Date modified: Thu 2023-05-11 . 11:51 AM
Date created: Sat 2023-10-07 . 09:02 PM
# Center of quarternion group. Consider the order 8 quarternion group $Q=\{\pm 1, \pm i \pm j, \pm k\}$ with the relations $i^2 = j^2 = k^2 = ijk = -1$, and $(-1)^2 = 1$. We remark its center is $\{\pm 1\}$, and the quotient $Q/Z(Q) \approx V_4$, the Klein-4 group. This isn't hard to see, as none of the $\pm i,\pm j,\pm k$ elements commute with each other, while $\pm 1$ does. So $Z(Q) = \{\pm1\}$. So the cosets are just $aZ = \{\pm a\}$, for $a=1,i,j,k$. The nontrivial elements are each order $2$, so this is precisely the Klein-4 group. #group-theory