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.

  • $\begingroup$ Have you proved that the center of $O_n(\mathbb{R})$ is $\{I_n, -I_n\}$ and same for $SO_n(\mathbb{R})$ ? If so, your (i) is correct. $\endgroup$ – Max Apr 28 '18 at 7:47
  • $\begingroup$ Yes, I have a proof of that. $\endgroup$ – Hadi K says thanks to Monica Apr 28 '18 at 7:52
  • $\begingroup$ The center of $SO_n(\mathbb{R})$ is certainly not $\{I_n,-I_n\}$ if $n$ is odd, but is reduced to $\{I_n\}$. $\endgroup$ – GreginGre Apr 28 '18 at 8:06
  • $\begingroup$ Yes, that is why we get the isomorphism for the odd case, but for the even case the centre is both of them so we don't get an isomorphism. $\endgroup$ – Hadi K says thanks to Monica Apr 28 '18 at 8:11

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.

  • $\begingroup$ Why is the intersection of $H,SO_n$ trivial? $\endgroup$ – Hadi K says thanks to Monica Apr 28 '18 at 8:17
  • $\begingroup$ 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. $\endgroup$ – GreginGre Apr 28 '18 at 8:19
  • $\begingroup$ 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. $\endgroup$ – Hadi K says thanks to Monica Apr 28 '18 at 8:25
  • $\begingroup$ 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... $\endgroup$ – GreginGre Apr 28 '18 at 8:29
  • $\begingroup$ Yes, please do, I haven't seen the inner direct product before. $\endgroup$ – Hadi K says thanks to Monica Apr 28 '18 at 8:33

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 $SO_n \times \{\pm I\}$?


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 $n=2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.