We look at the group of all real matrices nxn with det(A) = 1 with matrix multiplication operation and the group of all real matrices nxn with det(A) = 1 or -1 also with multiplication. We need to prove there's no isomorphism here. I need a general proof for matrices of any size.
So far I was only able to prove this for the matrices of nxn where n is odd. We take a matrix nxn with det(A)=1 which has 2^n square roots. It has 2^(n-1) roots which have det(B) = 1 and 2^(n-1) roots with det(B) = -1. So if we look at an image of an equation A=BB it has to have 2^(n-1) square roots. Suppose it does, then we look at f(A), it has det(f(A)) = 1 and 2^(n-1) roots. 2^(n-2) of them have det = 1. We look at an image of f(A)=CC. It has to have 2^(n-2) roots. Let's repeat this process until we find no matrix with 2^k square roots. At this point equation A=B*B and it's image have different amount of roots, which implies there's no isomorphism.
This method does not work for matrices of even size though, since I can't claim that if a matrix has k square roots, then exactly half of them are positive. So I believe we need to investigate how many square roots can a matrix have. I have made a different question for that, the link is here How many real square roots can a real invertible matrix have?
