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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.

Asaf Shachar
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    did you try conjugating some orthogonal matrices and see if you still get an orthogonal matrix? – Ittay Weiss May 28 '15 at 12:22
  • Some of the easiest examples (to write and multiply) of orthogonal matrices are permutation matrices. Try conjugating such a matrix. – P Vanchinathan May 28 '15 at 12:45
  • @Ittay Weiss: Yes I did. (before asking the question...). I got the result stated above (I have edited the question and showed my compoutation). The point is that I do not know how to proceed from there. – Asaf Shachar May 28 '15 at 13:06
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    @AsafShachar, almost any choice of $S$ and something in $O(n)$ would have shown that. To show that a universal claim does not hold you go for counter examples, not generic proofs. – Ittay Weiss May 28 '15 at 13:15
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    Take $Q = \left[\begin{smallmatrix}0 & 1\-1 & 0\end{smallmatrix}\right] \in O(2)$ and $S = \left[\begin{smallmatrix}2 & 0\0 & 1\end{smallmatrix}\right]$, then $T = SQS^{-1} = \left[\begin{smallmatrix}0 & 2\-\frac12 & 0\end{smallmatrix}\right]$. Since $T^t T = \left[\begin{smallmatrix}\frac14 & 0\0 & 4\end{smallmatrix}\right] \ne I_2$, $T \notin O(2)$. You can extend this counter-example from $O(2)$ to other $O(n)$ by padding $n-2$ copy of $1$ along the diagonals of $Q$ and $S$. – achille hui May 28 '15 at 13:16

2 Answers2

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

Alex Fok
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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 $O(n)$.

zjk
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