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Jun 7, 2014 at 14:47 vote accept Ken W. Smith
Jun 7, 2014 at 14:46 vote accept Ken W. Smith
Jun 7, 2014 at 14:47
Jun 7, 2014 at 13:34 history edited Ben Webster CC BY-SA 3.0
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Jun 7, 2014 at 13:30 comment added Ben Webster To any future readers: the comments above refer to my (wrong) previous answer.
Jun 7, 2014 at 9:01 history edited Ben Webster CC BY-SA 3.0
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Jun 7, 2014 at 0:59 comment added Ken W. Smith Yes, I would have thought that every representation of the quaternion group of order 8 would be "linear" over the quaternions. If G = $\langle x, y : x^4=y^4=1, yxy^{-1}=x^3, x^2=y^2 \rangle$ then I have four reps over the rationals. In addition I guess I could map $x$ to $\pm i$, $y$ to $\pm j$ but are there more than four of these options possible? Do I lose control of my set of irreducibles? (I am so comfortable with the complex field that I have no good intuition with the strange noncommutative quaternion division ring!)
Jun 6, 2014 at 21:48 comment added Mark Wildon Maybe I have misunderstood the question, but isn't every representation of the quaternion group of order 8 linear over the quaternions?
Jun 6, 2014 at 20:52 history answered Ben Webster CC BY-SA 3.0