Timeline for (A very limited instance of) Lagrange's Theorem's converse and A_5
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 18, 2011 at 4:57 | comment | added | Junkie | My strategy was essentially to look at (known) maximal subgroups, as Derek Holt did. I am not as facile with the subject as he. | |
| Apr 16, 2011 at 20:48 | comment | added | Charles Matthews | My strategy for disposing of the claim having failed, it seems I'm out of my depth. | |
| Apr 16, 2011 at 19:48 | comment | added | Jim Humphreys |
If my arithmetic is correct (not guaranteed), 5 doesn't divide the order of $PSL_3(3)$; the comment by Junkie should dispose of the other minimal simple groups. Still, as David points out there are further possibilities. It seems that 60 does divide the order of all but one of the 26 sporadic simple groups, so they need more scrutiny. Even assuming CFSG is in hand, what is the strategy for deciding whether or not $A_5$ occurs as a subgroup when 60 divides?
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| Apr 16, 2011 at 16:56 | comment | added | DavidLHarden | It's not that easy. If the minimal simple groups are checked, and that says that the only one that has order equal to a multiple of 60 is A_5 itself, then it is still possible that some group larger than A_5 is not minimal among simple groups, but minimal among those whose order is a multiple of 60 (and hence a counterexample). Also, the list of minimal simple groups is incomplete as given here, since it doesn't include $PSL_{3}(3)$. | |
| Apr 16, 2011 at 12:45 | history | edited | Charles Matthews | CC BY-SA 3.0 |
correct
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| Apr 16, 2011 at 7:50 | comment | added | Junkie | $PSL_2(2^p)$, $PSL_2(3^p)$, $PSL_2(p)$ with $p=2,3$ mod 5, $Sz(2^p)$ all have order not divisible by 60. | |
| Apr 16, 2011 at 7:43 | history | answered | Charles Matthews | CC BY-SA 3.0 |