Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

What is the group $O(4)/H$?

Here $O(4)$ is the group of orthogonal matrices and H is the center of $O(4)$.

share|improve this question
1  
It's the group of inner automorphisms of $O(4)$. –  David Roberts Feb 26 '11 at 4:01
4  
This looks a bit like homework to me (mathoverflow.net/faq#whatnot). What is your motivation for considering this question? –  David Roberts Feb 26 '11 at 4:04
    
Someone can make a nice question out of nice applications of this post! –  Romeo Feb 27 '11 at 2:00
    
If this answer is sufficient for your purposes, why not accept it? –  Yemon Choi Oct 2 '11 at 1:58
add comment

1 Answer

This group is isomorphic to $(SO(3) \times SO(3)) \rtimes \{\pm 1\}$, as is discussed in this question and its comments. (The order two factor comes from the fact that $O(4) = SO(4) \rtimes \{\pm 1\},$ and that $H \subset SO(4)$.) [Added as per Scott Carnahan's comment below: The order two factor shouldn't be a direct factor, but a semi-direct factor. I leave it as an exercise to determine its action on $SO(3) \times SO(3)$. If someone reading this knows the correct LaTeX command for a semi-direct product, feel free to make the appropriate edit.]

share|improve this answer
2  
Is it really a group-theoretic direct product? This seems like a situation where one should use the semidirect product symbol $\rtimes$ (and possibly specify the involution). –  S. Carnahan Feb 26 '11 at 7:08
1  
Dear Scott, No, of course you're correct; it's a semi-direct product. Thanks! Best wishes, Matthew –  Emerton Feb 26 '11 at 21:57
    
Added the \rtimes as needed. –  David Roberts Feb 27 '11 at 0:30
    
Dear David, Thanks! Regards, Matthew –  Emerton Feb 27 '11 at 20:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.