# No simple group of order $96$

I have to show that there is no simple group of order $96$ using the sylow theorems.

I know that $96 = 2^5\cdot 3$, and from the third sylow theorem $n_2 = 1$ or $3$ and $n_3 = 1$ or $4$ or $16$.

I have seen some proofs of this statement that use a so called index factorial theorem but I haven't learned about this. Is there a way to prove this using just the sylow theorems?

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$n_3$ may also be equal to $4$. –  Olivier Bégassat May 8 '12 at 21:38
One approach is to use my argument here: if $G$ were simple and had $3$ $2$-Sylows, it would embed in $S_3$. –  Chris Eagle May 8 '12 at 21:39
Or, by the same argument, if $G$ were simple and had a subgroup of index $3$ (such as a $2$-Sylow), it would embed in $S_3$. This may be what you call the index factorial theorem. –  Chris Eagle May 8 '12 at 21:42
Thanks Olivier, I missed that. @ChrisEagle: That was one of the proofs I had seen but I don't really understand it. Isn't there a simpler way? –  rt93 May 8 '12 at 21:50
@jb88, the simpler way is for you to spent a little time and try to understand Chris's argument, which is actually quite simple! –  Mariano Suárez-Alvarez May 8 '12 at 22:27

We can use the following result about the number of Sylow subgroups (this is Theorem 1.16 in Isaacs's Finite Group Theory book) :

Theorem: Suppose that $G$ is a finite group and $p$ a prime such that $n_p(G) > 1$, and choose distinct Sylow $p$-subgroups $S$ and $T$ of $G$ such that the order $|S \cap T|$ is as large as possible. Then $n_p(G) \equiv 1 \pmod{|S : S \cap T|}$.

Applying this for $|G| = 96$ and $p = 2$ (assuming $n_2(G) = 3$, otherwise we are done), we see that $|S : S \cap T| = 2$. Let $D = S \cap T$, so $D$ is normal in $S$ and $T$. Hence both $S$ and $T$ lie in the normalizer $\mathbf{N}_G(D)$. But $S$ and $T$ are distinct, hence $|\mathbf{N}_G(D)| > |S| = 2^5$. This forces $\mathbf{N}_G(D) = G$, i.e. $D \vartriangleleft G$. We found a nontrivial proper normal subgroup.

Note: If you haven't learnt about index factorial theorem you probably haven't seen the above theorem either and the index factorial argument is actually much simpler here. So although this answer isn't probably what you wanted, I just wanted to mention this method as it can be very useful in other situations.

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I believe this is also in his Algebra: A Graduate Course. –  Dylan Moreland May 8 '12 at 22:31
It appears that any book by Issacs is in a somewhat alien fashion, as if he was from outer space... Of course I am kidding here. –  awllower Jun 3 '12 at 17:12

There is a following generalization of one of the Sylow's theorems (see for example here), which is equivalent to the theorem in Cihan's answer.

Let $P_1, P_2, \ldots, P_k$ be the Sylow $p$-subgroups of $G$. If $[P_i : P_i \cap P_j] \geq p^d$ whenever $i \neq j$, then $n_p \equiv 1 \mod p^d$.

Taking $d = 1$ gives you one of Sylow's theorems, and the contrapositive of this theorem in the case $d = 2$ is useful here. If $n_p \not\equiv 1 \mod p^2$, then there exist different Sylow $p$-subgroups $P$ and $Q$ with $[P: P \cap Q] < p^2$. This implies that $[P : P\cap Q] = [Q : P \cap Q] = p$, and thus the intersection $P \cap Q$ is normal in both $P$ and $Q$.

In your problem, we can assume $n_2 = 3$. Then $3 \not\equiv 1 \mod 4$, so we find different Sylow $2$-subgroups $P$ and $Q$ with $P \cap Q$ normal in both $P$ and $Q$. Therefore the intersection $P \cap Q$ is normal in the subgroup $\langle P, Q \rangle$ generated by $P$ and $Q$. The order of $\langle P, Q \rangle$ is a multiple of $2^5$ and larger than $2^5$, so it has to equal $G$. Thus $P \cap Q$ is a nontrivial proper normal subgroup of $G$.

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