Is $O(n)$ normal in $GL(n)$?

Question

Is the orthogonal group $O(n)$ normal in $GL(n)$?

Here is what I did so far:

Let $Q\in O(n),S\in GL(n)$ we want to check if $S^{-1}QS\in O(n)$:

$(S^{-1}QS)^T=(S^{-1}QS)^{-1}\iff S^TQ^T(S^{-1})^T=S^{-1}Q^{-1}S $

$\iff S^TQ^{-1}(S^T)^{-1}=(S^{-1}Q^{-1}S)\iff SS^TQ^{-1}=Q^{-1}SS^T$

$\iff Q(SS^T)=(SS^T)Q$.

So the normality of $O(n)$ in $GL(n)$ is equivalent to the following claim: $\forall S\in GL(n)$ $SS^T$ commutes with any element of $O(n)$.

Update: Here is an easy way to continue (suggested by Alex Fok):

Focus upon symmetric matrices $S$. Then if $O(n)$ is normal, for every symmetric $S\in GL(n)$, its square $S^2$ must satisfy: $QS^2=S^2Q $ $\forall Q\in O(n)$.

Now take $S=\begin{pmatrix} x & 0 \\\ 0 & y \end{pmatrix}$ for $0\neq x\neq y\neq 0$. Then $S^2=\begin{pmatrix} x^2 & 0 \\\ 0 & y^2 \end{pmatrix}$. Now we can check which matrices commutes with $S^2$:

Let $Q= \begin{pmatrix} a & b \\\ c & d \end{pmatrix}$. Then: $QS^2=\begin{pmatrix} ax^2 & by^2 \\\ cx^2 & dy^2 \end{pmatrix}$, $S^2Q=\begin{pmatrix} ax^2 & bx^2 \\\ cy^2 & dy^2 \end{pmatrix}$. Hence the two products are equal iff $b=c=0$ that is $Q$ is of the form $\begin{pmatrix} a & 0 \\\ 0 & d \end{pmatrix}$. Since not every orthogonal matrix is of this form, this implies $O(n)$ is not normal.

Answer 1

Choose a diagonal matrix $D$ with distinct eigenvalues. Then for $A\in O(n)$, $DAD^{-1}\notin O(n)$.

Answer 2

Wikipedia says $ O(n) $ is

... is the group of distance-preserving transformations of a Euclidean space of dimension n that preserve a fixed point

Let's call this fixed point $P$. Let $t \in GL(n)$ be an thing that moves $P$ to another point $Q$.

Then $t O(n) t^{-1}$ (in which $Q$ is the fixed point) is different from $O(n)$.