Group of order $432$ is not simple How do we prove that a group $G$ of order $432=2^4\cdot 3^3$ is not simple?
Here are my attempts:
Let $n_3$ be the number of Sylow 3-subgroups.
Then, by Sylow's Third Theorem, $n_3\mid 16$ and $n_3\equiv 1 \pmod 3$.
Thus, $n_3=1, 4$ or 16.
Next, let $Q_1$ and $Q_2$ be two distinct 3-subgroups such that $|Q_1\cap Q_2|$ is maximum.
If $|Q_1\cap Q_2|=1$, then we can conclude $n_3\equiv 1\pmod {3^3}$, which will force $n_3=1$ and we are done.
If $|Q_1\cap Q_2|=3$, similarly we can conclude $n_3\equiv 1\pmod {3^2}$, and we are done.
The problem occurs when $|Q_1\cap Q_2|=3^2$, we can only conclude $n_3\equiv 1\pmod 3$ which is of no help.
Thanks for help!

Let $G$ be a simple group of order $432$. Note that $|G|$ does not divide $8!$, thus no proper subgroup of $G$ has index at most $8$. Sylow’s Theorem forces $n_3(G) = 16 \not\equiv 1 \bmod 9$, so that there exist $P_3, Q_3 \in \mathsf{Syl}_3(G)$ such that $|N_G(P_3 \cap Q_3)|$ is divisible by $3^3$ and another prime. Thus $|N_G(P_3 \cap Q_3)| \in \{ 3^3 \cdot 2, 3^3 \cdot 2^2, 3^3 \cdot 2^2, 3^3 \cdot 2^3, 3^3 \cdot 2^4 \}$. In each case either $P_3 \cap Q_3$ is normal in $G$ or its normalizer has sufficiently small index.
Remark. For more details on the steps see here.
