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Suppose that $G$ is a permutation group of degree $11$ and order $7920=11 \cdot 10 \cdot 9 \cdot 8$. Prove that $G$ is simple.

A hint to the problem is to show that $|N_G(P)|=55$ where $P \in{\rm Syl}_{11}(G)$. I've shown this. But I don't know how to link that with simplicity of $G$.

Shaun
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itticantdomath
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  • Isaacs has written more than one book; please indicate which one you are talking about. I am guessing it's Finite Group Theory, but still. – Arturo Magidin Jul 13 '21 at 19:52
  • The only result in that section that isn't specific to An, Sn, SL, or PSL is Iwasawa's theorem, 8.30. This can be used, but from the hint, I assume he is talking about Robin Chapman's proof in the monthly (which has been referenced a few times on this site). – Jack Schmidt Jul 13 '21 at 19:58
  • https://math.stackexchange.com/a/2482480/583 and https://math.stackexchange.com/q/4146685/583 reference the article – Jack Schmidt Jul 13 '21 at 20:05

1 Answers1

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Let $N$ be a nontrivial normal subgroup of $G$. Then, since $G$ has prime degree, it is primitive, and so $N$ is transitive. So $|N|$ is divisible by $11$ and we can assume that $P \le N$.

By the Frattini Argument, $G = NN_G(P)$. If $N_G(P) \not\le N$, then, since $|N_G(P)|=55$, we must have $P = N_N(P)$. But then $N$ has a normal $p$-complement $K$ by Burnside's Transfer Theorem, and $K$ is characteristic in $N$ and hence normal in $G$, and since $K$ cannot be transitive, we must have $K=1$. But then $N = P$ is normal in $G$, which we know is false.

So $N_G(P) \le N$ and $N=G$, and we have proved that $G$ is simple.

Derek Holt
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