# 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!

Note first the following: if $G$ is a simple group of and $H$ is subgroup of index $n$, then $|G|$ divides $n!$ (see I.M. Isaacs, Finite Group Theory, Corollary 1.3).

This implies that if $G$ is a simple group of order $432$ and $H$ is a proper subgroup of $2$-power index, one must have $|G:H|=16$.

Now we proceed where you left off: let $Q_1$, $Q_2 \in Syl_3(G)$, $|Q_1 \cap Q_2|=9$. Then $Q_1 \cap Q_2 \lhd Q_1$ (the index of $Q_1 \cap Q_2$ is the smallest prime dividing $|Q_1|$). Hence $Q_1 \subseteq N_G(Q_1 \cap Q_2)$. Note that $N_G(Q_1 \cap Q_2)$ must be proper, since $Q_1 \cap Q_2$ cannot be normal in $G$. It follows by the observation above that $|G:N_G(Q_1 \cap Q_2)|=16$. But then we conclude that actually $Q_1=N_G(Q_1 \cap Q_2)$. By the same reasoning, $Q_2=N_G(Q_1 \cap Q_2)$. So $Q_1=Q_2$, a contradiction to $|Q_1 \cap Q_2|=9$.

• Brilliant and well presented answer! I managed to understand your explanation. Thanks! – yoyostein Sep 22 '15 at 12:29
• Thank you, you are welcome! – Nicky Hekster Sep 22 '15 at 12:30

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.

• May I ask what is Lemma 13? I found the crazy project website but did not understand it. If you are the creator of crazy project I must thank you though as it has enlightened me many times. – yoyostein Sep 21 '15 at 14:07
• No, I have nothing to do with this project. You are right, Lemma $13$ is missing. You can find help about the proof given in the "crazy project" also here; it works the same way for $n=432$. See also here. – Dietrich Burde Sep 21 '15 at 18:33