## Groups of order p^n*(p+1) are not simple

Posted: September 23, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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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. $\Box$