Team 4 (Kevin, Ian, Anthony, Ally, Christin)

# The Sylow Theorems

Abstract Algebra: Theory and Applications
http://www.cl.cam.ac.uk/~lp15/papers/Kammueller/sylow.pdf

# History

The Sylow Theorems are a collection of theorems that are named after Peter Ludwig Mejdell Sylow (12 December 1832 – 7 September 1918) and they form a partial converse to the theorem of Lagrange. Legrange's theorem states that if $H$ is a subgroup of a finite group $G$, then the order of $H$ divides the order of $G$. Sylow was a Norwegian high school teacher and a substitute lecturer at Christiania University. In his lectures Sylow explained Abel's and Galois's work on algebraic equations. In 1898, he was appointed a special chair of the University that was created by Lie specifically for him.
Sylow's Theorems were published in 1872 in his Théorèmes sur les groupes de substitutions paper, with the focus on 3 main proofs. They are often referred to as the "existence", "development" and "conjugate" theorems. Some say that what Sylow proved is the most profound result in the theory of finite groups because almost all work with finite groups uses Sylow's theorems.

# Definitions

• A Normal Subgroup $H$ of a group $G$ is on which the left and right cosets are exactly the same.
• Example:
The identity group $\{e_G\}$ and $G$ are always normal subgroups of the group G.
• The Centralizer of an element $a \in G$ (expressed as $C_{G}(a)$) is the set of elements in $G$ that commute with $a$. $C_{G}(a) = {x \in G : xa =ax}$.
• Example:
The centralizer for each element in $\mathbb R$ is $\mathbb R$ under the operations multiplication and addition.
• A Sylow p-subgroup P of a group $G$ is a maximal $p$-subgroup of $G$. It is written as $Syl_{p}(G)$.
• $N(H)$ is the largest subgroup of $G$ in which $H$ is normal and is said to be the normalizer of $H$ in $G$.
• Example:
The normalizer in a group $G$ is as small as $H$ and as big as $G$.
• A group is said to be simple if it has no non-trivial, normal subgroups.
• Example:
$\mathbb Z_2$.

# Uses

Sylow Theory provides information about what structures are guaranteed to exist within a group, based only on the group's order.

# Cauchy Theorem

Let $G$ be a finite group and $p$ a prime such that $p$ divides the order of $G$. Then $G$ contains a subgroup order $p$.

### Case 1

The order of the centralizer subgroup, $C(x_i)$, is divisible by $p$ for some $i$, $i=1,\dots,k$.

### Case 2

The order of no centralizer subgroup is divisible by $p$.

# First Sylow Theorem

Let $G$ be a finite group and $p$ a prime such that $p^r$ divides $|G|$. Then $G$ contains a subgroup of order $p^r$.

Proof: We induct on the order of $G$ once again. If $|G| = p$, then we are done. Now suppose that the order of $G$ is $n$ with $n > p$ and that the theorem is true for all groups of order less than $n$. We shall apply the class equation once again:

(1)
\begin{align} |G| = |Z(G)| + [G : C(x_1)] + \dots + [G : C(x_k)]. \end{align}

First suppose that $p$ does not divide $[G : C(x_i)]$ for some $i$. Then $p^r | |C(x_i)|$, since $p^r$ divides $|G| = |C(x_i)| \centerdot [G : C(x_i)]$. Now we can apply the induction hypothesis to $C(x_i)$. Hence, we may assume that $p$ divides $[G : C(x_i)]$ for all $i$. Since $p$ divides $|G|$, the class equation says that $p$ must divide $|Z(G)|$; hence, by Cauchy's Theorem, $Z(G)$ has an element of order $p$, say $g$. Let $N$ be the group generated by $g$. Clearly, $N$ is a normal subgroup of $Z(G)$ since $Z(G)$ is abelian; therefore, $N$ is normal in $G$ since every element in $Z(G)$ commutes with every element in $G$. Now consider the factor group $G=N$ of order $|G|=p$. By the induction hypothesis, $G=N$ contains a subgroup $H$ of order $p^{r-1}$. The inverse image of $H$ under the canonical homomorphism $\phi : G \rightarrow G/N$ is a subgroup of order $p^r$ in $G$. $\blacksquare$

