Example. the Hurwitz action on the quaternion group [fgap-000C]
Example. the Hurwitz action on the quaternion group [fgap-000C]
The subgroup \(C\) acts on \(Q\) by left conjugation: \[ C\longrightarrow \operatorname {Aut}(Q),\qquad c\longmapsto (q\longmapsto cqc^{-1}). \] Indeed, conjugation by a Hurwitz unit gives \[ \omega i\omega ^{-1}=k,\qquad \omega k\omega ^{-1}=j,\qquad \omega j\omega ^{-1}=i. \] Thus conjugation by the generator preserves all 8 elements of \(Q\). Its conjugation powers preserve \(Q\) as well, so every \(c\in C\) gives the stated automorphism. The generator's action can be read from the two oriented cycles \[ i\longmapsto k\longmapsto j\longmapsto i, \qquad -i\longmapsto -k\longmapsto -j\longmapsto -i, \] while \(1\) and \(-1\) are fixed.
Conjugation preserves products: \[ c(q_1q_2)c^{-1}=(cq_1c^{-1})(cq_2c^{-1}). \] It is therefore an automorphism of \(Q\); its inverse is conjugation by \(c^{-1}\). The action law follows from \[ (c_1c_2)q(c_1c_2)^{-1} =c_1(c_2qc_2^{-1})c_1^{-1}. \] This verifies the action directly rather than inferring it from the picture. Voight records the normality of \(Q\) and the cyclic rotation in [voight2021quaternion, sec. 11.2.4, p. 168].