# Normalizer in matrix groups

I have the problem of calculating the normalizer of $$\begin{bmatrix} \lambda & 0 \\ 0 & \lambda^{- 1} \end{bmatrix}$$ in the group $$\begin{bmatrix} \cos (\theta) & -\sin (\theta) \\ \sin (\theta) & \cos (\theta) \end{bmatrix}$$

this, to solve the problem of:

For all $$g \in SL_2 (\mathbb {R})$$ there is a decomposition $$g = \begin{bmatrix} \cos (\theta_1) & -\sin (\theta_1) \\ \sin (\theta_1) & \cos (\theta_1) \end{bmatrix} \begin{bmatrix} \lambda & 0 \\ 0 & \lambda^{- 1} \end{bmatrix} \begin{bmatrix} \cos (\theta_2) & -\sin (\theta_2) \\ \sin (\theta_2) & \cos (\theta_2) \end{bmatrix}$$ which is unique except conjugation by elements of the normalizer.

I know that the normalizer $$N$$ is generated by $$N = \langle \begin {bmatrix} 0 & 1 \\ -1 & 0 \end {bmatrix} \rangle$$

I was able to show $$\langle \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \rangle \subseteq N$$

Does anyone have an idea of how to show the other contention?

• The normlizer of a single matrix is the same as its centralizer. Is that what you meant to ask? – Derek Holt May 21 at 7:35

You can do this directly by definition. Write $$R(\theta)$$ for your rotation matrix. You need $$R(\theta)\begin{bmatrix} \lambda&0\\0&\lambda^{-1}\end{bmatrix}=\begin{bmatrix}\lambda&0\\0&\lambda^{-1}\end{bmatrix} R(\theta).$$ This is $$\begin{bmatrix} \lambda \cos (\theta) & -\lambda^{-1}\sin (\theta) \\ \lambda\sin (\theta) & \lambda^{-1}\cos (\theta) \end{bmatrix}=\begin{bmatrix} \lambda\cos (\theta) & -\lambda\sin (\theta) \\ \lambda^{-1}\sin (\theta) & \lambda^{-1}\cos (\theta) \end{bmatrix}.$$ The diagonal entries give you nothing, they are always equal. The non-diagonal entries give you the equality $$\tag1 \lambda\sin\theta=\lambda^{-1}\sin\theta.$$ So, if $$\lambda=\pm1$$, then any $$\theta$$ works and the normalizer consists of all matrices in the group. When $$\lambda^2\ne1$$, the equality in $$(1)$$ forces $$\sin\theta=0$$, so $$\theta=0$$ and the normalizer is $$\{I\}$$.