What is the group $O(4)/H$?
Here $O(4)$ is the group of orthogonal matrices and H is the center of $O(4)$.
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1$\begingroup$ It's the group of inner automorphisms of $O(4)$. $\endgroup$– David Roberts ♦Commented Feb 26, 2011 at 4:01
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4$\begingroup$ This looks a bit like homework to me (mathoverflow.net/faq#whatnot). What is your motivation for considering this question? $\endgroup$– David Roberts ♦Commented Feb 26, 2011 at 4:04
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$\begingroup$ Someone can make a nice question out of nice applications of this post! $\endgroup$– RomeoCommented Feb 27, 2011 at 2:00
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$\begingroup$ If this answer is sufficient for your purposes, why not accept it? $\endgroup$– Yemon ChoiCommented Oct 2, 2011 at 1:58
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1 Answer
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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.]
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2$\begingroup$ 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). $\endgroup$– S. Carnahan ♦Commented Feb 26, 2011 at 7:08
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1$\begingroup$ Dear Scott, No, of course you're correct; it's a semi-direct product. Thanks! Best wishes, Matthew $\endgroup$– EmertonCommented Feb 26, 2011 at 21:57
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$\begingroup$ Added the \rtimes as needed. $\endgroup$– David Roberts ♦Commented Feb 27, 2011 at 0:30
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$\begingroup$ Dear David, Thanks! Regards, Matthew $\endgroup$– EmertonCommented Feb 27, 2011 at 20:36