Finite Subgroups of O_n(?) from SO_n(?)

extensionfinite-groupslie-groupsnormal-subgroupsriemannian-geometry

Let $G$ be a finite subgroup of $\mathrm{SO}_n(\mathbb R)$. We also assume $G$ to be "maximal" in the sense that for every $g\in\mathrm{SO}_n(\mathbb R)\setminus G$, we have that $\overline{\langle G,g\rangle}=\mathrm{SO}_n(\mathbb R)$. Let $H:=\{h\in\mathrm{O}_n(\mathbb R)\setminus\mathrm{SO}_n(\mathbb R): |\langle G,h\rangle|<\infty\}$. Is it true that for any $h,h'\in H$, if $h\overset{\mathrm{O}_n(\mathbb R)}\sim h'$, then $\langle G,h\rangle\overset{\mathrm{O}_n(\mathbb R)}\sim\langle G,h'\rangle,$
where $\overset{\mathrm{O}_n(\mathbb R)}\sim$ is the conjugate relation in $\mathrm{O}_n(\mathbb R)$?
Notice that for any $h\in H$, $|\langle G,h\rangle|=2|G|$, because $G=\ker\det$ is a normal subgroup of $\langle G,h\rangle$ of index $2.$

Best Answer

Now that you've added the maximality assumption, $\langle G,h\rangle$ has to be the normaliser of $G$ in $\mathop{\rm O}_n(\mathbb{R})$, and so does $\langle G,h'\rangle$. So they are not just conjugate, but equal. The set $H$ is the other coset of $G$ in its normaliser.

Here's why: Since $\langle G,h\rangle$ is finite, so is $\langle G,G^h\rangle$. This is then a finite subgroup of $\mathop{\rm SO}_n(\mathbb{R})$ containing $G$, and therefore by maximality we have $G^h=G$, so $h$ normalises $G$.

To expand the argument a bit, the centraliser of $G$ in $\mathop{\rm SO}_n(\mathbb{R})$ is $Z(G)$, since otherwise $G$ can be enlarged to a larger finite subgroup. So the centraliser in $\mathop{\rm O}_n(\mathbb{R})$ is also finite. The normaliser mod the centraliser is a group of automorphisms of a finite group, so is finite. Therefore the normaliser is finite.

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