Is it true that $R^2 = E(2)/U(1)$? (Just so we're clear: that the Lie group of planar translations $R^2$ is isomorphic to a quotient of the 2D Euclidean Lie group $E(2)$ and the circle group $U(1)$.) 
I am trying to prove that $R^2 = E(2)/U(1)$ directly but not getting very far. It seems as if it should be true, and trivially so, but I am new to Lie groups and am not quite sure what implies what yet. For instance, it is known that $E(2)$ is the semidirect product of $R^2$ and $U(1)$ (or $O(2)$ if you like); and I reckon that implies that E(2) is a principal fibre bundle with base $R^2$ and fibre $U(1)$, is any of that true? And does it imply there is a quotient relationship?
If so - can you give explicit homomorphisms $f$ and $g$ such that the following sequence is exact:
$1\rightarrow U(1)\xrightarrow f E(2)\xrightarrow g R^2\rightarrow1$?
(i.e. Im $f$ = Ker $g$) 
I thought that $f$ is the map that takes each element of $U(1)$ to the element of $E(2)$ with the same rotation but $\underline{0}$ translation - i.e. if a general element of E(2) is expressed as $(M,\underline{v})$ for a rotation matrix $M$ and translation vector $\underline{v}$ [with $(M,\underline{a}).(N,\underline{b})=(MN,\underline{a}+M\underline{b})]$ then $f$ embeds $U(1)$ in E(2) as
$f:R\mapsto (R, \underline{0})$
for $R\in U(1)$, $R$ a rotation matrix. But presumably that would mean that $g$ would map the whole of $E(2)$ to $R^2$ by taking $(R,\underline{v})\mapsto (0,\underline{v})$ (Thaking Im $F$, the pure rotations, to the identity element $e_{R^2}=\underline{0}$ - unfortunately, this doesn't appear to be a homomorhpism.
 A: The group of distance preserving applications from the Euclidean plane to itself is exactly the group of applications of the form $x \mapsto Ax + b$ with $A \in O(2)$ and $b \in \mathbb{R}^2$. So you should have $R^2 \simeq E(2)/O(2)$ not $E(2)/U(1)$ where $R^2$ is the translation group. The problem with only quotienting by $U(1) \simeq SO(2)$ is that you still have the reflections in your group.
A: Your main problem is that in the semidirect product $E(2)=\mathbf R^2\rtimes O(2)$, you are trying to quotient by the non-normal subgroup $O(2)$ rather than by the normal subgroup $\mathbf R^2$. The subgroup $O(2)$ is the stabiliser of the origin $0$; its conjugate by $g\in E(2)$ is the stabiliser of $g(0)$, which is a different subgroup whenever $g(0)\neq0$, so it really is a non-normal subgroup. Any attempt to understand the nonexistent quotient $E(2)/O(2)$ is doomed to fail.
What does hold, and is easy to understand, is $O(2)\cong E(2)/\mathbf R^2$, through the usual map that associates to an affine transformation a linear transformation (its "differential") of the (tangent) vector space associated to the affine space.
