Chapter 5 - Hall's Theorem for Soluble Groups Flashcards
For π={p1,…,pn} a finite set of primes define a π-group, a π-subgroup, and a Hall π-subgroup
A π-group is a group whose order is powers of primes in π
A π-subgroup is a subgroup of G which is a π-group
A Hall π-subgroup of G is a π-subgroup H whose index in G is coprime to all primes in π. In this respect it is the largest π-subgroup of G.
What is Halls Theorem for soluble groups?
Theorem:
Let G be a soluble group and π a set of primes. Then:
1) G has a Hall π-subgroup
2) Any two Hall π-subgroups are conjugate
3) Any π-subgroup of G is contained in a Hall π-subgroup
In the proof of Halls theorem for soluble groups, how do we subdivide the proof?
Let G be a soluble group of order m•n where m is a product of primes in π and n is a product of primes not in π. Then a Hall subgroup is one of order m. We split the proof into the cases:
Case (1): there exists a minimal normal subgroup M of order dividing M. Then use induction on order of G.
Case (2): there does not exist such a normal subgroup of G. Then we have to split into two cases depending on the quotient of G/M when M is any minimal normal subgroup. We use induction again.
Let p’ be the set of all primes not p. Then if G is a soluble group of order p^a•m where m is coprime to p, show we can split G into the product of a Hall p’ subgroup and one other group.
Well let H be a Hall p’ group of G (exists as G is soluble) and P a Sylow p-sbgp. Then P^H=1 as their orders are coprime, and thus
P•H =G
Let G be a group, H, K subgroups of G. Let a, b be the coprime index’ of H, K in G respectively. Then how that the index of H^K in G is a•b.
Let c be the index of H^K in G. Then show that a and b divide c. Since a, b are coprime, a•b also divides c.
Then define a map from cosets of H^K to the direct product of cosets of H and cosets of K. Showing this is injective shows c=<a•b
Define a Sylow basis of a group. If one exists then how can we directly extract all the possible Hall π-subgroups using this?
Let G be a group and p1,…,pk be the list of primes dividing the order of G. A Sylow basis for G is a set of Sylow pj subgroups Pj such that:
Pi•Pj=Pj•Pi for all i, j.
Noting that for any subgroups H,K of G, HK is a subgroup IFF HK=KH, we see that multiplying any of he Sylow basis subgroups together gives the respective Hall π- subgroup where π is the list of primes to which we have Sylow p-groups in our product. This given a basis we can construct a Hall π-subgroup for all possible orders.
What THM does Hall give us about Sylow bases?
Can we prove it?
THEOREM:
If G is soluble then it has a Sylow basis and any two such bases are conjugate by a single element in G.
PROOF:
Let pi^ai be the highest power of pi dividing the order of G for all i. And the ai are non zero only up to i=k.
For any i let Qi be the Hall pi’ subgroup of G. So that the index of Qi in G is pi^ai. An obvious induction argument shows that the intersection of any of the Qt gives another Hall π-subgroup (for which π?).
In particular, if we intersect all Qj with j not i, we see this is a Sylow pi-group called Pi. Then take any non equal i,j and their Pi, Pj. Then their intersection is trivial by orders, and so Pi•Pj has the same order as Pj•Pi. Then also, each group is contained in the intersection of all Qt, where we don’t have t=i or j. But this obviously has the same order as Pi•Pj, and so all three of these are equal. So these Pi give a Sylow basis for G.
Now suppose Bi is another Sylow basis for G. Then let Ct be the product of all Bi except Bt. Show, by induction on the number of t such that Qt=Ct, that there is a g such that for all t, Ct is conjugate under g to Qt.
Using this info, Bi is then the intersection of all Ct with i not t, which is the intersection of all the conjugate Qt, which is just the conjugate of Pi. This is true for each i.
Define a Sylow system
This is a collection of Hall subgroups, one of each possible order, obtained by a Sylow basis for G.
