Let $A,B$ be two finite subgroups of $SO(4)$ such that $A$ and $B$ are isomorphic as abstract groups. Can we find a $g \in SO(4)$ such that $$ gAg^{-1}=B? $$ If it is the case, does the same conclusion hold for $SO(n)$?


$\newcommand{\mat}[1]{\left(\begin{matrix}#1\end{matrix}\right)}$ Identifying $SO(4)$ with the group of deteminant-one matrices $Q\in \mathrm{M}_4(\mathbb R)$ such that $Q^TQ=QQ^T=\mathrm{Id}$, take

$$A=\lbrace\mathrm{Id},\mat{-1\\&-1\\&&-1\\&&&-1}\rbrace\quad\text{and}\quad B=\lbrace\mathrm{Id},\mat{1\\&-1\\&&1\\&&&-1}\rbrace.$$

Then both are subgroups of order $2$, and they cannot be conjugate, since their non-trivial elements have different characteristic polynomials.

Edited: The group $A$ was changed following a comment by TastyRomeo, the original group was a subgroup of $O(4)$, rather than of $SO(4)$.


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.