6

What i deduced is that $n_5=1,6$ or $36$. We are done if $n_5=1$.

If $n_5=36$ we $N_G(P)=P$ for any Sylow $5$-subgroup P as $|N_G(P)|=\frac{180}{36}=5$ and $P$ is abelian cyclic so by Burnside p-complement theorem there exist a normal subgroup of order $36$ which is complement to $P$ , fine, and we are done, not simple.

For $n_5=6$, $|N_G(P)|=\frac{180}{6}=30$. From here I can see that by $N/C$- theorem $|N_G(P)/C_G(P)|\le 2$. In this case I am stuck, Any help please.

Thanks in advance!

Nicky Hekster
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Bhaskar Vashishth
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1 Answers1

2

Assume $G$ is simple and that $n_5=6$, so index$[G:N_G(P)]=6$, $P \in Syl_5(G)$. But then $G$ embeds homomorphically into $A_6$ (note that core$_G(N_G(P))=1$, since $G$ is simple!). But index$[A_6:G]=360/180=2$, implying $G$ is normal in $A_6$, which contradicts the simplicity of $A_6$.

Nicky Hekster
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  • how are you embedding $G$ in $A_6$. I can embed it homomorphically in $S_6$ by considering action of $G$ on set of sylow 5-subgroups. – Bhaskar Vashishth Nov 11 '14 at 22:55
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    If you are trying to solve problems of this type, then one of the basic things that you need to know is that every nonabelian simple group with a subgroup of index $n \ge 5$ embeds into $A_n$. – Derek Holt Nov 11 '14 at 23:14
  • @Bashkar - if $G$ is non-abelian simple and is a subgroup of $S_n$, then the commutator subgroup $G'=G \subseteq [S_n,S_n]=A_n$. – Nicky Hekster Nov 12 '14 at 06:44