4
$\begingroup$

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!

$\endgroup$
4
$\begingroup$

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 $|Q_1 \cap Q_2|=9$.

$\endgroup$
  • $\begingroup$ Brilliant and well presented answer! I managed to understand your explanation. Thanks! $\endgroup$ – yoyostein Sep 22 '15 at 12:29
  • $\begingroup$ Thank you, you are welcome! $\endgroup$ – Nicky Hekster Sep 22 '15 at 12:30
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – yoyostein Sep 21 '15 at 14:07
  • 1
    $\begingroup$ 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. $\endgroup$ – Dietrich Burde Sep 21 '15 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.