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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.

trisct
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1 Answers1

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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 $SO(n)$.

YCor
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Angina Seng
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    For $n$ even, you implicitly use that an isomorphism $O(n)\to SO(n)\times Z/2$ should map $SO(n)$ onto $SO(n)$. This is true for $n\neq 2$ since then $SO(n)$ is the derived subgroup at both sides. For $n=2$ however one can directly argue that $O(2)$ is non-abelian while the product group is abelian. Or alternatively restate the argument (for arbitrary positive even $n$) saying that in $O(n)$ the center is contained in the derived subgroup, and not in $SO(n)\times Z/2$. – YCor Oct 19 '19 at 09:19