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Let $O_n(\mathbb{R})$ be the group of real orthogonal $n\times n$ matrices and $SO_n(\mathbb{R})$ be the group of real orthogonal matrices with determinant $1$.

(i) Show that $O_n(\mathbb{R}) = SO_n(\mathbb{R}) × \{\pm I_n\}$ if and only if $n$ is odd.

(ii) Show that if $n$ is even, then $O_n(\mathbb{R})$ is not the direct product of $SO_n(\mathbb{R})$ with any normal subgroup.

Here is the progress I have made so far:

(i) If $n$ is odd then consider the map $\phi: SO_n(\mathbb{R}) × \{\pm I_n\} \to O_n : (A,B) \to AB$. This is a homomorphism as all the elements of $\{\pm I_n\}$ commute with the elements of $SO_n(\mathbb{R})$. Furthermore it is injective as is $AB = CD$ then since $A,C$ have determinant $1$ we get that $B,D$ have the same determinant so $B,D$ are the same matrix so $A,C$ are the same as well. It is also surjective as if $E \in O_n(\mathbb{r})$ then either $E$ or $-E \in SO_n(\mathbb{R})$ and so either $(E,I_n)$ or $(-E,-I_n)$ maps to $E$. Hence we have an isomorphism and they are the same.

If $n$ is even then note that $O_n(\mathbb{R})$ has center of order $2$ while $ SO_n(\mathbb{R}) × \{\pm I_n\}$ has center of order $4$ so they are not isomorphic.

(ii) I am having trouble with this bit. I can't even manage to show $O_n(\mathbb{R})$ is not isomorphic to $SO_n(\mathbb{R})$ for even $n$.

Any help is much appreciated.

Hadi Khan
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3 Answers3

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Hint for $(ii)$. Show that we always have $O_n(\mathbb{R})\cong SO_n(\mathbb{R})\rtimes C_2$, with the semidirect product. When is this a direct product? Consider possible homomorphisms $\{\pm I_n\}\rightarrow SO_n(\mathbb{R})$. Furthermore, see the questions on MSE, e.g., here:

Is $O_n$ isomorphic to $SO_n \times \{\pm I\}$?

Dietrich Burde
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For the last part, let $H$ be a subgroup(necessarily normal) such that $O_n=SO_n\odot H$. We then have $H\cap SO_n=\{I_n\}$.

Let $h\in H$ which is not $I_n$, so $h\notin SO_n$. Since $SO_n$ has index $2$ in $O_n$, $O_n$ is generated by $h$ and elements of $SO_n$.

Now $h$ commutes with every element of $SO_n$ (direct product properties), and commutes with itself, so $h$ commutes with any element of $O_n$. Hence $h=- I_n$. But since $n$ is even, $h\in SO_n$, a contradiction.

GreginGre
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  • Why is the intersection of $H,SO_n$ trivial? – Hadi Khan Apr 28 '18 at 08:17
  • By definition of a direct product of two subgroups. We have $G=H_1\odot H_2$ if and only if , any element of $G$ can be written in a unique way as a product of an element of $G_1$ and an element of $G_2$ , and any element of $G_1$ commutes wit hany element of $G_2$. The uniquness part implies easliy that intersection of the two subgroups must be trivial. – GreginGre Apr 28 '18 at 08:19
  • I'm a bit confused, I thought $H_1 \times H_2$ means elements of the form $(h_1,h_2)$ with $h_i \in H_i$ and $(h_1,h_2)\times (h_3,h_4) = (h_1\cdot h_3,h_2 \cdot h_4)$. We need an isomorphism to go from this to the group $G$ and I know if certain conditions are satisfied (trivial intersection, product is whole group, elements commute etc) then we have such an isomorphism so a direct product but I don't see how having a direct product gives us these properties. – Hadi Khan Apr 28 '18 at 08:25
  • I use the inner version of direct product. I'm talking about direct product of two subgroups, (that i suspect you don't know) while you are talking about the external version.

    I can translate the proof every in terms of external direct products, but the resulting proof will be less natural...

     – GreginGre Apr 28 '18 at 08:29
  • Yes, please do, I haven't seen the inner direct product before. – Hadi Khan Apr 28 '18 at 08:33
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Fix $n\ge 1$. Since $SO(n)$ has index 2, if it's part of a direct product decomposition, the direct summand $P$ has order 2, and hence is central. The center $Z$ of $O(n)$ is reduced to $\{\pm 1\}$, so the only candidate for $P$ is $Z$. If $n$ is odd then indeed it's a direct product decomposition, while if $n$ is even, $Z\subset SO(n)$ so the only candidate is discarded and hence $SO(n)$ is not part of a direct product decomposition.

A little harder: (1) for $n\ge 2$ even there no nontrivial direct decomposition of $O(n)$ at all. (2) for $n\ge 5$ odd there's no other nontrivial direct decomposition. It's immediate for $n\geq 2,4$ using that the only normal subgroups of $O(n)$ are $1$, $\{\pm 1\}$, $SO(n)$ and $O(n)$, almost as immediate for $n=4$ (where there are a few more normal subgroup), and requires a specific argument for $n=2$.

YCor
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