Showing group of order 9555 is not simple by using following Hint

Question

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.

Answer 1

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.

Answer 2

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.