how to show that there is no simple group of order $1755 = 3^3 \cdot 5 \cdot 13$? Thank you very much.
questions about simple groups
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$\begingroup$
group-theory
simple-groups
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3Use Sylow's theorem and counting elements of order 5 and 13. – 2011-04-06
1 Answers
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using the sylow theorems we see that:
the number of 5-subgroups must be 351 or 1 (the only divisors of 13*27 which are 1 mod 5).
the number of 13-subgroups must be 1 or 27 (the only divisors of 5*27 which are 1 mod 13).
the number 27-subgroups must be 1 or 13 (the only divisors of 13*5 which are 1 mod 3).
if the number of 5 subgroups and 13 subgroups were both not equal to 1 then the number of non-identity elements of these subgroups would be: $(5-1)\cdot(351)+(13-1)\cdot(27)=1728=1755-27$ (noting that the 5 and 13 subgroups all intersect trivially) so that the remainder of the group would have to consist of exactly the 27-subgroup. hence there is a normal sylow subgroup.
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0@user: Seems someone bipolar... As to saying things "when I do not know the truth", I do know for a fact that it is hard to believe that someone can ask if a Sylow p-subgroup can intersect nontrivial with a Sylow q-subgroup if they've taken (and presumably passed) a group theory course that covered the Sylow Theorems. While it is, perhaps, *possible*, it is nonetheless hard to believe (and it indicates, at least to me, that the person did not learn the material). – 2011-04-09