How to show that the extension of group $\mathbb Z_{2}$ by $\operatorname{SO}(n)$: $$\operatorname{Id} \to \operatorname{SO}(n) \to \operatorname{O}(n) \xrightarrow{\det} \mathbb Z_2 \to 1$$ is a direct product for odd $n$ and semidirect product for even $n$ ?


Such a short exact sequence splits if and only if you can find a group morphism :

$$\phi: \mathbb{Z}_2\rightarrow O(n)$$

With the condition that $det\circ \phi$ is the identity of $\mathbb{Z}_2$. In the following I will denote $\mathbb{Z}_2=\{\pm 1\}$ multiplicatively (it is more coherent with the determinant).

Since $\mathbb{Z}_2$ is a very small group, in order to define properly $\phi$ you just need to send $1$ (the neutral element) to the identity matrix and $-1$ to some matrix $A$. We need :

$$A\text{ to be in } O(n)\text{ of order }2\text{ so that } \phi\text{ is a group morphism.} $$

$$det(A)=-1\text{ so that we do have a section, i.e. we have the relation }det\circ\phi=Id $$

When you have $n$ odd you now that $det(-I_n)=-1$, since it is clear that $(-I_n)^2=I_n$, $A:=-I_n$ is clearly what we are looking for. Furthermore, since $-I_n$ is central, it is clear that not only we have :

$$O(n)=SO(n)\rtimes \langle -I_n\rangle $$

But we actually have a direct product :

$$O(n)=SO(n)\times \langle -I_n\rangle $$

In the case $n$ even, we cannot choose $A:=-I_n$ anymore since $det(-I_n)=1$. Anyway you may choose the following matrix :


Then you have :

$$O(n)=SO(n)\rtimes \langle A\rangle $$

Finally, if you wanted to show that no $A$ can be chosen so that it gives a direct product, it is easy, if it were the case then the center of $O(n)$ would be of order $4$ (isomorphic to $\{\pm I_n\}\times \langle A\rangle$), but it is easy to prove that $Z(O(n))=\{\pm I_n\}$


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.