Question 1131289: Prove that no group of order 96 is simple.
Answer by MathLover1(20850)   (Show Source):
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Theorem: Suppose that G is a finite group and p a prime such that np(G)>1, and choose distinct Sylow p-subgroups S and T of G such that the order |S∩T| is as large as possible. Then np(G)≡1(mod|S:S∩T|)
Applying this for |G|=96 and p=2 (assuming n[2](G)=3, otherwise we are done), we see that
|S:S∩T|=2.
Let D=S∩T, so D is normal in S and T.
Hence both S and T lie in the normalizer NG(D).
But S and T are distinct, hence |NG(D)|>|S|=25.
This forces NG(D)=G, i.e. D⊲G. We found a nontrivial proper normal subgroup.
or
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  , ,…, be the Sylow p-subgroups of G. If [ :∩ ]≥  whenever i≠j, then  ≡1mod
.Taking  gives you one of Sylow's theorems, and the contrapositive of this theorem in the case  is useful here.
If ≢1mod , then there exist different Sylow p-subgroups  and with [:∩ ]<.
This implies that [:∩]=[:∩]= , and thus the intersection ∩ is normal in both and .
In your problem, we can assume  . Then  ≢1mod , so we find different Sylow 2-subgroups and with ∩ normal in both and .
Therefore the intersection ∩ is normal in the subgroup ⟨,⟩ generated by and . The order of ⟨,⟩ is a multiple of  and larger than , so it has to equal  .
Thus ∩ is a nontrivial proper normal subgroup of .
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