Equivalence of definitions of weakly closed conjugacy functor

This article gives a proof/explanation of the equivalence of multiple definitions for the term weakly closed conjugacy functor
View a complete list of pages giving proofs of equivalence of definitions

Statement

Suppose $G$ is a finite group, $p$ a prime number, and $W$ a conjugacy functor on $G$ with respect to $p$. The following are equivalent:

1. Either of these:
   • There exists a $p$-Sylow subgroup $P$ of $G$ such that $W(P)$ is a weakly closed subgroup of $P$ relative to $G$.
   • For every $p$-Sylow subgroup $P$ of $G$, $W(P)$ is a weakly closed subgroup of $P$ relative to $G$.
2. Either of these:
   • There exists a $p$-Sylow subgroup $P$ such that, for every $p$-Sylow subgroup $Q$ containing $W(P)$, $W(P)=W(Q)$.
   • For every $p$-Sylow subgroup $P$, and for every $p$-Sylow subgroup $Q$ containing $W(P)$, $W(P)=W(Q)$.
3. Either of these:
   • There exists a $p$-Sylow subgroup $P$ of $G$ such that for any $p$-Sylow subgroup $Q$ of $G$ containing $W(P)$, $W(P)$ is a normal subgroup of $Q$ (the fancy jargon for this is that $W(P)$ is a conjugation-invariantly relatively normal subgroup of $P$ in $G$).
   • For every $p$-Sylow subgroup $P$ of $G$, it is true that for any $p$-Sylow subgroup $Q$ of $G$ containing $W(P)$, $W(P)$ is a normal subgroup of $Q$ (the fancy jargon for this is that $W(P)$ is a conjugation-invariantly relatively normal subgroup of $P$ in $G$).

Related facts

Similar facts

• Characteristic subgroup of Sylow subgroup is weakly closed iff it is normal in every Sylow subgroup containing it

Applications

• Equivalence of normality and characteristicity conditions for conjugacy functor
• Equivalence of definitions of characteristic p-functor whose normalizer generates whole group with p'-core

Facts used

1. Sylow implies order-conjugate
2. Sylow implies WNSCDIN (used only in (3) implies (1) proof)
3. Conjugacy functor gives normalizer-relatively normal subgroup (used only in abstract version of (3) implies (1) proof)
4. WNSCDIN implies every normalizer-relatively normal conjugation-invariantly relatively normal subgroup is weakly closed (used only in abstract version of (3) implies (1) proof)

Proof

Preliminary notes

The equivalence between both versions of (1), the equivalence between both versions of (2), and the equivalence between both versions of (3), follow from Fact (1) (Sylow implies order-conjugate): any two $p$-Sylow subgroups are conjugate, and the conjugating automorphism preserves all properties including weak closure.

(1) implies (2)

Given: A finite group $G$, a prime number $p$, a $p$-conjugacy functor $W$ in $G$ and $p$-Sylow subgroups $P,Q$ of $G$ such that $W(P)$ is weakly closed in $P$ (with respect to $G$) and also $W(P)$ is contained in $Q$.

To prove: $W(P)=W(Q)$

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	There exists $g\in G$ such that $gQg^{-1}=P$	Fact (1)	$P,Q$ are $p$-Sylow subgroups of $G$		Given-fact direct
2	$gW(Q)g^{-1}=W(P)$		$W$ is a conjugacy $p$-functor, $P,Q$ are Sylow subgroups	Step (1)	Follows from definition of conjugacy functor and the preceding step
3	$W(Q)$ is a weakly closed subgroup of $Q$ with respect to $G$.		$W(P)$ is weakly closed in $P$ with respect to $G$	Steps (1), (2)	Apply the automorphism $x\mapsto g^{-1}xg$. This sends $P$ to $Q$ and $W(P)$ to $W(Q)$. Thus, $W(P)$ being weakly closed in $P$ implies that $W(Q)$ is weakly closed in $Q$.
4	$W(P)=W(Q)$		$W(P)\leq Q$	Steps (2), (3)	By Step (2) and the given, $gW(Q)g^{-1}=W(P)\leq Q$. By Step (3), this implies that $gW(Q)g^{-1}=W(Q)$, so that $W(P)=W(Q)$.

(2) implies (3)

Given: A finite group $G$, a prime number $p$, a $p$-conjugacy functor $W$ in $G$, $p$-Sylow subgroups $P,Q$ satisfying $W(P)=W(Q)$.

