Normalizer of a subgroup of $GL_2(\mathbb{R})$

Question

I have following subgroup of $GL_2(\mathbb{R}):$ $$A=\Bigg\{\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right),\left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right),\left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right),\left( \begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array} \right)\Bigg\}.$$

It's basically the rotation group of a square centered at the origin in $\mathbb{R}^2.$

I need to compute the normalizer of $A$.

Let's denote the normalizer by $A_N.$ Then, by definition, $$A_N:=\{g\in GL_2(\mathbb{R}): gA=Ag\}.$$ $$\equiv \{g\in GL_2(\mathbb{R}): gag^{-1}\in A,\ \forall \ a\in A\}.$$

I am using the second definition to compute all such $g$ by hand, but the computation is leading me nowhere. I have a feeling that all matrices that have $1$ or $-1$ on one diagonal and $0$ in the other diagonal satisfy the normalization condition. For example: all elements of $A$ are clearly in $A_N$. When I check for other matrices with $1$ or $-1$ on one diagonal and $0$ in the other, they all satisfy the condition. How to make this idea more precise? I am sure there must be a nice way to arrive at such $g$ through computation but I am yet to find one.

Answer 1

$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\in N_{GL_2(\Bbb R)}(A)\implies $$

$$\begin{align*} \bullet&\begin{pmatrix}b&\!\!-a\\d&\!\!-c\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&\!\!-1\\1&0\end{pmatrix}=\begin{pmatrix}0&\!\!-1\\1&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}\!\!-c&\!\!-d\\a&b\end{pmatrix}\iff\begin{cases}a=d\\{}\\b=-c\end{cases}\\{}\\\bullet&\begin{pmatrix}\!\!-b&a\\\!\!-d&c\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&1\\\!\!-1&0\end{pmatrix}=\begin{pmatrix}0&1\\\!\!-1&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}c&d\\\!\!-a&\!\!-b\end{pmatrix}\iff\begin{cases}a=d\\{}\\b=-c\end{cases}\end{align*}$$

The other two cases are useless as those are scalar matrices and thus they commute with everything...

We thus get that the general form of an element of the normalizer is

$$\begin{pmatrix}a&b\\\!\!-b&a\end{pmatrix}\;,\;\;\text{and of course}\;\;a^2+b^2\neq0$$

Answer 2

Hint: It suffices to find the set of matrices $g$ which satisfy $gag^{-1} = b$ where $a,b$ are the rotation matrices by $\pm90$ degrees. You only need to consider two cases $a=b$ and $a \neq b$ where you fix one choice of $a,b$ for the two cases. All matrices satisfy $g(\pm I)g^{-1} = \pm I$. And the normalizer certainly contains more matrices than just $0,1$ matrices: If $M$ is in the normalizer, then so is $kM$ for any constant $k \neq 0$. And $kM$ is automatically in the normalizer if $M \in A$.

Answer 3

Notice that $A$ only contains two elements of order $4$. Anything which normalizes $A$ must ether fix these two elements (in which case it commutes with everything in $A$ ), or else interchange these two elements of order $4.$ As indicated in my comment above (and as you noticed) there are elements outside $A$ of order two which normalize $A.$ It seems that you can complete the calculation from what you said above.