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Mar 14 at 11:45 comment added Sean Eberhard As for your last question, there are no other examples, but I may offend you by using representation theory again. We know that $B$ acts reguarly on the nonzero points of $V = \mathbb F_p^\ell$, in particular irreducibly, so by Schur's lemma, $K = \mathrm{End}_B(V)$ is a finite division ring, hence a field, and $B \le K^\times$. Since $K^\times$ is abelian and transitive on $V \setminus \{0\}$, $|K^\times| = |V| - 1$. This implies that $B = K^\times$ and $V \cong K$.
Mar 14 at 0:22 comment added tomasz This is a nice argument, thanks!
Mar 13 at 22:24 history edited KhashF CC BY-SA 4.0
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Mar 13 at 21:59 history answered KhashF CC BY-SA 4.0