Hall subgroups exist in finite solvable group
Statement
Suppose is a finite solvable group, and
is a prime set. Then, there exists a
-Hall subgroup (?) of
: a subgroup whose order and index are relatively prime, and with the property that the set of prime divisors of its order is within
.
Related facts
- Hall's theorem is a converse of sorts.
- Sylow's theorem: Sylow subgroups exist, Sylow implies order-conjugate, Sylow implies order-dominating, congruence condition on Sylow numbers
- ECD condition for pi-subgroups in solvable groups: Hall subgroups exist in finite solvable, Hall implies order-conjugate in finite solvable, Hall implies order-dominating in finite solvable, congruence condition on factorization of Hall numbers
- Pi-Hall subgroups exist in pi-separable
Facts used
- Solvability is subgroup-closed
- Solvability is quotient-closed
- Minimal normal implies elementary Abelian in finite solvable
- Sylow subgroups exist
- Normality is quotient-transitive
- Frattini's argument
- Product formula
Proof
Given: A finite solvable group , a prime set
.
To prove: has a
-Hall subgroup.
Proof: We prove the claim by induction on the order of . The base case of the trivial group is clear.
Suppose the result is true for all finite solvable groups of order strictly less than the order of . Then, by facts (1) and (2), the result is true for all proper subgroups and all proper quotients of
.
Let be a minimal normal subgroup of
. By fact (3),
is elementary Abelian, and in particular, is a
-group for some prime divisor of
. Observe that:
- Case
: By applying induction to
, we conclude that
has a
-Hall subgroup. Taking the inverse image of this in
, we get a
-Hall subgroup of
.
- Case
: By applying induction to
, we conclude that
has a
-Hall subgroup. The inverse image of this gives a subgroup, say
, of
. If
is not itself a
-group, then
is proper in
, and
is a proper subgroup of
, then
has a
-Hall subgroup
. Note that
is relatively prime to
, and
is relatively prime to
, so
is relatively prime to
. Thus,
is a Hall
-subgroup of
.
Thus, the only case of concern is where but
is a
-group. Note that in this case,
must be a
-Sylow subgroup (since
has order relatively prime to
).
Let be a subgroup of
such that
is a minimal normal subgroup of
. Then, since the order of
is not a multiple of
, and
is solvable, fact (3) yields that
is an elementary Abelian
-group for some prime
.
By fact (5),
is itself a normal subgroup of
.
Let be a
-Sylow subgroup of
(using fact (4)).
Now, if is normal in
, then argue as before replacing
by
. Note that by our assumption,
, so we land up in case (1).
Let's now consider the case that is not normal in
. By Frattini's argument (fact (6)),
. The product formula yields:
, so
divides
, and is hence a power of
. Thus,
is a power of
. Thus,
is a proper subgroup of
whose index is a power of
. By induction,
has a
-Hall subgroup, and this must also be a
-Hall subgroup for
.
References
Textbook references
- Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Page 200, Exercise 33, Section 6.1 (
-groups, nilpotent groups and solvable groups)