I am doing Exercise 4-16 in Armstrong's Basic Topology. The question is : are $SO(n)\times Z_2$ and $O(n)$ isomorphic as topological groups? (I have proved the homeomorphic part).
The problem is, I can prove that the map I constructed is merely a homeomorphism but not an isomorphism, but I cannot prove that there exists no isomorphism.
I am aware of this question, where an answer says to consider the map $O(n)\to SO(n)\times Z_2,\ M\mapsto(M\det M,-1)$. But this is not well defined if $n$ is even. For example, when $n=2$, $\left(\matrix{1&0\\0&-1}\right)$ is mapped to $\left(\matrix{-1&0\\0&1}\right)$, which does not lie in $SO(2)$. Any help is appreciated.
Yes when $n$ is odd. No when $n$ is even.
When $n$ is odd $\{\pm I\}$ is a complementary subgroup to $SO(n)$ in $O(n)$.
But when $n$ is even, the centre of $O(n)$ consists of $\pm I$ which are both in $SO(n)$. But in $SO(n)\times Z_2$ there is a central element not in $SO(n)$.