and completes the proof of the theorem. q.e.d.


