Order is product of three distinct primes implies normal Sylow subgroup
- Number of elements of prime order for multiplicity-free prime divisor equals Sylow number times one less than the prime
- Congruence condition on Sylow numbers
- Divisibility condition on Sylow numbers
Given: Three distinct primes . A group of order .
To prove: Either or .
Proof: Let denote the Sylow numbers for the primes ; in other words, is the number of -Sylow subgroups of and is the number of -Sylow subgroups of .
- By fact (1), the number of elements of order is and the number of elements of order is . Since both these sets of elements are disjoint and they are all non-identity elements, we get .
- or : By fact (2), . Thus, . By fact (3), . Thus, if , , so .
- or : By fact (2), . Thus, . By fact (3), . Thus, if , . Thus . In either event, .
- If and , we have a contradiction: If , , and if , . Thus, . Also, note that . Thus, , a contradiction to the conclusion of step (1).
- 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