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how to show that there is no simple group of order $1755 = 3^3 \cdot 5 \cdot 13$? Thank you very much.

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Use Sylow's theorem and counting elements of order 5 and 13. – Jack Schmidt Apr 6 '11 at 23:37
up vote 3 down vote accepted

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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13*27 = 351 ≡ 1 mod 5 – Jack Schmidt Apr 7 '11 at 0:01
@jack my bad... – yoyo Apr 7 '11 at 0:26
@yoyo, thank you very much. I don't know why 5 and 13 subgroups all intersect trivially. Let G and H be a 5 and 13 subgroup of the original group A respectively. Let $M = G \cap H$. Is M a subgroup of A? Does $|M|$ divide $|G|$ and $|H|$? – user Apr 7 '11 at 2:21
@Jianrong: Yes $M$ is a subgroup of $A$. That is true for the intersection of any subgroups. In particular $G$ and $H$ intersect trivially because all non-trivial elements of the two subgroups have order 5 and 13, respectively. – Brandon Carter Apr 7 '11 at 2:32
@Jianrong: Seems hard to believe you would ask if a Sylow $p$-subgroup can intersect a Sylow $q$-subgroup, $p\neq q$, nontrivially, if you've indeed already taken the course and are merely reviewing, as you clamed here: – Arturo Magidin Apr 8 '11 at 18:46

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