04.5.シローの定理

Theorem

シローの定理

Statement

$G$:Group
$|G| \lt \infty$
$|G| = p^a m \quad (p,m \in \mathbb{Z}, p:\text{Prime Number}, \gcd (m,p) = 1)$
then
I
$\exists P:\text{Subgroup of }G, |P| = p^a$
II
$P$:p-Sylow Subgroup of $G$
$Q$:Subgroup of $G$
$|Q| = p^b \quad (1 \le b \le a)$
then
$\exists g \in G, Q \subset g P g^{-1}$
III
$P,Q$:p-Sylow Subgroup of $G$
then
$P,Q$ is 共役 each other

Proof

I
Let
$\mathbb{X} = \{X \subset G | |X| = p^a\}$
$|\mathbb{X}| = _{p^am}C_{p^a}$
generally
$_m C_k =$ coefficient of $X^k$ in $(X + 1)^m$
$\therefore _{p^a m} C_{p^a} =$ coefficient of $X^{p^a}$ in $(X + 1)^{p^a m}$
$(X + 1)^p = X^p + _p C_{p-1} X^{p -1} + \cdots + _p C_1 X + 1$
$\equiv X^p + 1 \quad (\because _p C_i \equiv 0 (\text{mod }p) \quad 0 \lt i \lt p)$
$\therefore (X + 1)^{p^a m} \equiv (X^{p^a} + 1)^m$
$\equiv X^{p^a m} + _mC_{m-1} X^{p^a (m - 1)} + \cdots + _mC_1 X^{p^a} + 1$
$\therefore _{p^a m} C_{p^a} \equiv m \quad (\text{mod }p)$
$\therefore |\mathbb{X}| \equiv m \quad (\text{mod }p)$
$\text{gcd }(|\mathbb{X}|, p) = 1$---(1)

(別)
$_{p^am}C_{p^a} = \prod_{i=1}^{p^a}\frac{p^am - (k-1)}{p^a - (k-1)}$
let $k-1 = p^b n \quad (\text{gcd }(p,n) = 1, 0 \le b \lt a)$
then $\frac{p^am - (k-1)}{p^a - (k-1)} = \frac{p^am - p^bn}{p^a - p^b n} = \frac{p^b(p^{a-b}m - n)}{p^b(p^{a-b} - n)} = \frac{p^{a-b}m - n}{p^{a-b} - n}$
$\therefore \text{gcd }(p^{a-b}m - n,p) = 1 \quad (\because \text{gcd }(p,n) = 1)$
$\therefore \text{gcd }(_{p^am}C_{p^a}, p) = 1$
$\text{gcd }(|\mathbb{X}|, p) = 1$---(1)

define action $G$ as
$G \times \mathbb{X} \rightarrow \mathbb{X}$
$X \mapsto g\cdot X = \{gx | x \in X\}$
$\mathbb{X} = \coprod_{\bar X \in \mathbb{X}/\sim} G \cdot X \quad (G\cdot X = \{g\cdot X|g \in G\} = \bar X)$
$\therefore |\mathbb{X}| = \sum_{\bar X \in \mathbb{X}/\sim} |G\cdot X| \quad (\because \text{Prop.軌道による分割})$---(2)

$\exists X \in \mathbb{X}, \text{gcd}(|G \cdot X|,p) = 1$---(3)

fix $X$, such that $\text{gcd }(|G \cdot X|,p) = 1$
$|G \cdot X| = \frac{|G|}{|G_X|} \quad (\because \text{Prop.安定化群による剰余類})$
$|G| = p^a m$ and (3) lead to $|G_X| = p^a l \quad (1 \le l \le m$)---(4)

fix $x \in X$
define $f_x$ as
$f_x:G_X \rightarrow X$
$f_x:g \in G_X \mapsto gx \in X$
then
$f_x$ is injective.
$\because f_x(g)=f_x(h)$
$\Leftrightarrow gx = hx$
$\Leftrightarrow g = h$
$\therefore |G_X| \le |X| = p^a$---(5)

(4),(5) leads to $|G_X| = p^a$

II
Let
$X = G/P$
Define
$\cdot:Q \times X \Rightarrow X$
$q \cdot gP:(q, gP) \mapsto qgP \in X$
then
$\cdot$ is an action over $X$

$|G| = |G/P||P|$
$|G| = p^a m, |P| = p^a$
$\therefore |X| = |G/P| = m \quad \text{gcd}(m,p) = 1$---(1)

Let
$x = gP \in X$
$\bar x = Q\cdot x := \{q \cdot x|q \in Q\} \subset X \quad$(orbit of $x$)
$|Q \cdot x||Q_x| = |Q| = p^b \quad (\because \text{Prop.安定化群による剰余類})$
$\therefore |Q \cdot x| = p^{b_x} \quad (0 \le b_x \le b)$---(2)

$X = \coprod Q \cdot x$
$|X| = \sum |Q \cdot x|$---(3)

$\exists x \in X,\text{ s.t. } |Q \cdot x| = 1 = p^0$
$\because (\forall x \in X,b_x \gt 0) \Rightarrow (\forall x \in X,\text{gcd}(|Q \cdot x|,p) = p)$
$\Rightarrow (\text{gcd}(|X|,p) = p)$ contradicts to (1)

