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For this blog post, we shall show that a group of order is not simple. We will be using several previous posts as lemmas to prove this nontrivial result.
Suppose to the contrary G is simple. By Sylow’s Third Theorem, , . This means that is 1, 4 or 16.
We recall that if , then the Sylow 3-subgroup is normal.
Let and be two distinct Sylow 3-subgroups of such that is maximum.
Using our previous lemma regarding index of intersection of Sylow subgroups, we split our analysis into three cases, the hardest of which is Case 3.
Case 1) If , . Thus , which allows us to conclude .
Case 2) If , . Thus . Similarly, we can conclude and we are done.
Case 3) If , then .
By another previous lemma regarding index of least prime divisor, . Thus, . Thus .
If , then which is a contradiction. Hence we suppose . Let . The possible values of k are .
Next, we use the fact that if is a simple group and is a subgroup of index , then divides .
Thus, , which forces k=16.
But then and similarly . Thus . This is a contradiction to .
Therefore all cases lead to contradiction and thus G is not simple.