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Prove that group of order $396=11\cdot2^2\cdot3^2$ is not simple. $n_{11}$ is $1$ or $12$, so I assumed $n_{11}=12$ and tried to look at the action of the group on $Syl_{11}\left(G\right)$ by conjugation. Thanks for help

daPollak
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    Do you know Burnside's Transfer Theorem? – Derek Holt Jan 25 '15 at 17:20
  • Negative.. Is it useful in these kind of problems ? – daPollak Jan 25 '15 at 17:24
  • Yes, by Burnside's Transfer Theorem, the fact $N_G(P)$ is abelian implies that $G$ has a normal $11$-complement - that is, a normal subgroup of order $36$, and hence is not simple. But Prahlad's solution is easier! – Derek Holt Jan 25 '15 at 18:09
  • Alternatively one can look at the centralizer of an element of order $3$ in the normalizer of an $11$-Sylow subgroup to get a subgroup of order 99 rsp. index 4. – j.p. Jan 26 '15 at 12:30

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So the action of $G$ on $Syl_{11}(G)$ gives a homomorphism from $G$ into $S_{12}$. Since $G$ is simple, this must be an injective map into $A_{12}$. If $P \in Syl_{11}(G)$, then $[G:N_G(P)] = 12$, so $|N_G(P)| = 33$. Any group of order 33 is cyclic (since both Sylow subgroups are normal). This means that $A_{12}$ must have an element of order 33, which it cannot since that would have to be a product of two disjoint cycles of orders 11 and 3 respectively.

Prahlad Vaidyanathan
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