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Problem. (Basic topology (M.A.Armstrong) Exercise 16 in chapter 4.3)

(1) Prove that $\mathrm{O}(n)$ is homeomorphic to $\mathrm{SO}(n)\times \mathbb{Z}_2$. (2) Are these two isomorphic as topological groups?

($\mathrm{O}(n)$ : Orthogonal group, $\mathrm{SO}(n)$ : Special orthogonal group)

My attempt for (1) :

Let's pick any element $C$ from $\mathrm{O}(n)-\mathrm{SO}(n)$, and let

\begin{aligned} &f:\mathrm{SO}(n)\to \mathrm{SO}(n) ; \; f(A)=A\\ &g:\mathrm{O}(n)-\mathrm{SO}(n)\to \mathrm{SO}(n) ; \; g(A)= CA \end{aligned}

Then $f$ and $g$ are homeomorphisms. Since $\mathrm{SO}(n)$ is compact, $g^{-1} (\mathrm{SO}(n))=\mathrm{O}(n)-\mathrm{SO}(n)$ is also compact. So, they are closed in the subspace topology on $\mathrm{O}(n)$. Thus, by the glueing lemma, the function $$\varphi:\mathrm{O}(n)\to \mathrm{SO}(n)\times \mathbb{Z}_2; \; \varphi(A)= \begin{cases} (f(A), 0)\quad&(\text{when }A\in \mathrm{SO}(n))\\ (g(A), 1)&(\text{otherwise}) \end{cases}$$ becomes continuous. We can easily confirm that $\varphi$ is bijective and $\varphi^{-1}$ is continuous.

Questions :

(1) I'm not sure I'm going in the right direction (in proving (1)). I used glueing lemma to justifying continuousness of the function $\varphi$, but I doubt whether I'm using that lemma in the right place.

(2) I tried to prove that two groups are NOT isomorphic, but I can't find any key for that. I tried to compare orders of elements in two groups, but I couldn't find some good elements to compare... Is there any simple way to check whether two groups are isomorphic?

Thanks.

hunminpark
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    Think about the centers of the two groups. The answer, if I'm right, depends on $n$. –  Oct 07 '15 at 04:29
  • Mike Miller // Thanks. I'll think about them. – hunminpark Oct 07 '15 at 06:16
  • OP I hope your question hasn't been closed as I don't think your question(1) has an answer in the aforementioned link. Yes the map you have constructed is a homeomorphism and you use the gluing lemma in the correct way. – Coco Oct 15 '23 at 19:55

2 Answers2

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Consider the mapping $f:O(n)→SO(n)×Z_22$ given by $M→(det(M)⋅M,detM)$

I think your answer is not right. Check det [det(M)M]= [det(M)]^{n+1}. Thus if $n$ is even, it will be wrong.

Nguyen Dang Son
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Hint: Consider the mapping $f: O(n)\to SO(n)\times \Bbb{Z}_2$ given by $$K \to \{\text{det}(M)\cdot M,\text{det}M\}$$ This is a well defined one to one and onto map. You can check that the map above is continuous.

Now since we know that $O(n)$ is compact and $SO(n)\times \Bbb{Z}_2$ is Hausdorff , the following map is a homeomorphism.