The automorphism group of the quaternion group is the symmetric group on four letters. Sounds fascinating to me. But once I figured it out it's rather easy.
The elements of Q8 are:
1:1, 2:-1, 3:i, 4:j, 5:k, 6:-i, 7:-j, 8:-k.
The generators are:
3:i, 4:j.
Then there are 24 isomorphisms, since there are 6 possible generators i, j, k, -i, -j, and -k but after 1 has been chosen, 4 remain ( choose i then -i can't be chosen anymore ).
So up to here I have shown that Aut(Q8)=24, not yet that it is isomorphic to S4. ( That's for later. )
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