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Abelian sylow fusion in its normalizer

Statement

Let $G$ be a finite group, $p$ a prime and $P \in {Syl}_{p} (G)$ an abelian Sylow $p$ -subgroup. If $x, y \in P$ are conjugate in $G$ , then $x$ and $y$ are conjugate in $N_{G} (P)$ : there is $u \in N_{G} (P)$ with $y = u x u^{- 1}$ .

Facts & Assumptions

Given: A finite group $G$ , a prime $p$ , an abelian $P \in {Syl}_{p} (G)$ , elements $x, y \in P$ and $g \in G$ with $y = g x g^{- 1}$ .

[F1]

Write $∣ G ∣ = p^{a} m$ with $p ∤ m$ ; then $∣ P ∣ = p^{a}$ , and a subgroup $H \leq G$ is a Sylow $p$ -subgroup of $G$ exactly when $∣ H ∣ = p^{a}$ ; a $p$ -subgroup of $G$ of order $p^{a}$ is a Sylow $p$ -subgroup (Sylow $p$ -subgroups of a finite group, Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[F2]

Every $p$ -subgroup of $G$ is contained in a Sylow $p$ -subgroup of $G$ , and any two Sylow $p$ -subgroups of $G$ are conjugate: for every Sylow $Q$ there is $u \in G$ with $Q = u P u^{- 1}$ (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).

[F3]

As $P$ is abelian, every element of $P$ commutes with $x$ and with $y$ ; that is, $P \leq C_{G} (y)$ and $P \leq C_{G} (x)$ , where $C_{G} (y) = {u \in G : u y = y u}$ (The centralizer $C_{G} (H)$ of a subgroup, The conjugacy class ${Cl}_{G} (x)$ and centralizer $C_{G} (x)$ of an element).

[F4]

For every $u \in G$ the map $c_{u} (z) = u z u^{- 1}$ is an automorphism of $G$ , so $c_{u} (z w) = c_{u} (z) c_{u} (w)$ , $c_{u} (z)^{- 1} = c_{u} (z^{- 1})$ and $c_{u}$ is injective; also $u \in N_{G} (P)$ exactly when $u P u^{- 1} = P$ (Conjugation $x \mapsto g x g^{- 1}$ is an automorphism, The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup).

[F5]

$C_{G} (y)$ and $N_{G} (P)$ are subgroups of $G$ ( $C_{G} (x)$ and $N_{G} (H)$ are subgroups of $G$ ).

Proof

technique · direct

1.1

$P^{g} = g P g^{- 1}$ is contained in $C_{G} (y)$ : for $w \in P$ one has $z : = g w g^{- 1} \in P^{g}$ and $y = g x g^{- 1}$ , so $z y = g w g^{- 1} g x g^{- 1} = g (w x) g^{- 1} = g (x w) g^{- 1} = y z$ by [F4] and the commutativity of $P$ in [F3]. Also $P \leq C_{G} (y)$ by [F3].

F3F4given

2.1

Both $P$ and $P^{g}$ are Sylow $p$ -subgroups of $C_{G} (y)$ : each has order $p^{a}$ by [F1] and [F4], and each is contained in $C_{G} (y)$ by step 1.1; since $∣ C_{G} (y) ∣$ divides $∣ G ∣ = p^{a} m$ by [F1], the exact power of $p$ dividing $∣ C_{G} (y) ∣$ is $p^{a}$ , so a subgroup of $C_{G} (y)$ of order $p^{a}$ is a Sylow $p$ -subgroup of $C_{G} (y)$ by [F1] applied to $C_{G} (y)$ .

F1F2F5step 1.1

3.1

By step 2.1 and Sylow conjugacy inside the finite group $C_{G} (y)$ [F2], there is $c \in C_{G} (y)$ with $(P^{g})^{c} = P$ ; since $(P^{g})^{c} = c (g P g^{- 1}) c^{- 1} = (c g) P (c g)^{- 1}$ , [F4] gives $c g \in N_{G} (P)$ .

F2F4step 2.1

4.1

For this $c$ one has $(x^{g})^{c} = y^{c} = y$ , the first equality by [F4] applied twice and the second because $c \in C_{G} (y)$ .

F4step 3.1

5.1

Since $(x^{g})^{c} = x^{c g} = y$ by step 4.1, $x$ and $y$ are conjugate by the element $c g \in N_{G} (P)$ , as claimed. ∎

step 3.1step 4.1

Depends on

Used by

Burnside normal p complement theorem Theorem

Dependency tree · two levels

31 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)