Is $O(2,\mathbb{R})$, the group of orthogonal $2\times2$ matrices a normal subgroup of $GL(2,\mathbb{R})$, the group of invertible $2\times2$ matrices?
My attempt: If $O(2,\mathbb{R})$ were a normal subgroup, then for any $A$ in $O(2,\mathbb{R})$ and any $G$ in $GL(2,\mathbb{R})$, we would have that $GAG^{-1}\in O(2,\mathbb{R})$, i.e. $(GAG^{-1})(GAG^{-1})^T=I$. This does not seem immediately apparent to me, so I'm trying to think of a simple counterexample. Any thoughts?
Any help appreciated!
$$ \begin{bmatrix} 1 & 0\\ 0 & -1\\ \end{bmatrix} $$ is orthogonal, but $$ \begin{bmatrix} 1 & 1\\ 0 & 1\\ \end{bmatrix}^{-1} \begin{bmatrix} 1 & 0\\ 0 & -1\\ \end{bmatrix} \begin{bmatrix} 1 & 1\\ 0 & 1\\ \end{bmatrix} = \begin{bmatrix} 1 & 2\\ 0 & -1\\ \end{bmatrix} $$ is not.
Try for $G$ a diagonal matrix like $\operatorname{diag}(2,1)$. You will find a counterexample.
The bigger picture is this: $O(2,\mathbb{R})$ is the subgroup of $G$ that leaves the dot product of two vectors invariant: if $A\in O(2,\mathbb{R})$, then $x\cdot y = Ax \cdot Ay$. But the dot product is only one among many possible inner products on $\mathbb{R}^2$. You can transform one inner product into another through a change of variables. For example, scaling the axes will change the dot product into another inner product. And the same change of variables will change $O(2,\mathbb{R})$ into a different subgroup of $GL(2,\mathbb{R})$, the one that leaves the new inner product invariant.