Question 1131289

{{{96=2^5*3}}}

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.


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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]}}},…,{{{P[k ]}}} be the Sylow p-subgroups of G. If [{{{P[i]}}}:{{{P[i]}}}∩{{{P[j]}}}]≥ {{{p^d}}} whenever i≠j, then {{{n[p]}}}≡1mod {{{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]}}} &#8801;&#824;1mod{{{p^2}}}, then there exist different Sylow p-subgroups {{{P}}} and{{{ Q}}} with [{{{P}}}:{{{P}}}&#8745;{{{Q}}}]<{{{p^2}}}. 
This implies that [{{{P}}}:{{{P}}}&#8745;{{{Q}}}]=[{{{Q}}}:{{{P}}}&#8745;{{{Q}}}]={{{p}}}, and thus the intersection {{{P}}}&#8745;{{{Q}}} is normal in both {{{P}}} and {{{Q}}}.

In your problem, we can assume {{{n[2]=3}}}. Then {{{3}}}&#8801;&#824;1mod{{{4}}}, so we find different Sylow 2-subgroups {{{P}}} and {{{Q}}} with{{{ P}}}&#8745;{{{Q}}} normal in both {{{P}}} and {{{Q}}}. 
Therefore the intersection {{{P}}}&#8745;{{{Q}}} is normal in the subgroup &#10216;{{{P}}},{{{Q}}}&#10217; generated by {{{P}}} and {{{Q}}}. The order of &#10216;{{{P}}},{{{Q}}}&#10217; is a multiple of {{{2^5}}} and larger than {{{2^5}}}, so it has to equal {{{G}}}. 
Thus {{{P}}}&#8745;{{{Q}}} is a nontrivial proper normal subgroup of{{{ G}}}.