As usual, for a subgroup H of a group G, we will denote by N(H) and C(H) the normalizer and the centralizer of H in G. If H = \langle a \rangle, then we will write N(a) and C(a) for N(H) and C(H) respectively.

Problem. Let p be a prime number and let G be a group of order p^n(p+1), \ n \geq 1. Prove that G is not simple.

Solution. If G is abelian, there is nothing to prove. So we suppose that G is non-abelian and simple and we will get a contradiction. By Sylow theorem, the number of Sylow p-subgroups is p+1 and so P=N(P) for every Sylow p-subgroup P of G. Now, we consider two cases.

Case 1 . n=1. Let P be a Sylow p-subgroup. Then |P|=p and so P \subseteq C(P). Suppose that P \neq C(P) and choose a \in C(P) \setminus P. Then P \subseteq C(a). Let Q=aPa^{-1}. We have P \cap Q = \{1\} because |P|=|Q|=p and a \notin P=N(P). Thus PQ \subseteq C(a) and so |C(a)| \geq |PQ|=p^2 which implies that C(a)=G because |C(a)| \mid |G|=p(p+1). Thus \langle a \rangle is in the center of G and hence it’s a normal subgroup of G, contradicting our assumption that G is simple. Therefore N(P)=P=C(P) and we are now done by the Burnside’s normal complement theorem.

Case 2.   n \geq 2. The idea for this case is similar to the one we used for case 2 in this problem. Let P and Q be two distinct Sylow p-subgroups of G and put H=P\cap Q. Then

\displaystyle \frac{p^{2n}}{|H|} = |PQ| \leq |G|=p^n(p+1)

which gives us |H|=p^{n-1} because |H| \mid |G|=p^n(p+1). As a result, H is a non-trivial normal subgroup of both P and Q. Therefore PQ \subseteq N(H) and so |N(H)| \geq |PQ|=p^{n+1}. But |N(H)| \mid |G|=p^n(p+1) and so N(H)=G, i.e. H is a normal subgroup of G. This contradicts our assumption that G is simple. \Box

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