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

Posted: May 13, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

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<q 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


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