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There is already another proof of this theorem, but I'm curious about solving this problem as written in Dummit&Foote(Ex.6.2 12)

Show there is no simple group $G$ with $|G|=9555$.

No simple groups of order 9555: proof

Hint: Let $Q \in Syl_{13}(G)$ and let $P\in Syl_7(N_G(Q))$. Argue that $Q \unlhd N_G(P)$. Why is this a contradiction?

It is quite similar to previous solution, but there are diffrences:

(1) They chose Sylow 7-subgroup as Q.

(2) No normality as $Q \unlhd N_G(P)$ does not require. Contradiction from Lagranges theorem.

(3) Unlikely another proof, in hint, $P$ is not 7-subgroup of $G$.

I'm in this progress: First $9555=3 \times 5 \times 7^2 \times 13$.

$n_{13}=105$ is uniquely determined, so $|N_G(Q)|=7\times 13$. Also from $P\leq N_G(Q)$, we can think $PQ$ is a subgroup of $G$.

Since $|P|=7$ and $|Q|=13$, $PQ$ is abelian so we can conclude $Q\leq N_G(P)$.

However, I'm facing in those problems:

(1) Strengthening $Q\leq N_G(P)$ to $Q \unlhd N_G(P)$

(2) Explaining why $Q \unlhd N_G(P)$ gives contradiction?

For above (2), I'm guessing $N_G(P)=G$ so it contradicts to simplicity.

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Maddy
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2 Answers2

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I solved with some uncertainty. Please verify it.

(1) Consider order of $N_G(P)$. By Lagrange's theorem, $|P|,|Q|$ divides $N_G(P)$. But Sylow's divisabilty condition gives $Q \unlhd N_G(P)$ in any case of order < $3 \times 5 \times 7^2 \times 13$ except $3 \times 5 \times 7 \times 13$. But latter case can deduce normality by element counting.

(2) $Q \unlhd N_G(P)$ gives $N_G(P) \cap N_G(Q)=N_{N_G(P)}(Q)=N_G(P)$ So $N_G(P) \subseteq N_G(Q)$. But order of normalizer of Sylow 7-subgroup of $G$ is already $7^2\times 13$ and $|N_G(P)|$ will be equal or greater. Contradiction by inclusion and order comparison.

Maddy
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Since I recently did this problem, I thought I would add my solution for reference. I am using the hint, and I also straighten out some of the ideas mentioned above:

Let $G$ be a group of order $9555=3\cdot5\cdot7^2\cdot13$, and let $n_p$ be the number of Sylow $p$-groups in $G$.

By Sylow's theorem, $n_{13}\mid3\cdot5\cdot7^2$ and $n_{13}\equiv1\pmod{13}$. This leaves us with $n_{13}=1, \text{ or } 105$. If $n_{13}=1$, then the Sylow $13$-group would be normal in $G$. Hence, assume $n_{13}=105$. This also gives us that $n_{13}=|G:N_G(Q)|=105$, where $Q\leq G$ is a Sylow $13$-group in $G$. Furthermore, $|N_G(Q)|=\frac{|G|}{|G:N_G(Q)|}=7\cdot13$.

Let $P\leq N_G(Q)$ be a Sylow $7$-group in $N_G(Q)$, and let ${n_p}'$ denote the number of Sylow $p$-groups in $N_G(Q)$. By Sylow's theorem, we have ${n_7}'\mid13$ and ${n_7}'\equiv 1\pmod{7}$. This only leaves ${n_7}'=1$ and thus $P\triangleleft N_G(Q)\Leftrightarrow N_G(Q)\leq N_G(P)$.

Now, by Sylow's theorem, we know that there exists a Sylow $7$-group in $G$, $P^*$, such that $P<P^*$. Furthermore, $|P^*:P|=7$, which is the lowest prime divisor of $|P^*|$, and hence $P\triangleleft P^*\Leftrightarrow P^*\leq N_G(P)$.

Since $N_G(Q)\leq N_G(P)$ and $P^*\leq N_G(P)$, we get that $\langle P^*, N_G(Q)\rangle\leq N_G(P)$ with $|P^*|=7^2\mid|N_G(P)|$ and $|N_G(Q)|=7\cdot13\mid|N_G(P)|$. This gives us $|N_G(P)|\geq 7^2\cdot 13$, $7^2\mid|N_G(P)|$ and $7\cdot13\mid|N_G(P)|$, which leaves us with the possibilities $|N_G(P)|=7^2\cdot13, 3\cdot7^2\cdot13, 5\cdot7^2\cdot13, \text{ or } 3\cdot5\cdot7^2\cdot13$.

Let ${n_p}''$ denote the number of Sylow $p$-groups in $N_G(P)$. Note that if $|N_G(P)|=3\cdot5\cdot7^2\cdot13$, then $N_G(P)=G$, which would give us $P\trianglelefteq G$. So let us consider the other cases.

Suppose $|N_G(P)|=7^2\cdot13$. By Sylow's theorem, we get ${n_{13}}''\mid 7^2$ and ${n_{13}}''\equiv 1\pmod{13}$. This only leaves ${n_{13}}''=1$; so, there is a unique Sylow $13$-group in $N_G(P)$, meaning the Sylow $13$-group is normal in $N_G(P)$.

Suppose $|N_G(P)|=3\cdot7^2\cdot13$. By Sylow's theorem, we get ${n_{13}}''\mid 3\cdot7^2$ and ${n_{13}}''\equiv 1\pmod{13}$. Again, this only leaves ${n_{13}}''=1$, which means there is a unique Sylow $13$-group in $N_G(P)$, which is also normal in $N_G(P)$.

Suppose $|N_G(P)|=5\cdot7^2\cdot13$. By Sylow's theorem, ${n_{13}}''\mid 5\cdot7^2$ and ${n_{13}}''\equiv 1\pmod{13}$. Again, this only leaves ${n_{13}}''=1$, which means there is a unique Sylow $13$-group in $N_G(P)$, which is also normal in $N_G(P)$.

In every case, we get a unique Sylow $13$-group in $N_G(P)$. Let $Q'\leq N_G(P)$ be this unique Sylow $13$-group in $N_G(P)$. Then $Q'\trianglelefteq N_G(P)$. Now, $|Q'|=13$, which is also the maximal power of $13$ dividing $|G|$; so, $Q'$ is also a Sylow $13$-group in $G$. Hence, we have $|N_G(Q')|=7\cdot13$, as computed in the beginning. However, $|N_G(Q')|=7\cdot13<7^2\cdot13\leq|N_G(P)|$, which is a contradiction to $N_G(Q')$ being the maximal subgroup in $G$ where $Q'$ is normal. Consequently, the three cases for $|N_G(P)|$ we considered cannot happen, which gives us that $G$ cannot be simple, as desired.

gogo
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