Normal subgroup of O(2, $\Bbb R$) Suppose that $N$ is a normal subgroup of $O(2, \Bbb R)$. Show that if $N$ contains a reflection then $N = O(2, \Bbb R).$
I've been trying to conjugate the reflection itself with an arbitrary element of $O(2, \Bbb R)$ but with little success. Is this the correct approach please?
 A: Yes it is the correct approach, although you could be a bit more picky in choosing the 'arbitrary' element to conjugate with.
It helps to think about $O(2, \mathbb{R})$ in terms of its action on the circle (or on $\mathbb{R}^2$, whichever you prefer). In geometric terms the group consists of all rotations around the origin and all reflections in lines through the origin. 
First convince yourself that there is nothing more!
Now underlying this exercise is the following fact:
Any subgroup of $O(2, \mathbb{R})$ that contains all reflections is the whole group $O(2, \mathbb{R})$.
Once you found a proof of this, a proof of the statement you want to prove is not so far: you just need to convince yourself that for any arbitrary reflection $r$ there is an element $g$ of $O(2, \mathbb{R})$ such that conjugating your original reflection with $g$ will map it to $r$. I hope you see how that in turn answers your question, otherwise we can discuss it a bit more.
A: Denote by $r_\theta$ the rotation of angle $\theta$ and $s_\theta$ the reflection across the axis of angle $\theta$ with the $x$-axis.
By hypothesis, an element $s_\alpha$ belongs to $N$.
Now, notice that:


*

*For any $\theta \in \mathbb R$, $r_\theta \circ s_\alpha \circ r_\theta^{-1} = s_{\alpha+\theta} \in N$ as $N$ is normal. Therefore all reflections belong to $N$.

*A rotation $r_\theta$ is the product $s_{\theta/2} \circ s_{0}$ which belongs to $N$ as $N$ is a group.


We've proven that all rotations and reflections belong to $N$, hence $N=O(2,\mathbb R)$.
