Groups of order pqr are not simple

Posted: January 18, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

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

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Comments
  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.

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