Sylow theory · Qual Corpus

Let $G$ be a finite group with $\abs{G}=p^km$, $p$ prime and $p\nmid m$. The Sylow theorems, proved with the actions of group actions and counting: $G$ has a subgroup of order $p^k$; every $p$-subgroup of $G$ is contained in one, and any two are conjugate; and their number $n_p$ divides $m$ and satisfies $n_p\equiv1\pmod{p}$. A Sylow $p$-subgroup is normal if and only if $n_p=1$. Frattini’s argument: if $N\trianglelefteq G$ and $P$ is a Sylow $p$-subgroup of $N$, then $G=N_G(P)\,N$.

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 $G$.

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 $p$.

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 $g\in G$.

Sylow's third theorem: numerical constraints (Tag T-3X5FF)

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

Let $G$ be a finite group and $p$ a prime, write $\abs G=p^a m$ with $p\nmid m$, and let $n_p$ be the number of Sylow $p$-subgroups of $G$. Then \[ n_p\mid m, \qquad n_p\equiv1\pmod p, \qquad n_p=[G:N_G(P)] \] for every Sylow $p$-subgroup $P$, where $N_G(P)$ is the normalizer of $P$.

Let $G$ act by conjugation on the set $\Syl_p(G)$ of Sylow $p$-subgroups. By Sylow's second theorem the action is transitive, and the stabilizer of $P$ is $N_G(P)$, so the orbit-stabilizer theorem gives $n_p=[G:N_G(P)]$. Since $P\leq N_G(P)$, $n_p$ divides $[G:P]=m$.

Now restrict the action to $P$. If $Q\in\Syl_p(G)$ is fixed by $P$, then $P\leq N_G(Q)$, so $P$ and $Q$ are Sylow $p$-subgroups of $N_G(Q)$ in which $Q$ is normal; by Sylow’s second theorem in $N_G(Q)$ they are conjugate there, so $P=Q$. Thus $P$ is the only fixed point, and every other $P$-orbit has size a positive power of $p$, so $n_p\equiv1\pmod p$.

Count Sylow subgroups before classifying a finite group (Tag ST-DKYXZ)

Let $G$ be a finite group. For each prime $p$ dividing $\abs G$, write $\abs G=p^am$ with $p\notdivides m$ and let $n_p$ be the number of Sylow $p$-subgroups of $G$.

List the divisors of $m$ that are congruent to $1$ modulo $p$; by the Sylow theorems these are the only possible values of $n_p$.
If $n_p=1$ is forced, the Sylow $p$-subgroup is normal.
If several values remain, count the nonidentity elements contributed by distinct Sylow subgroups. When $a=1$, distinct Sylow $p$-subgroups have order $p$, intersect trivially, and contribute $n_p(p-1)$ elements of order $p$; when $a\geq2$, bound the intersections of distinct Sylow $p$-subgroups separately before counting.
Use normal Sylow subgroups to form products of subgroups, and determine whether the conjugation action of one factor on another is trivial.

In overlapping element counts, count only nonidentity elements and add the identity once.

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 $G$.

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.

A group of order $pq$ with $p<q$ has a normal Sylow $q$-subgroup (Tag L-WROBY)

In order $pq$ with $p<q$, the Sylow $q$-subgroup is unique and normal.Suggest a slogan

Let $p<q$ be primes and let $G$ be a group with $\abs G=pq$. Then $G$ has a normal subgroup of order $q$.

Let $n_q$ be the number of Sylow $q$-subgroups of $G$. By Sylow's third theorem, $n_q\equiv1\pmod q$ and $n_q\divides p$. The divisors of $p$ are $1$ and $p$, and $1<p<q$ gives $p\not\equiv1\pmod q$, so $n_q=1$ and $G$ has a unique Sylow $q$-subgroup $Q$. For every $g\in G$, the conjugate $gQg^{-1}$ is again a Sylow $q$-subgroup, so $gQg^{-1}=Q$. Therefore $Q\trianglelefteq G$, and $\abs Q=q$.

Frattini's argument (Tag L-6QBOJ)

If $N\normal G$ and $P$ is Sylow in $N$, then $G=N_G(P)N$.Suggest a slogan

Let $G$ be a finite group, let $N$ be a normal subgroup of $G$, let $p$ be a prime, and let $P$ be a Sylow $p$-subgroup of $N$. Then $G=N_G(P)N$, where $N_G(P)$ is the normalizer of $P$ in $G$.

Let $g\in G$. Since $N$ is normal, $gPg^{-1}\subseteq N$, and it is a Sylow $p$-subgroup of $N$. By Sylow's second theorem in $N$, there is $n\in N$ with $ngPg^{-1}n^{-1}=P$, so $ng\in N_G(P)$. Hence $g=n^{-1}(ng)\in NN_G(P)=N_G(P)N$, the last equality because $N$ is normal.

Frattini's argument (Tag T-4XKGD)

A normal subgroup together with the normalizer of one of its Sylow subgroups generates the whole group.Suggest a slogan

Let $G$ be a finite group, $H\normal G$ a normal subgroup, $p$ a prime, and $P \in \Syl_p(H)$ a Sylow $p$-subgroup of $H$. Then $G=H\,N_G(P)$, and $[G: H]$ divides $\abs{N_G(P)}$.

Let $g\in G$. Since $H\normal G$, $gPg^{-1}\leq H$ is a Sylow $p$-subgroup of $H$, so by Sylow's second theorem in $H$ there is $h\in H$ with $hgPg^{-1}h^{-1}=P$. Then $hg\in N_G(P)$, so $g\in h^{-1}N_G(P)\subseteq H\,N_G(P)$. Hence $G=H\,N_G(P)$, and by the second isomorphism theorem $G/H\cong N_G(P)/(N_G(P)\cap H)$, so $[G:H]$ divides $\abs{N_G(P)}$.

$p$-groups (Tag D-FIB7S)

Let $p$ be a prime. A finite group $G$ is a $p$-group if $\abs{G} = p^k$ for some integer $k\geq 0$.

$p$-group (Tag FD-OR3BW)

Let $p$ be a prime. A $p$-group is a group of order $p^n$ for some integer $n\geq 1$.

What is a $p\dash$group?

Classification of groups of order $p^2$ (Tag PR-IGLFV)

Every group of order $p^2$ is abelian: cyclic or $C_p\times C_p$.Suggest a slogan

Let $p$ be a prime. Every group $G$ of order $p^2$ is abelian, and hence $G \cong C_{p^2}$ or $G \cong C_p \times C_p$.

Finite $p$-groups are solvable (Tag L-Q44MT)

Finite $p$-groups are solvable because their centers are nontrivial.Suggest a slogan

Let $p$ be a prime. Every finite $p$-group is solvable.

Induct on $\abs G$; the trivial group is solvable. A nontrivial finite $p$-group $G$ has nontrivial center $Z(G)$ by the class equation. The quotient $G/Z(G)$ is a $p$-group of smaller order, hence solvable by induction, and $Z(G)$ is an abelian normal subgroup. Pulling back a normal series of $G/Z(G)$ with abelian factors to $G$ and appending $Z(G)\trianglerighteq\theset{e}$ gives a normal series of $G$ with abelian factors.

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