$\Leftrightarrow \exists Q \cdot x = \{x\}$
$\Leftrightarrow \exists x \text{ s.t. } \forall q \in Q, q \cdot x = x \in X = G/P$
$\Leftrightarrow \exists gP \text{ s.t. } \forall q \in Q, q \cdot gP = gP$
$\Leftrightarrow \exists g \text{ s.t. } \forall q \in Q, q \cdot gP = gP$
$\therefore Qg \subset QgP = gP$
$\therefore Q \subset gPg^{-1}$

III

$\exists g \in G, Q \subset gPg^{-1}$
$|P| = |Q| = p^a$
$|P| = |gPg^{-1}|$
$\therefore |Q| = |gPg^{-1}|$
$\therefore Q = gPg^{-1}$
QED

Proposition

pシロー群と正規化群

Statement

$G$:Group
$|G| = p^a m\quad(p:\text{Prime Number}, \text{gcd}(p,m) = 1)$
$P$:p-Sylow subgroup $\quad(|P| = p^a)$
then
I
$P \triangleleft G \Leftrightarrow \nexists H:\text{subgroup of }G \text{ s.t. }|H| = p^a, H \neq P$

II
$H$:p-Sylow subgroup of $N(P) \Rightarrow H = P$

Proof

I
$\Rightarrow$
$P \triangleleft G \Rightarrow \forall g \in G, gPg^{-1} = P$
Let $H$:subgroup,$|H| = p^a$
then $\exists g \in G, H = g P g^{-1} = P \quad(\because \text{Prop.シローの定理 III})$
$\therefore \nexists H:\text{subgroup of }G \text{ s.t. }|H| = p^a, H \neq P$

$\Leftarrow$
$\forall g \in G, gPg^{-1}$:subgroup of $G, |gPg^{-1}| = p^a$
$\therefore gPg^{-1} = P$
$\therefore P \triangleleft G$

II
$N(P)$:subgroup of $G$
$P \subset N(P)$
$P \triangleleft N(P) \quad (\because N(P) = \{g \in G|gPg^{-1} = P\}$
$\therefore P$ is a only p-Sylow subgroup of $N(P) \quad(\because (I))$
QED

Theorem

シロー部分群の数

Statement

$G$:Group
$|G| \lt \infty$
$|G| = p^a m \quad (p,m \in \mathbb{Z}, p:\text{Prime Number}, \gcd (m,p) = 1)$
then
$\mathbb{P} = \{P \subset G| P \text{ is p-Sylow Subgroup of }G\}$
$=\{P_1,\cdots,P_s\}$
then
$s = |\mathbb{P}| = \frac{|G|}{|N(P_i)|} \equiv 1 \quad (\mod p)\quad(1 \le i \le s,)$

Proof

Define
$\cdot:G \times \mathbb{P} \rightarrow \mathbb{P}$
$g \cdot P_i = g P_i g^{-1} \in \mathbb{P}$
then
$\cdot$ is an Action over $\mathbb{P}$

$\forall P_i \in \mathbb{P}, \forall g \in G, g \cdot P_i = g P_i g^{-1} \in \mathbb{P}$
$\forall P_i \in \mathbb{P}, \exists g_i \in G \text{ s.t. }P_i = g_i P_1 g_i^{-1} \quad(\because\text{Prop.シローの定理 III})$
$\therefore G \cdot P_1 = \mathbb{P}$

$\forall P \in \mathbb{P}$
$g \in G_P \Leftrightarrow g \cdot P = gPg^{-1} = P \Leftrightarrow g \in N(P)$
$\therefore G_P = N(P)$
$\therefore s = |\mathbb{P}| = |G \cdot P| = |G/G_P| = |G|/|G_P| = |G|/|N(P)|$

Define
$P = P_1$
$\cdot:P \times \mathbb{P} \rightarrow \mathbb{P}$
$g \cdot P_i = g P_i g^{-1} \in \mathbb{P}$
then
$\cdot$ is an Action over $\mathbb{P}$

$\mathbb{P} = \coprod P \cdot P_i \quad(\because \text{Prop.軌道による分割})$
$|\mathbb{P}| = \sum |P \cdot P_i| \quad(\because \text{Prop.軌道による分割})$
$|P\cdot P_i| = |P|/|P_{P_i}| = p^a/|P_{P_i}| = p^b \quad(0 \le b \le a) \quad(\because \text{Prop.安定化群による剰余類})$

Suppose $|P \cdot P_i| = 1$
then
$|P \cdot P_i| = 1$
$\Leftrightarrow \forall h \in P, h \cdot P_i = h P_i h^{-1} = P_i$
$\Leftrightarrow \forall h \in P, h \in N(P_i) = \{g \in G|g P_i g^{-1} = P_i\}$
$\Leftrightarrow P \subset N(P_i)$

$|P| = p^a$
$P$ is a p-Sylow subgroup of $N(P_i)$

$P_i$ is an only p-Sylow subgroup of $N(P_i) \quad(\because \text{Prop.pシロー群と正規化群II})$

$\therefore P = P_i$
$1 \lt i \Rightarrow 1 \lt |P \cdot P_i| = p^b \quad (1 \le b \le a)$
$s = |\mathbb{P}| = 1 + \sum_{1 \lt i} |P \cdot P_i| = 1 + p n \quad (n \in \mathbb{Z})$
$\therefore s \equiv 1 \quad (\mod p)$
QED