Order is product of three distinct primes implies normal Sylow subgroup

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Suppose p < q < r are three distinct prime numbers and G is a group of order pqr. Then, either the q-Sylow subgroup or the r-Sylow subgroup of G is normal in G. In particular, G has a normal Sylow subgroup.

Facts used

  1. Number of elements of prime order for multiplicity-free prime divisor equals Sylow number times one less than the prime
  2. Congruence condition on Sylow numbers
  3. Divisibility condition on Sylow numbers


Given: Three distinct primes p < q < r. A group G of order pqr.

To prove: Either n_q = 1 or n_r = 1.

Proof: Let n_q, n_r denote the Sylow numbers for the primes q,r; in other words, n_q is the number of q-Sylow subgroups of G and n_r is the number of r-Sylow subgroups of G.

  1. By fact (1), the number of elements of order q is n_q(q-1) and the number of elements of order r is n_r(r-1). Since both these sets of elements are disjoint and they are all non-identity elements, we get n_q(q-1) + n_r(r-1) \le pqr - 1.
  2. n_r = 1 or n_r = pq: By fact (2), n_r|pq. Thus, n_r = 1, p, q, pq. By fact (3), n_r \equiv 1 \mod r. Thus, if n_r \ne 1, n_r > r, so n_r = pq.
  3. n_q = 1 or n_q \ge r: By fact (2), n_q|pr. Thus, n_q = 1,p,r,pr. By fact (3), n_q \equiv 1 \mod q. Thus, if n_q \ne 1, n_q > q. Thus n_q = r, pr. In either event, n_q \ge r.
  4. If n_q \ne 1 and n_r \ne 1, we have a contradiction: If n_q \ne 1, n_q \ge r, and if n_r \ne 1, n_r = pq. Thus, n_q(q-1) + n_r (r-1) \ge r(q - 1) + pq(r- 1) = pqr + r(q-1) - pq. Also, note that r(q - 1) = p(q - 1) + (r-p)(q-1) \ge p(q-1) + 2(q-1) > p(q-1) + p = pq. Thus, n_q(q-1) + n_r(r-1) > pqr, a contradiction to the conclusion of step (1).


Textbook references

  • Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 147, Section 4.5 (Sylow's theorem), Exercise 16, More info