## Groups of order pqr are not simple

Posted: January 18, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Problem. Suppose that $p,q, r$ are three prime numbers which are not necessarily distinct. Prove that if $G$ is a group of order $pqr,$ then $G$ is not simple.

Proof. I proved in this post that a group of order $p^2q,$ where $p,q$ are (not necessarily distinct) primes is not simple. So I may assume that $p,q,r$ are distinct and, say, $p < q < r$. For any prime number $s$ let $n_s$ be the number of Sylow $s$-subgroups of $G.$ Now suppose that $G$ is simple. So $n_r > 1$ and, by the Sylow theorem, $n_r \mid pq.$  But we cannot have $n_r = p$ or $n_r = q$ because $n_r \equiv 1 \mod r$ and hence $n_r > r > q > p.$ Thus $n_r = pq$ and so, since every Sylow $r$-subgroup of $G$ has exactly $r-1$ elements of order $r,$ the number of elements of order $r$ in $G$ is $pq(r-1)=|G|-pq.$ We also have $n_q > 1$ and $n_q > q > p.$ Thus there are more than $p(q-1)=pq-p$ elements of order $q.$ Hence the number of elements of order $p$ is less than $|G|-(|G|-pq+pq-p)=p,$ which is impossible because $n_p > 1.$ $\Box$

1. Nicholas says:

Quick question: Why does assuming that G is simple give n_s > 1 ? I know that for a group to be simple, it has no proper and non-identity normal subgroups, but how does that lead us to say n_s >1 ?

• Yaghoub Sharifi says:

In general, If a (finite) group G has only one subgroup H of a given order, then H is normal in G because for every g in G we have |H|=|gHg^{-1}| and so H=gHg^{-1}.

2. Ivan Malison says:

Why is it that you can assume that the intersection of all the conjugate Sylow-r subgroups is only the identity element. (You make that assumption when you say that the number of elements of order r in G is pq(r-1))

• Yaghoub says:

If H and K are two distinct r-Sylow subgroups of G, then |H| = |K| = r. Let L be the intersection of H and K. Then L is a subgroup of both H and K. So |L| divides r = |H| = |K|. But r is a prime number and so either |L| = 1 or |L| = r = |H| = |K|. If |L| = 1, we are done. Otherwise, since L is a subgroup of H and K and |L|= |H| = |K|, we get L = H = K, which is false because H and K were distinct.