I would like to know if O(n) isomorphic to the product group SO(n) ×{ħI}?

where O(n) is orthogonal group

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- Sep 24th 2012, 07:13 AM #1

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- Sep 24th 2012, 10:54 AM #2
## Re: Isomorphsim

$\displaystyle O(n)$ has two connected components, namely, the set of matrices of determinant $\displaystyle 1$ and those of determinant $\displaystyle -1$. So you can define $\displaystyle f(A)=(A,I)$ if $\displaystyle \det A$ and $\displaystyle f(A)=(-A,-I)$ otherwise.

- Sep 24th 2012, 03:38 PM #3

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## Re: Isomorphsim

yes. SO(n) is normal in O(n), being of index 2 (because det:O(n)-->{-1,1} is a surjective group homomorphism).

it should be clear that we thus get two cosets: SO(n), and -SO(n) (since if a matrix A = BU is not in SO(n), where U is in SO(n), then:

det(A) = det(B)det(U) = det(B) = -1, so we have det(-A) = det(-IA) = det(-I)det(A) = (-1)(-1) = 1, so -A is in SO(n), and A = -(-A)).

this shows we can write any matrix in O(n) UNIQUELY as a product of a matrix in SO(n) and either I or -I.

therefore O(n) is the (internal) direct product of SO(n) and {-I,I} (the above shows it is the semi-direct product of SO(n) and {-I,I}. but let A be an orthogonal nxn matrix. then:

IAI^{-1}= IAI = A, and:

(-I)A(-I)^{-1}= IAI = A,

which shows that {-I,I} acts trivially on O(n) by conjugation, so we have a direct product. alternatively, you can show that {-I,I} is normal in O(n) by direct computation:

AIA^{-1}= AA^{-1}= I, and A(-I)A^{-1}= -AA^{-1}= -I).