## Groups of order p^2*q^n are not simple

Posted: May 13, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Notation and Remark. Let $G$ be a finite group and let $s$ be a prime number which divides $|G|.$ We let $n_s$ be the number of Sylow $s$-subgroups of $G.$ Note that if a Sylow $s$-subgroup is not normal, then $n_s > s$ because $n_s > 1$ and $n_s \equiv 1 \mod s.$ So if $G$ is simple, then $n_s > s$ for all primes $s \mid |G|.$

Problem. Prove that a group $G$ with $|G|=p^2q^n, \ n \geq 0$ and $p \leq q,$ where $p,q$ are primes, is not simple.

Solution.  If $p=q,$ then $|G|=p^{n+2}$ and we know that every group of order $p^m,$ where $p$ is a prime and $m \geq 1,$ has a subgroup of order $p^{m-1}$ and all such subgroups are normal. So in this case $G$ is not simple. Thus we may assume that $p < q.$ The case $n=0$ is obvious. So we will consider two cases:

Case 1. $n=1$: [we do not need the condition $p for this case.] Suppose that $G$ is simple. Then $n_p=q$ and either $n_q=p$ or $n_q=p^2.$ If $n_q = p^2,$ then $G$ will have $p^2(q-1)=|G|-p^2$ elements of order $q$ because every Sylow $q$-subgroup of $G$ has exatly $q-1$ elements of order $q.$ But then we will only have $p^2$ elements left and so $n_p=1,$ which is a contradiction. If $n_q=p,$ then $q=n_p > p$ and $p=n_q > q,$ which is nonsense.

Case 2. $n \geq 2$: Suppose that $G$ is simple. Let $Q_1 \neq Q_2$ be two $q$-Sylow  subgroups of $G$ and put $H=Q_1 \cap Q_2.$ Then

$\displaystyle |G|=p^2q^n \geq |Q_1Q_2|=\frac{|Q_1| \cdot |Q_2|}{|H|}=\frac{q^{2n}}{|H|}.$

Thus $|H| \geq \frac{q^n}{p^2} > q^{n-2}.$ Therefore, since $|H| \mid q^n$ and $|H| < q^n,$ we’ll get $|H|=q^{n-1}.$ So $H \neq \{1\}$ is a normal subgroup of $Q_1$ and $Q_2.$ Hence $N_G(H)$ contains both $Q_1,Q_2$ and so $Q_1Q_2 \subseteq N_G(H).$ As a result $|N_G(H)|=p^kq^n,$ for some $k \leq 2.$ But then

$p^kq^n=|N_G(H)| \geq |Q_1Q_2|=q^{n+1},$

which gives us $p^k \geq q.$ Since $p < q,$ the only possibility would be $k=2.$ Thus $N_G(H)=G,$ i.e. $H \lhd G. \ \Box$