As usual, for a subgroup of a group we will denote by and the normalizer and the centralizer of in If then we will write and for and respectively.

**Problem**. Let be a prime number and let be a group of order Prove that is not simple.

**Solution**. If is abelian, there is nothing to prove. So we suppose that is non-abelian and simple and we will get a contradiction. By Sylow theorem, the number of Sylow -subgroups is and so for every Sylow -subgroup of Now, we consider two cases.

*Case 1* . Let be a Sylow -subgroup. Then and so Suppose that and choose Then Let We have because and Thus and so which implies that because Thus is in the center of and hence it’s a normal subgroup of contradicting our assumption that is simple. Therefore and we are now done by the Burnside’s normal complement theorem.

*Case 2*. The idea for this case is similar to the one we used for case 2 in this problem. Let and be two distinct Sylow -subgroups of and put Then

which gives us because As a result, is a non-trivial normal subgroup of both and Therefore and so But and so i.e. is a normal subgroup of This contradicts our assumption that is simple.