Order is product of three distinct primes implies normal Sylow subgroup
From Groupprops
Statement
Suppose are three distinct prime numbers and
is a group of order
. Then, either the
-Sylow subgroup or the
-Sylow subgroup of
is normal in
. In particular,
has a normal Sylow subgroup.
Facts used
- 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
Proof
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).
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