Quaternion Group

jujab · #1 $Old$ November 3rd, 2009, 05:14 PM

G={+/- 1, +/- i, +/- j, +/- k}
H={1, -1}

How do I construct the cayley table for G/H?

tonio · #2 $Old$ November 4th, 2009, 06:37 AM

Quote:

Originally Posted by jujab $View Post$

G={+/- 1, +/- i, +/- j, +/- k}
H={1, -1}

How do I construct the cayley table for G/H?

The basic laws are: $i^2=j^2=k^2=-1\,,\,\,ij=k\,,\,\,jk=i\,,\,\,ki=j$ . Try now to have some fun proving that if you reverse the order of the above products you get minus the result, i.e. $ji=-k\,\,and\,\,etc.$

Tonio

jujab · #3 $Old$ November 4th, 2009, 12:26 PM

Ok, thanks!