To prove: $W(P)$ is normal in $Q$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$W(Q)$ is normal in $Q$		$W$ is a $p$-conjugacy functor		For any $g\in G$, we must have, by the definition of conjugacy functor, that $W(hQg^{-1})=hW(Q)h^{-1}$. In particular, this is also true for any $h\in Q$, giving that for any $h\in Q$, $W(Q)=hW(Q)h^{-1}$. Hence, $W(Q)$ is normal in $Q$.
2	$W(P)$ is normal in $Q$.		$W(P)=W(Q)$	Step (1)	given-step direct

(3) implies (1) (concrete proof)

Given: A finite group $G$, a prime number $p$, a conjugacy functor $W$ and a $p$-Sylow subgroup $P$ of $G$ such that for any $p$-Sylow subgroup $Q$ of $G$ containing $W(P)$, $W(P)$ is normal in $Q$. $g\in G$ is such that $gW(P)g^{-1}\leq P$.

To prove: $gW(P)g^{-1}=W(P)$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Let $Q=g^{-1}Pg$. Then, $Q$ is a $p$-Sylow subgroup of $G$.		$P$ is $p$-Sylow in $G$.
2	$W(Q)=g^{-1}W(P)g$		$W$ is a conjugacy functor for the prime $p$	Step (1)	By the given, $W(g^{-1}Pg)=g^{-1}W(P)g$, and by Step (1), the left side is $W(Q)$.
3	$W(P)$ is contained in $Q$.		$gW(P)g^{-1}\leq P$	Step (1)	$gW(P)g^{-1}\leq P$ rearranges to $W(P)\leq g^{-1}Pg=Q$.
4	$W(P)$ is normal in $Q$.		For any $p$-Sylow subgroup $Q$ of $G$ containing $W(P)$, $W(P)$ is normal in $Q$.	Step (3)	Step-given direct
5	$W(Q)$ is normal in $Q$		$W$ is a conjugacy functor for the prime $p$.		For any <mathhg \in G</math>, we must have, by the definition of conjugacy functor, that $W(hQg^{-1})=hW(Q)h^{-1}$. In particular, this is also true for any $h\in Q$, giving that for any $h\in Q$, $W(Q)=hW(Q)h^{-1}$. Hence, $W(Q)$ is normal in $Q$.
6	$W(P)$ and $W(Q)$ are both normal subgroups of the $p$-Sylow subgroup $Q$ of $G$that are conjugate in $G$.			Steps (2), (4), (5)	Steps (4) and (5) give the normality of the two subgroups. Step (2) shows that they are conjugate in $G$.
7	There exists $u\in N_{G}(Q)$ such that $uW(Q)u^{-1}=W(P)$.	Fact (2)		Step (6)	Fact-step combination direct
8	$uW(Q)u^{-1}=W(Q)$		$W$ is a conjugacy functor for the prime $p$.	Step (7)	Since $W$ is a conjugacy functor, $uW(Q)u{-1}=W(uQu^{-1})$. By the preceding step, $u\in N_{G}(Q)$, so the right side simplifies as $W(Q)$.
9	$W(P)=W(Q)$			Steps (7), (8)	Step-combination direct

(3) implies (1) (abstract proof)

Given: Finite group $G$, prime $p$, $p$-conjugacy functor $W$, $p$-Sylow subgroup $P$ of $G$. $W(P)$ is conjugation-invariantly relatively normal in $P$.

To prove: $W(P)$ is weakly closed in $P$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$W(P)$ is normalizer-relatively normal in $P$ with respect to $G$, i.e., $W(P)$ is normal in $N_{G}(P)$.	Fact (3)	$W$ is a $p$-conjugacy functor, $P$ is a $p$-Sylow subgroup.
2	$W(P)$ is weakly closed in $P$ with respect to $G$.	Facts (2), (4)	$P$ is $p$-Sylow in $G$, $W(P)$ is conjugation-invariantly relatively normal in $P$.	Step (1)	By Fact (2), $P$ is WNSCDIN in $G$ (what this means does not matter here). By the given and Step (1), $W(P)$ is conjugation-invariantly relatively normal and normalizer-relatively normal in $P$ with respect to $G$. Fact (4) yields that $W(P)$ is weakly closed in $P$.