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Apr 7, 2019 at 15:38 history edited Stefan Kohl CC BY-SA 4.0
Fixed a typo in the title of the cited paper.
Jan 17, 2013 at 13:51 comment added Emil Jeřábek @Derek: According to the Brandl and Wujie paper, every such group is isomorphic to $[\mathit{PSL}(3,4)]\langle\beta\rangle$, where $\beta$ is a unitary automorphism of $\mathit{PSL}(3,4)$.
Jan 17, 2013 at 13:10 comment added Derek Holt You are right, $A_7$ has spectrum $\{1,2,3,4,5,6,7\}$. Can you think of an example with $\{1,\ldots,8\}$?
Jan 17, 2013 at 10:34 comment added Stefan Kohl @Derek: Sorry -- but how does an element of $A_7$ of order 8 look like?
Jan 17, 2013 at 9:07 comment added Derek Holt Since the proof of this result involves considering all finite simple groups using the classification, it is unlikely that there is any easily described reason for this result. Note that the group $A_7$ has spectrum $\{1,\ldots,8\}$. As $n$ gets larger, it gets harder to avoid commuting elements resulting in elements of order higher than $n$.
Jan 17, 2013 at 8:17 history edited José Hdz. Stgo. CC BY-SA 3.0
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Jan 17, 2013 at 1:19 comment added Martin Brandenburg I like these results about quite general and abstract questions where suddenly such an explicit bound pops up. For example the automorphism group of a smooth algebraic curve of genus $g>1$ has order at most ${\bf 84}(g-1)$. It always makes me wonder if there is any deeper reason behind this. For example, your answer says that there groups with spectrum $\{1,\dotsc,8\}$, but the spectrum $\{1,\dotsc,9\}$ is not possible. Why on earth ...
Jan 16, 2013 at 19:27 history edited José Hdz. Stgo. CC BY-SA 3.0
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Jan 16, 2013 at 19:19 history edited José Hdz. Stgo. CC BY-SA 3.0
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Jan 16, 2013 at 18:49 history edited José Hdz. Stgo. CC BY-SA 3.0
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Jan 16, 2013 at 18:37 history answered José Hdz. Stgo. CC BY-SA 3.0