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By Sylow's theorems, $G$ has a unique $11$-sylow subgroup of order $11$ and then cyclic, say this $11$-sylow subgroup called $N$. And $N$ contains every element of order $11$ in $G$. Then how can I complete the proof since I am stuck here.

Jyrki Lahtonen
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Student
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2 Answers2

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Let $N=\left<a\right>$ and let $g\in G$. Then $g^{-1}ag=a^r$ for some $r$. This way $G$ maps to $\text{Aut}(N)$ by conjugation. As $\text{Aut}(N)$ has order $10$ and $G$ has order $231$, which is coprime to $10$, this map is trivial, so $g^{-1}ag=a$ for all $g\in G$.

Angina Seng
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    Thanks. But why Aut(N) has order 10, I cannot figure it out? – Student Sep 28 '17 at 14:51
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    @Student If $|G| = p$ and $p$ is prime, then $|\operatorname{Aut}(G)| = p-1$. This follows from a more general fact that if $|G| = n$ (not necessarily prime) and $G$ is cyclic, then $|\operatorname{Aut}(G)| = \phi(n)$ where $\phi$ is the Euler totient function. This is because $G$ has $\phi(n)$ generators, and any automorphism must map a generator to another generator and is completely defined by this mapping. See here for more details. –  Sep 28 '17 at 16:09
  • @Bungo Thank you too much, I got it now. – Student Sep 28 '17 at 18:09
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Let $n_7$ be the number of Sylow 7-subgroup. We have $n_7 \equiv 1$ (mod 7) and $n_7$ divides $33$. So there is a unique Sylow 7-subgroup, named $L$.

In the same process, there is a unique Sylow 11-subgroup, named $N$. They are both normal.

Let $P$ be a Sylow 3-subgroup. Because $L$ is normal, then $PL$ is a subgroup of order 21 of $G$.

Consider its normalizer $N_G(PL)$. One has $PL \leq N_G(PL) \leq G$. But the order of $N_G(PL)$ cannot be 21. So $N_G(PL) = G$. It means that $PL \trianglelefteq G$.

So, $N$ and $PL$ are normal subgroup, then $NPL$ is a subgroup of order 231, or $NPL = G$. Moreover, $N$ commute with $PL$ because if $n\in N$, $p \in PL$, then by normality of $N$ and $PL$, $npn^{-1}p^{-1} \in N \cap PL = \{e\}$. So $N$ commute with $G$ since every $g\in G$ can be written as $n'p'$.

GAVD
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    Why cannot the order of $N_G(PL)$ be $21$? (I'm not saying it's false, but you didn't say why it is true.) –  Sep 28 '17 at 16:05