# Abstract Algebra/Group Theory/The Sylow Theorems

In this section, we will have a look at the Sylow theorems and their applications. The Sylow theorems are three powerful theorems in group theory which allow us for example to show that groups of a certain order $d \in \N$ are not simple.

The proofs are a bit difficult but nonetheless interesting. Important remark: Wikipedia also has proofs of the Sylow theorems, see Wikipedia article on the Sylow theorems, which are shorter and more elegant. But here you can find other proofs. This is because the author wanted to avoid redundancy. So you can choose the proof you like, or read both :-)

## The Sylow theorems

Definition 1: Let $G$ be a finite group of order $mp^a$, where $p$ is a prime, $a,m \in \mathbb{N}$ and $m$ is coprime to $p$. We say that a subgroup of $G$ is a Sylow $p$-subgroup iff it has order $p^a$.

Definition 2: Let H be a subgroup of a group G. We define the normalizer N[H] of H as follows:

$N[H] := \{g \in G: gHg^{-1} = H\}$

Theorem 3 (Cauchy's theorem) Let G be a group and $p$ be a prime number such that $p$ divides $|G|$. Then there exists an element of G which has order p. In particular, there is a subgroup of order p of G, namely $

$

.

Theorem 4 (Sylow I): Let $G$ be a finite group of order $mp^a$, where $p$ is a prime, $a,m \in \mathbb{N}$ and $m$ is coprime to $p$. For every $0 \le i \le a$, there is a subgroup of order $p^a$ of G. In particular, there exists a Sylow $p$-subgroup of G.

Proof: For this proof, we use induction. Let H be a p-subgroup of G, i. e. $|H| = p^i$ for some natural i. H acts on the sets of left cosets G/H by left multiplication. By corollary 23 from the section about group actions, we obtain that $|G/H| \equiv |Z| \mod p (*)$, where $Z = \{gH \in G/H : \forall h \in H: hgH = gH\}$. But also the following equivalences are true:

$\forall h \in H: hgH = gH \Leftrightarrow \forall h \in H: hg \in gH \Leftrightarrow \forall h \in H : g^{-1}hg \in H$ $\Leftrightarrow gHg^{-1} \subseteq H \Leftrightarrow gHg^{-1} = H$ because $|H| = |gHg^{-1}|$ $\Leftrightarrow g \in N[H]$

But from this we can conclude that $Z = \{gH \in G/H : g \in N[H]\} = N[H]/H$. Therefore, (*) becomes $[G:H] \equiv [N[H]:H] \mod p$. From this we can conclude, that if $i < a$, and therefore p divides $[G:H]$ by the theorem of Lagrange, that then also p divides $[N[H]:H]$. And also: Since $H$ is a normal subgroup of $N[H]$, we know that $N[H]/H$ is a group. Therefore we can apply Cauchy's theorem: $N[H]/H$ has a subgroup $N$ of order p. But if we set $H' = \bigcup_{gH \in N} gH$, then $H'$ is a subgroup of $G$ of order $p^{i+1}$, because
a) the intersection of two different cosets is the empty set, and
b) $H'$ is a subgroup of G because for $gh, g'h' \in H'$ $gh(g'h')^{-1} = ghh'^{-1}g'^{-1} = gg'^{-1}h''h''' \in H'$ for some $h'', h''' \in H$, because $g \in N[H]$, the normaliser. QED.

Lemma 5 (order of the conjugate): Let G be a group with identity $Id_G$, and $g \in G$ an element of that group. Then $\forall x \in G: ord(g) = ord(xgx^{-1})$

Proof: First, we observe that $(xgx^{-1})^n = xg^nx^{-1}$ by induction: For n = 1, the claim is obviosly true, and the calculation

$(xgx^{-1})^{n+1} = xg^nx^{-1} xgx^{-1} = x g^n x^{-1}$

shows the induction step.

Therefore, $(xgx^{-1})^{ord(g)} = x g^{ord(g)} x^{-1} = Id_G$, which shows that $ord(xgx^{-1}) \le ord(g)$.

Let furthermore $k \in \N$. Then $(xgx^{-1})^n = x g^k x^{-1} = Id_G \Rightarrow g^k x^{-1} = x^{-1} \Rightarrow g^k = Id_g$, where the first implication is true because the inverse is uniquely determined, and the second implication is true because the identity is uniquely determined. Therefore $ord(xgx^{-1}) \ge ord(g)$, implying with the former inequality that $ord(g) = ord(xgx^{-1})$ and finishing the proof. QED.

Lemma 6: Let $X = \{gPg^{-1} : g \in G\}$, and let G act on X by conjugation. Then $G_p$ is a subgroup of G, and any p-subgroup of $G_P$ is contained in P.

Proof: Conjugation of G on X is a transitive action: If $g, g' \in G$ are arbitrary, $h * gPg^{-1} = g'Pg'^{-1}$ by choosing $h = g'g^{-1}$. By the section about group actions, transitivity implies $G_P$ is really a group.

By the definition of X, we have that P is even a normal subgroup of $G_P$. Let now Q be an arbitrary p-subgroup of $G_P$. Then $QP$ is a subgroup due to the section about normal subgroups. Due to the second isomorphism theorem, we have that $QP/P \approx Q/(Q \cap P)$. Therefore, we also have by Lagrange's theorem, that $|QP/P| = |Q/(Q \cap P)| = \frac{|Q|}{|Q \cap P|} = p^s$ for some $s \in \N_0$, because Q is a Sylow p-subgroup. Furthermore, Lagrange's theorem also assures that $|QP/P| = \frac{|QP|}{|P|}$. Since P is a Sylow p-subgroup and QP is a subgroup of G and therefore $|QP|$ divides |G|, we know that p does not divide $|QP/P|$. Therefore, $|QP/P|$ must be the trivial subgroup, and therefore also $Q/(Q \cap P)$, which implies $Q \cap P = Q$ because $gH = H \Leftrightarrow g \in H$ due to the section about subgroups, QED.

Theorem 7 (Sylow II): If P is a Sylow p-subgroup of G, and Q is an arbitrary p-group of G, then $\exists g \in G: Q \subseteq gPg^{-1}$. In particular, all Sylow $p$-subgroup of $G$ are conjugate.

Proof: Let's choose $X = \{gPg^{-1} : g \in G\}$. P acts on X by conjugation. By the orbit-stabilizer theorem (corollary 19 of the section on group actions), we have that $\forall x \in X: |P * x| = \frac{|P|}{|P_x|}$. But since P is a Sylow p-group, we know that $|P * x| \equiv 0 \mod p$ or $|P * x| \equiv 1 \mod p$. Since $\forall h \in P: hPh^{-1} = P$, we furthermore have $P * P = \{P\}$ and therefore $|P * P| = 1$, because P is a single element in X (not finished - i will finish this tomorrow not finished again - sry, too many lemmas :-/ :-).).

Theorem 8 (Sylow III): Let $G$ be a finite group of order $mp^a$, where $p$ is a prime, $a,m \in \mathbb{N}$ and $m$ is coprime to $p$. If $n_p$ is the number of Sylow $p$-subgroups of $G$, then $n_p \mid m$ and $n_p \equiv 1 \mod p$.

Proof:

Theorem 9 (Sylow III*): Let again $n_p$ be the number of Sylow $p$-groups of $G$. Then $n_p = (G: N[P])$, where $P$ is any Sylow $p$-group.

Proof: