Sylow theory · Qual Corpus

Throughout, $G$ is a finite group, $p$ is a prime, $\size G = p^k m$ with $p\notdivides m$, $\Syl_p(G)$ is the set of Sylow $p$-subgroups of $G$, and $n_p = \size{\Syl_p(G)}$.

Sylow $p\dash$subgroup (Tag D-7TQ2M)

Let $G$ be a finite group and $p$ a prime, and write $\abs G = p^a m$ with $a\geq 0$ and $p \nmid m$. A subgroup $S \leq G$ is a Sylow $p\dash$subgroup of $G$ if $\abs S = p^a$.

Let $G$ be a finite group and $p$ a prime. A subgroup $S\leq G$ is a Sylow $p\dash$subgroup if and only if $S$ is a $p\dash$group that is maximal under inclusion among the $p\dash$subgroups of $G$.

(Counting Sylow subgroups)

$n_p\divides m$ and $n_p\equiv1\pmod p$; the case $n_p=1$ is the normality criterion in Conjugacy.
If $k=1$, distinct Sylow $p$-subgroups intersect trivially, so they contain exactly $n_p(p-1)$ elements of order $p$. For $k>1$, two distinct Sylow $p$-subgroups can intersect nontrivially.

Fixed-point congruence for actions of $p$-groups (Tag L-354HC)

A finite $p$-group action has total points congruent mod $p$ to its global fixed points.Suggest a slogan

Let $p$ be a prime, let $G$ be a finite $p$-group, and let $G$ act on a finite set $X$. Let $\Fix(G)\coloneqq\theset{x\in X\suchthat gx=x\text{ for all }g\in G}$. Then \[ \abs X\equiv\abs{\Fix(G)}\pmod p. \]

The orbits of $G$ partition $X$. A point of $\Fix(G)$ is an orbit of size $1$. Every other orbit $Gx$ has size $[G:G_x]>1$ by the orbit-stabilizer theorem, and this index divides $\abs G$, so it is a positive power of $p$. Reducing $\abs X=\abs{\Fix(G)}+\sum\abs{Gx}$, the sum over the orbits of size greater than $1$, modulo $p$ gives the congruence.

The Sylow theorems

Sylow theorems (Tag FT-ZENUU)

Sylow $p$-subgroups exist, are conjugate, and their number divides the prime-to-$p$ part while equaling $1$ modulo $p$.Suggest a slogan

Let $G$ be a finite group and let $p$ be a prime. Write $\abs G=p^am$ with $a\ge0$ and $p\nmid m$, and let $n_p$ be the number of Sylow $p$-subgroups of $G$.

$G$ has a Sylow $p$-subgroup, that is, a subgroup of order $p^a$.
Any two Sylow $p$-subgroups of $G$ are conjugate in $G$.
$n_p\divides m$, $n_p\equiv1\pmod p$, and $n_p=[G:N_G(P)]$ for every Sylow $p$-subgroup $P$, where $N_G(P)$ is the normalizer of $P$.

State the three Sylow theorems.

Existence

Sylow's first theorem: existence of prime-power subgroups (Tag T-WRMBM)

Every power of $p$ dividing $\abs G$ is the order of a subgroup, and every $p$-subgroup lies in a Sylow $p$-subgroup.Suggest a slogan

Let $G$ be a finite group and $p$ a prime, and write $\abs G=p^a m$ with $p\nmid m$. Then $G$ has a subgroup of order $p^n$ for every $0\leq n\leq a$; in particular, $G$ has a Sylow $p$-subgroup, of order $p^a$. Moreover, every $p$-subgroup of $G$ is contained in a Sylow $p$-subgroup [@DF04].

For $a\geq1$ this strengthens Cauchy's theorem, which gives a subgroup of order $p$.

For every $\beta \leq k$, $G$ has a subgroup of order $p^{\beta}$; in particular Sylow $p$-subgroups exist. Every $p$-subgroup of $G$ is contained in a Sylow $p$-subgroup.

Conjugacy

Sylow's second theorem: Sylow subgroups are conjugate (Tag T-EF2MZ)

All Sylow $p$-subgroups are conjugate; uniqueness is exactly normality.Suggest a slogan

Let $G$ be a finite group and $p$ a prime. Any two Sylow $p$-subgroups of $G$ are conjugate: if $P,Q\in\Syl_p(G)$, then there exists $g\in G$ with $gPg^{-1}=Q$. In particular, a Sylow $p$-subgroup $P$ is normal in $G$ if and only if it is the unique Sylow $p$-subgroup of $G$ [@DF04].

