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

where O(n) is orthogonal group

Results 1 to 3 of 3

- Sep 24th 2012, 08:13 AM #1

- Joined
- Sep 2012
- From
- AUS
- Posts
- 1

- Sep 24th 2012, 11:54 AM #2

- Sep 24th 2012, 04:38 PM #3

- Joined
- Mar 2011
- From
- Tejas
- Posts
- 3,546
- Thanks
- 842

## 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).