Frobenius' normal p-complement theorem
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 .
View other normal p-complement theorems
Let be a finite group and be a prime number. Then, the following are equivalent:
- has a -Sylow subgroup that is conjugacy-closed in : any two elements of that are conjugate in are conjugate in .
- There is a normal p-complement in : a normal subgroup whose index is a power of and whose order is relatively prime to . In other words, any -Sylow subgroup is a retract of .
- For every non-identity -subgroup of , the subgroup has a normal p-complement
- For every non-identity -subgroup , the quotient is a -group.
- Conjugacy-closed and Sylow implies retract: This is the equivalence of (1) and (2) above.
- Conjugacy-closed Abelian Sylow implies retract: This is a weaker version of the equivalence of (1) and (2) above, that has a more direct proof.
- Equivalence of definitions of Sylow direct factor: This states that a conjugacy-closed normal Sylow subgroup is a direct factor. Note that the statement is again a weakening of the fact that any conjugacy-closed Sylow subgroup is a retract, and it again has a more direct proof.
- Burnside's normal p-complement theorem: A corollary of the fact that any conjugacy-closed Abelian Sylow subgroup is a retract.
- Index is multiplicative
- Second isomorphism theorem
- Retract implies conjugacy-closed
- Conjugacy-closed and Sylow implies retract
(2) implies (3)
(Note: This implication uses nothing about the special nature of normalizers of -subgroups, and works for all subgroups).
Given: A finite group , a prime , a -Sylow subgroup of . A normal subgroup of such that and is trivial.
To prove: If is a non-identity -subgroup of , then has a normal -complement.
Proof: Let . By the second isomorphism theorem, we have:
Since divides , and equals the order of , is a power of . Thus, so is the order of the right side, .
Thus, is a normal subgroup of whose order is relatively prime to (since it is also a subgroup of ) and whose index is a power of . Thus, it is a normal -complement in .
(3) implies (4)
Given: A finite group , a prime , a -Sylow subgroup of . For any non-identity -subgroup , has a normal -complement.
To prove: If is a non-identity -subgroup of , then is a -group.
Proof: Proof: Let . Let be a normal -complement in .
Now, and are both normal subgroups of and since the order of is relatively prime to , is trivial. Thus, . Thus, by fact (1):
Since the left side is a power of , so are both terms of the right side. In particular, is a -group.
(2) implies (1)
This follows from fact (3).
(1) implies (2)
This follows from fact (4).