1. ## Isomorphsim

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

2. ## Re: Isomorphsim

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

3. ## 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).