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!
Best Answer
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 $E(n)$."