----- > [!proposition] Proposition. ([[conjugacy classes of the quaternion group]]) > Let $Q$ denote the [[quaternion group]]. Its [[conjugate|conjugacy classes]] are... > [!proof]- Proof. ([[conjugacy classes of the quaternion group]]) > The [[center of a group|center]] of $Q$ consists of $\v 1$ and $- \v 1$, so [[conjugate|center iff conjugacy classes are singletons]] we have $[\v 1]=\{ \v 1 \}$ and $[- \v 1]=\{ - \v 1 \}$. Next let $\b \ell \in Q \cut \{ \v 1, -\v 1 \}$, and consider the [[conjugate|conjugation]] of $\b \ell$ by an arbitrary $\v g \in Q$. \ If $\v g = \pm \v 1$ we get $\v 1 \b \ell \v 1=(-\b 1) \b \ell (-\b 1)=\b \ell$, so assume more interestingly that $\v g \notin \{ \v 1, -\v 1 \}$. Recall the relation discussed in [[quaternion group is nonabelian, but all its subgroups are normal|this proof]]: in general, $\v g \b \ell = - \b \ell \v g=\b \ell^{3} \v g$. Applying it we find $\v g \b \ell \v g ^{-1}=\b \ell^{3}\v g \v g^{-1}=\b \ell^{3}=-\b \ell.$ Thus we see that for all $\b \ell \in Q \cut \{ \v 1, - \v 1 \}$, $\b \ell \sim -\b \ell$ as witnessed by all $\v g \notin \{ \v 1, -\b 1 \}$. From this we conclude the [[conjugate|conjugacy classes]] of $Q$ are $\{ \v 1 \}, \{ - \v 1 \}, \{ \v i, - \v i \}, \{ \v j, - \v j \}, \{ \v k, -\v k \}.$ Compare to [[complex conjugate|complex conjugation]]. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```