Center of the Orthogonal Group and Special Orthogonal Group How can I prove that the center of $\operatorname{O}_n$ is $\pm I_n$?
I have that $AM = MA$, $\forall M \in \operatorname{O}_n$ and $A^{-1} = A^T$, $M^{-1} = M^T$.
Then $M = A^{-1}MA = A^{T}MA$. 
I see that since conjugating by $A$ must leave the determinant of $M$ unchanged then the result of conjugation must be a rotation. But now I'm stuck. 
How to proceed? Thanks in advance.
 A: Hints:


*

*By considering $AD=DA$ for some diagonal matrices $D$ with diagonal entries in $\{-1,1\}$, show that $A$ is a diagonal matrix.

*By considering $AR=RA$ for some Givens rotation matrices $R$, show that all diagonal entries of $A$ are equal to each other.

A: Alright, I answered my own question:
Suppose $A$ commutes with every element in $O_n$.
Then $A$ must commute with the elementary orthogonal matrices.
These are  


*

*the identity

*the row switching/column switching matrices

*the matrices that are identical to the identity but with a -1 for one of the entries on the main diagonal.
Then $AE = EA$ implies  $A = EAE^{-1}$.
Now, conjugation by Type 2 matrices shows that all the elements on the diagonal must be equal.
And conjugation by Type 3 matrices shows that all the off-diagonal elements must be zero.  Suppose we have that the $a_{ii}$ entry of Type 3 is -1 then conjugation leaves $a_{ii}$ unchanged but reverses the signs of all the elements in the same row and column as $a_{ii}$.
Since $A$ must also have determinant $\pm 1$, then the only matrices in the center must be $\pm I$.
The center of $SO_n$ is $\{ \pm I \}$ for $n > 3$ and $SO_2$ for $n=2$.
Suppose A commutes with every element in $SO_n$. Then $A$ must commute with the following matrices,


*

*a row switching transformation where one of the switched rows is multiplied by -1.

*a double row multiplying transformation where the multiplier is -1 in each case.
Now conjugation by Type I, shows that all the elements on the main diagonal must be equal, and that $a_{ij} = -a_{ji}$ for $i \neq j$.
And conjugation by Type 2 matrices shows that for $n > 2$ all the non-main-diagonal elements must be zero.
Since A must also have determinant $1$, then the only matrices in the center must be


*

*$SO_2$ if $n=2$. (All matrices in $SO_2$ meet the first condition.
This is easily verified by taking arbitrary matrices in $SO_2$ and using sum of angle identities.)

*$I$ if $n$ is odd.

*$\pm I$ if $n$ is even.
