Frobenius' normal p-complement theorem

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This article gives the statement, and possibly proof, of a normal p-complement theorem: necessary and/or sufficient conditions for the existence of a Normal p-complement (?). In other words, it gives necessary and/or sufficient conditions for a given finite group to be a P-nilpotent group (?) for some prime number p.
View other normal p-complement theorems

Statement

Let G be a finite group and p be a prime number. Then, the following are equivalent:

  1. G has a p-Sylow subgroup P that is conjugacy-closed in G: any two elements of P that are conjugate in G are conjugate in P.
  2. There is a normal p-complement in G: a normal subgroup whose index is a power of p and whose order is relatively prime to p. In other words, any p-Sylow subgroup is a retract of G.
  3. For every non-identity p-subgroup Q of G, the subgroup N_G(Q) has a normal p-complement
  4. For every non-identity p-subgroup Q, the quotient N_G(Q)/C_G(Q) is a p-group.

Related facts

Facts used

  1. Index is multiplicative
  2. Second isomorphism theorem
  3. Retract implies conjugacy-closed
  4. Conjugacy-closed and Sylow implies retract

Proof

(2) implies (3)

(Note: This implication uses nothing about the special nature of normalizers of p-subgroups, and works for all subgroups).

Given: A finite group G, a prime p, a p-Sylow subgroup P of G. A normal subgroup M of G such that MP = G and M \cap P is trivial.

To prove: If Q is a non-identity p-subgroup of G, then N_G(Q) has a normal p-complement.

Proof: Let N = N_G(Q). By the second isomorphism theorem, we have:

NM/M \cong N/(N \cap M).

Since [NM:M] divides [G:M], and [G:M] equals the order of P, [NM:M] is a power of p. Thus, so is the order of the right side, [N:N \cap M].

Thus, N \cap M is a normal subgroup of N whose order is relatively prime to p (since it is also a subgroup of M) and whose index is a power of p. Thus, it is a normal p-complement in N.

(3) implies (4)

Given: A finite group G, a prime p, a p-Sylow subgroup P of G. For any non-identity p-subgroup Q, N_G(Q) has a normal p-complement.

To prove: If Q is a non-identity p-subgroup of G, then N_G(Q)/C_G(Q) is a p-group.

Proof: Proof: Let N = N_G(Q). Let L be a normal p-complement in N.

Now, Q and L are both normal subgroups of N and since the order of L is relatively prime to p, Q \cap L is trivial. Thus, L \le C_G(Q). Thus, by fact (1):

[N:L] = [N:C_G(Q)][C_G(Q):L].

Since the left side is a power of p, so are both terms of the right side. In particular, N_G(Q)/C_G(Q) is a p-group.

(2) implies (1)

This follows from fact (3).

(1) implies (2)

This follows from fact (4).

References

Textbook references

  • Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page Theorem 4.5, Chapter 7 (Fusion, transfer and p-factor groups),
    "Theorem" can not be assigned to a declared number type with value 4.5.
    "Theorem" can not be assigned to a declared number type with value 4.5.
    "Theorem" can not be assigned to a declared number type with value 4.5.
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