Hall subgroups exist in finite solvable group

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Statement

Suppose G is a finite solvable group, and \pi is a prime set. Then, there exists a \pi-Hall subgroup (?) of G: a subgroup whose order and index are relatively prime, and with the property that the set of prime divisors of its order is within \pi.

Related facts

Facts used

  1. Solvability is subgroup-closed
  2. Solvability is quotient-closed
  3. Minimal normal implies elementary Abelian in finite solvable
  4. Sylow subgroups exist
  5. Normality is quotient-transitive
  6. Frattini's argument
  7. Product formula

Proof

Given: A finite solvable group G, a prime set \pi.

To prove: G has a \pi-Hall subgroup.

Proof: We prove the claim by induction on the order of G. 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 G. Then, by facts (1) and (2), the result is true for all proper subgroups and all proper quotients of G.

Let N be a minimal normal subgroup of G. By fact (3), N is elementary Abelian, and in particular, is a p-group for some prime divisor of G. Observe that:

  1. Case p \in \pi: By applying induction to G/N, we conclude that G/N has a \pi-Hall subgroup. Taking the inverse image of this in G, we get a \pi-Hall subgroup of G.
  2. Case p \notin \pi: By applying induction to G/N, we conclude that G/N has a \pi-Hall subgroup. The inverse image of this gives a subgroup, say K, of G. If G/N is not itself a \pi-group, then K/N is proper in G/N, and K is a proper subgroup of G, then K has a \pi-Hall subgroup H. Note that [G:K] = [G/N:K/N] is relatively prime to \pi, and [K:H] is relatively prime to \pi, so [G:H] is relatively prime to \pi. Thus, H is a Hall \pi-subgroup of G.

Thus, the only case of concern is where p \notin \pi but G/N is a \pi-group. Note that in this case, N must be a p-Sylow subgroup (since G/N has order relatively prime to p).

Let M be a subgroup of G such that M/N is a minimal normal subgroup of G/N. Then, since the order of G/N is not a multiple of p, and G/N is solvable, fact (3) yields that M/N is an elementary Abelian q-group for some prime q \ne p. By fact (5), M is itself a normal subgroup of G.

Let Q be a q-Sylow subgroup of M (using fact (4)).

Now, if Q is normal in G, then argue as before replacing N by Q. Note that by our assumption, q \in \pi, so we land up in case (1).

Let's now consider the case that Q is not normal in G. By Frattini's argument (fact (6)), MN_G(Q) = G. The product formula yields:

\frac{|G|}{|N_G(Q)|} = \frac{|M|}{|M \cap N_G(Q)|}

Q \le M \cap N_G(Q), so |M|/(|M \cap N_G(Q)|) divides |M|/|Q|, and is hence a power of p. Thus, |G|/|N_G(Q)| is a power of p. Thus, N_G(Q) is a proper subgroup of G whose index is a power of p. By induction, N_G(Q) has a \pi-Hall subgroup, and this must also be a \pi-Hall subgroup for G.

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 (p-groups, nilpotent groups and solvable groups)