Set of rotations and translations in $\mathbb{R}^2$ is a normal subgroup of isometries group Let $\mathcal{M}$ be the group of motions in $\mathbb{R}^2$, and $\mathcal{M}_+$ the subgroup of direct (also called "rigid", I think) motions.
I'm trying to show that $$[\mathcal{M} : \mathcal{M}_+]=2$$ and to conclude that $$\mathcal{M}_+  \triangleleft \mathcal{M}.$$
I thought about using Lagrange's theorem for the first part, from where I know $$|\mathcal{M}|=[\mathcal{M}:\mathcal{M}_+]|\mathcal{M}_+|,$$ but I'm not quite sure how to use it in order to get the result I'm looking for.
For the other part, I don't really know how to proceed.
Any help is more than welcome, thanks in advance!
 A: $\mathcal{M}$ is the set of products of finitely many reflections about lines(not necessarily through the origin). 
$\mathcal{M_+}$ is the set of all products of an even number number of reflections about lines. What you have to do is take a member $f$ of $\mathcal{M}$ and member $g$ of
$\mathcal{M_+}$ and show that there is a member $g'$ of
$\mathcal{M_+}$ such that $f o g=g' o f$. This result is an easy conequence of the fact that if $\ell_1$ and $\ell_2$ are lines there exists a line $\ell_3$ in the pencil determined by $\ell_1 \text { and }\ell_2$ such that
$$\omega_{\ell_1}o\omega_{\ell_2}=\omega_{\ell_3}o\omega_{\ell_1}$$ where $\omega_{\ell}$ denotes reflection about the line $\ell$.
A: In general, the subgroup of rigid motions in the Euclidean isometry group $E(n)=O(n) \ltimes \Bbb R^n$ is given by
$E^+(n)=SO(n) \ltimes \Bbb R^n$, see here. Now $SO(n)$ is a normal subgroup of $O(n)$ of index $2$, see here. 
See also wikipedia, where an argument is given before saying: "It follows that the subgroup $E^+(n)$ is of index $2$ in $E(n)$." 
