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May 21, 2010 at 11:52 comment added Someone No. Unfortunately Jim's answer is gone. Half of it was an even better (compared to Xani's) explanation, when the matrices are diagonalizable. But at the end Jim made a claim that was simply wrong.
May 21, 2010 at 9:07 comment added Guntram OK, but this doesn't contradict Xandi, right? A representation preserves diagonalizable iff $k$ contains $\mathbb F_q$.
May 21, 2010 at 8:29 comment added Someone As with Jim's answer also the comment I added there vanished, I'll repeat the simple counterexample here: take the embedding $SL_2(q) \to GL_{2e}(p)$ by simply considering a $2$-dimensional $F_q$-vector space as $2e$-dimensional $F_p$-vector space.
May 20, 2010 at 11:54 comment added Victor Protsak Sorry, I was carried away by your algebraic group language - I meant "semisimple". Indeed, semisimple matrices aren't diagonalizable in general. Nonetheless, an element of order dividing $q-1$ will a fortiori be diagonalizable over any field containing $\mathbb{F}_q$ (and if it doesn't, it cannot possibly be true), by a root of unity argument. But it looks like Jim has already answered your question without the lifting assumption.
May 20, 2010 at 11:29 comment added Guntram Thank you for your answer. I don't see, though, where you used the specific group $SL_2$. For example, a cyclic group has a representation over the reals which is not diagonalizable over the reals.
May 20, 2010 at 11:04 comment added Victor Protsak If the representation lifts to characteristic zero, the answer is "yes", because the group is finite and every finite order matrix in char 0 is semisimple.
May 20, 2010 at 11:00 answer added Jim Humphreys timeline score: 7
May 20, 2010 at 10:58 answer added Xandi Tuni timeline score: 6
May 20, 2010 at 9:48 history asked Guntram CC BY-SA 2.5