LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27

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Fusion control and centralizer transitivity are equivalent

Statement

Let $H$ be a finite group, $p$ a prime, and $P \in {Syl}_{p} (H)$ (Sylow $p$ -subgroups of a finite group). The following are equivalent.

(i) $N_{H} (P)$ controls fusion in $P$ with respect to $H$ : whenever $x, y \in P$ and $y = h x h^{- 1}$ for some $h \in H$ , there is $u \in N_{H} (P)$ with $y = u x u^{- 1}$ . (ii) For every $x \in P$ with $x \neq e$ , the centralizer $C_{H} (x) = {h \in H : h x = x h}$ acts by conjugation transitively on the set ${Syl}_{p} (H; x) : = {T \in {Syl}_{p} (H) : x \in T}$ of Sylow $p$ -subgroups of $H$ containing $x$ ; that is, for any $T_{1}, T_{2} \in {Syl}_{p} (H; x)$ there is $c \in C_{H} (x)$ with $T_{2} = T_{1}^{c} = c T_{1} c^{- 1}$ .

Facts & Assumptions

Given: A finite group $H$ , a prime $p$ , a Sylow $p$ -subgroup $P \leq H$ , and the notation $x^{h} = h x h^{- 1}$ of The conjugacy class ${Cl}_{G} (x)$ and centralizer $C_{G} (x)$ of an element.

[F1]

Sylow $p$ -subgroups of $H$ are conjugate, and the conjugate of a Sylow $p$ -subgroup by any element of $H$ is again a Sylow $p$ -subgroup (Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class, Sylow $p$ -subgroups of a finite group).

[F2]

Conjugation $z \mapsto h z h^{- 1}$ is an automorphism, and $(z^{a})^{b} = z^{b a}$ , equivalently $z^{a b} = (z^{b})^{a}$ ; also $(z^{- 1})^{a} = (z^{a})^{- 1}$ (Conjugation $x \mapsto g x g^{- 1}$ is an automorphism, In a group $e^{- 1} = e$ , $(g^{- 1})^{- 1} = g$ and $(g h)^{- 1} = h^{- 1} g^{- 1}$ , the order of the last product being essential, The conjugacy class ${Cl}_{G} (x)$ and centralizer $C_{G} (x)$ of an element).

[F3]

$C_{H} (x)$ and $N_{H} (P)$ are subgroups of $H$ ; $z \in C_{H} (x)$ satisfies $x^{z} = x$ , and $u \in N_{H} (P)$ satisfies $P^{u} = P$ ( $C_{G} (x)$ and $N_{G} (H)$ are subgroups of $G$ , The centralizer $C_{G} (H)$ of a subgroup, The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup, Subgroup).

[F4]

If $T \in {Syl}_{p} (H)$ , $x \in T$ and $h \in H$ , then $T^{h} \in {Syl}_{p} (H)$ and $x^{h} \in T^{h}$ ; hence conjugation by $h$ carries ${Syl}_{p} (H; x)$ into ${Syl}_{p} (H; x^{h})$ , and if $c \in C_{H} (x)$ then $T^{c} \in {Syl}_{p} (H; x)$ for every $T \in {Syl}_{p} (H; x)$ (Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class, Sylow $p$ -subgroups of a finite group, Conjugation $x \mapsto g x g^{- 1}$ is an automorphism).

Proof

technique · direct

1.1

(ii) implies (i). Assume (ii), and let $x, y \in P$ , $h \in H$ with $y = x^{h} = h x h^{- 1}$ . If $x = e$ then $y = e = e^{e}$ with $e \in N_{H} (P)$ , so assume $x \neq e$ . Then $x = y^{h^{- 1}}$ by [F2], so $x \in P \cap P^{h^{- 1}}$ ; by [F1] and [F4] both $P$ and $P^{h^{- 1}}$ lie in ${Syl}_{p} (H; x)$ , so (ii) provides $c \in C_{H} (x)$ with $P^{h^{- 1}} = P^{c}$ .

F1F2F4assume-hyp

1.2

(i) implies (ii). Assume (i). Let $x \in P$ with $x \neq e$ and let $T \in {Syl}_{p} (H; x)$ . By [F1] there is $g \in H$ with $T = P^{g}$ ; then $x \in P^{g}$ gives $a : = x^{g^{- 1}} = g^{- 1} x g \in P$ , and $a^{g} = x$ by [F2], so $a$ and $x$ are $H$ -conjugate elements of $P$ and (i) provides $u \in N_{H} (P)$ with $a^{u} = x$ .

F1F2assume-hyp

2.1

The equality $P^{h^{- 1}} = P^{c}$ says $h^{- 1} P h = c P c^{- 1}$ . Multiplying on the left by $h$ and on the right by $h^{- 1}$ gives $P = (h c) P (h c)^{- 1}$ , so $n : = h c$ satisfies $P^{n} = P$ , that is $n \in N_{H} (P)$ .

F2F3step 1.1

3.1

Moreover $x^{n} = (h c) x (h c)^{- 1} = h (c x c^{- 1}) h^{- 1} = h x h^{- 1} = y$ , since $c$ centralizes $x$ . So $y$ is conjugate to $x$ by the element $n \in N_{H} (P)$ , which proves (i).

F3step 2.1

4.1

Put $v : = u g^{- 1}$ , so that $v^{- 1} = g u^{- 1}$ and $g = v^{- 1} u$ . The identity $a^{u} = x$ reads $u (g^{- 1} x g) u^{- 1} = x$ , that is $x^{v} = x$ by [F2]; hence $v \in C_{H} (x)$ . Moreover $T = P^{g} = P^{v^{- 1} u} = (P^{u})^{v^{- 1}} = P^{v^{- 1}}$ by [F2], since $u \in N_{H} (P)$ ; and $v^{- 1} \in C_{H} (x)$ . So every member $T$ of ${Syl}_{p} (H; x)$ equals $P^{c}$ for the element $c : = v^{- 1} \in C_{H} (x)$ , which is (ii). ∎

F2F3step 1.2

Depends on

Used by

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Paul Flavell, An Introduction to Transfer and Fusion in Finite Groups, §§2–5 (standard reference, not scraped)
Hans Kurzweil and Bernd Stellmacher, The Theory of Finite Groups, §§7.1–7.2 (standard reference, not scraped)