Order is product of three distinct primes implies normal Sylow subgroup

Statement

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

Proof

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).

References

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}