Write $\abs G=p^a m$ with $p\nmid m$, and let $P$ and $Q$ be Sylow $p$-subgroups of $G$. Let $P$ act on the set $G/Q$ of left cosets of $Q$ by left translation, $x\cdot gQ=xgQ$. Since $\abs{G/Q}=m$ is prime to $p$, the fixed-point congruence gives $\abs{(G/Q)^P}\equiv m\not\equiv 0\pmod p$, so some coset $gQ$ is fixed by $P$. Then $xgQ=gQ$ for every $x\in P$, that is, $g^{-1}Pg\subseteq Q$, and since both groups have order $p^a$, $g^{-1}Pg=Q$.

For the second statement, every conjugate $gPg^{-1}$ has order $p^a$ and is therefore a Sylow $p$-subgroup. If $P\normal G$, every conjugate of $P$ equals $P$, so by the first statement $P$ is the only Sylow $p$-subgroup. Conversely, if $P$ is the only Sylow $p$-subgroup, then $gPg^{-1}=P$ for every $g\in G$.

A Sylow $p$-subgroup $P$ is normal if and only if $n_p = 1$.

Every Sylow $p$-subgroup is conjugate to $P$, so $\Syl_p(G) = \ts{gP\inverseof{g} \st g\in G}$, and this set is $\ts P$ if and only if $P\normal G$.

The number of Sylow subgroups

Sylow's third theorem: the number of Sylow subgroups (Tag T-S5T5C)

The Sylow count divides the prime-to-$p$ part, is $1$ modulo $p$, and equals a normalizer index.Suggest a slogan

Let $G$ be a finite group and $p$ a prime, write $\abs G = p^a m$ with $p\nmid m$, let $n_p$ be the number of Sylow $p$-subgroups of $G$, and let $S_p$ be one of them. Then:

$n_p \divides m$, and in particular $n_p \leq m$.
$n_p \equiv 1 \pmod p$.
$n_p = [G : N_G(S_p)]$, where $N_G(S_p)$ is the normalizer of $S_p$ in $G$.

(The conditions on $n_p$)

For $P\in\Syl_p(G)$, \[ n_p \equiv 1 \pmod p, \qquad n_p \divides m, \qquad n_p = [G : N_G(P)]. \] The equality $n_p = [G : N_G(P)]$ is orbit-stabilizer for the transitive conjugation action of $G$ on $\Syl_p(G)$. The congruence follows from the conjugation action of $P$ on $\Syl_p(G)$: its only fixed point is $P$, and every other orbit has size divisible by $p$.

Applications

Arguments producing normal subgroups from Sylow counts are collected on Show $G$ is not simple.

Classifying groups of a given order. If $n_p = 1$, then $P\in\Syl_p(G)$ is normal, $\size{G/P} = m$ is coprime to $\size P$, and by the Schur–Zassenhaus theorem $P$ has a complement $H$, so $G \cong P\semidirect H$.
Induction on the order. A normal Sylow subgroup $P$ gives the quotient $G/P$ of order $m < \size G$.
Nilpotence. A finite group is nilpotent if and only if every Sylow subgroup is normal, in which case it is the direct product of its Sylow subgroups.

Exercises

Normal Sylow $p$-subgroup when $|G|=p^{e}v$ with $p>v$ (Tag E-K3OJW)

Let $G$ be a finite group of order $p^e v$, where $e$ and $v$ are positive integers, $p$ is prime, $p>v$, and $p\nmid v$. Show that $G$ has a normal Sylow p-subgroup.

Solution

Let $n_p$ denote the number of Sylow $p$-subgroups of $G$.

<1>1. Sylow’s theorem gives \[ n_p\mid v, \qquad n_p\equiv1\pmod p. \]

Since \[ |G|=p^e v \] with $p\nmid v$, a Sylow $p$-subgroup has order $p^e$. The third Sylow theorem says that the number of such subgroups divides the complementary factor $v$ and is congruent to $1$ modulo $p$.

<1>2. One has $n_p=1$.

Because $n_p\mid v$, we have \[ 1\le n_p\le v<p. \] The only positive integer strictly less than $p$ that is congruent to $1$ modulo $p$ is $1$. Thus $n_p=1$.

<1>3. Therefore the Sylow $p$-subgroup is normal.

All Sylow $p$-subgroups are conjugate. Since there is exactly one, it is fixed by conjugation by every element of $G$, hence is normal.

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