Difference between $\mathbb{R}^2$ and $SE(2)$ I would like to have a good explanation of which is the difference between the Euclidean Group $SE(2)$ and the Euclidean space $\mathbb{R}^2$. 
From what I understood in $SE(2)$ there is also a rotation but I don't understand if this is the only difference. 
EDIT: First, I want to know if is it correct that the $\mathbb{R}^2$ is a SPACE and the $SE(2)$ is a group. There is a difference between those two, right?
Then I would like to know if is it correct that $\mathbb{R}^2$ can be seen as a subset of $SE(2)$. Because from what I understood we can write :
$SE(2)= \mathbb{R}^2 \times S^1$ where $S^1$ is what? A space in which the only moevement is a rotation? 
Thanks for your time,
F.
 A: Edit To address the question's edit here are additional details to address the new format.


*

*Any set can be made into a space, a space is a set with other structure, like a topology on it. You can endow $E(2)$ with a topology to also make it into a topological space, or not, but this is not a fundamental difference with $\Bbb R^2$, since you can also choose not to endow $\Bbb R^2$ with other structure. If you want anything more specific, you have to be more specific than just "space."

*Yes, you can see $\Bbb R^2\subseteq E(2)$ The description below is from my original answer and tells you explicitly how to find the inclusion:

There is a clear difference between the two as groups. Let an arbitrary action of $E(2)$ acting on $\Bbb R^2$ be denoted by $(x,y)\mapsto (ax+by+c, dx+ey+f)$. Then $E(2)$ has a proper subgroup isomorphic to $\Bbb R^2$ when $a=e=1, b=d=0$ and we let $c,f$ be free parameters. Since this is also a subset of $SE(2)$ we get the result you ask for.



*

*$S^1$ is called the "circle group" and means all elements of $SE(2)$ where $a=e=\cos\theta$ and $-b=d=\sin\theta$ and $c=f=0$. It is called this because the set of complex numbers $S^1=\{z\in\Bbb C : |z|=1\}$ is a group under multiplication of complex numbers, since they are all complex numbers of the form $e^{i\theta}$. This set looks like a circle if you plot it, since $|z|=1\iff |z|^2=1\iff x^2+y^2=1$ where we write $z=x+iy$ with $x,y\in\Bbb R$. The group I describe explicitly with the sines and cosines is isomorphic to this group, that's where it's coming from. The notation $S^1$ means $1$-dimensional sphere, which is a circle, that's why the notation has an $S$ and a $1$.