### Try These:

Find the possible orders of Sylow $p$-subgroups of groups of the following orders:
(a) $12$
Prime factorization of $|a|$:$3,2,2$
Sylow subgroups of $(a)$:$2,3,4$
(b) $24$
Prime factorization of $|b|$:$2,2,2,3$
Sylow subgroups of $(b)$:$2,3,4,8$
(c) $30$
Prime factorization of $|c|$:$2,3,5$
Sylow subgroups of $(c)$:$2,3,5$
(d) $36$
Prime factorization of $|d|$:$3,3,4$
Sylow subgroups of $(d)$:$3,4,9$
(e) $40$
Prime factorization of $|e|$:$2,2,2,5$
Sylow subgroups of $(e)$:$2,4,5,8$
(f) $42$
Prime factorization of $|f|$:$2,3,7$
Sylow subgroups of $(f)$:$2,3,7$
(g) $45$
Prime factorization of $|g|$:$3,3,5$
Sylow subgroups of $(g)$:$3,5,9$
(h) $48$
Prime factorization of $|h|$:$2,2,2,2,3$
Sylow subgroups of $(h)$:$2,3,4,8,16$
(i) $56$
Prime factorization of $|i|$:$2,2,2,7$
Sylow subgroups of $(i)$:$2,4,7,8$
(j) $60$
Prime factorization of $|j|$:$2,2,3,5$
Sylow subgroups of $(j)$:$2,3,4,5$

# Second Sylow Theorem

Let $G$ be a finite group and $p$ a prime dividing $|G|$. Then all Sylow $p$-subgroups of $G$ are conjugate. That is, if $P_1$ and $P_2$ are two Sylow $p$-subgroups, there exists a $g\in G$ such that $g(P_1)g^{-1} = P_2$

Proof. Let $P$ be a Sylow $p$-subgroup of $G$ and suppose that $|G| = p^rm$ and $|P| = p^r$. Let

(2)
\begin{align} \mathcal P = \{ P = P_1, P_2, \dots , P_k \} \end{align}

consist of the distinct conjugates of $P$ in $G$. By Lemma 13.5, $k = [G : N(P)]$. Notice that

(3)
\begin{align} |P| = p^rm = |N(P)|\centerdot [G : N(P)] = |N(P)| \centerdot k. \end{align}

Since $p^r$ divides $|N(P)|$, $p$ cannot divide $k$. Given any other Sylow $p$- subgroup $Q$, we must show that $Q \in P$. Consider the $Q$-conjugacy classes of each $P_i$. Clearly, these conjugacy classes partition $\mathcal P$. The size of the partition containing $P_i$ is $[Q : N(P_i) \cap Q]$. Lagrange's Theorem tells us that this number is a divisor of $|Q| = p^r$. Hence, the number of conjugates in every equivalence class of the partition is a power of $p$. However, since $p$ does not divide $k$, one of these equivalence classes must contain only a single Sylow $p$-subgroup, say $P_j$ . Therefore, for some $P_j , x^{-1}P_jx = P_j$ for all $x \in Q$. By Lemma 13.4, $P_j = Q$ . $\blacksquare$

### Try This:

Given the following Sylow $3$-subgroups of $S_4$, show they are all conjugate.

(4)
\begin{align} P_1=\{e,(123),(132)\}, P_2=\{e,(124),(142)\}, P_3=\{e,(134),(143)\}, P_4=\{e,(234),(243)\} \end{align}

# Third Sylow Theorem

Let $G$ be a finite group and let $p$ be a prime dividing the order of $G$. Then the number of Sylow $p$-subgroups is congruent to $1\mod{p}$ and divides $|G|$.

Proof: Let $P$ be a Sylow $p$-subgroup acting on the set of Sylow $p$-subgroups, $\mathcal P = \{P = P_1, P_2, \dots, P_k\}$ by conjugation. From the proof of the Second Sylow Theorem, the only $P$-conjugate of $P$ is itself and the order of the other $P$-conjugacy classes is a power of $p$. Each $P$-conjugacy class contributes a positive power of $p$ toward $|\mathcal P|$ except the equivalence class $\{P\}$. Since $|\mathcal P|$ is the sum of positive powers of $p$ and $1, |\mathcal P| \equiv 1$ (mod $p$). Now suppose that $G$ acts on $P$ by conjugation. Since all Sylow $p$- subgroups are conjugate, there can be only one orbit under this action. For $P\in P$, $|\mathcal P| = |\text{orbit of } P| = [G:N(P)]$. But $[G:N(P)]$ is a divisor of $|G|$; consequently, the number of Sylow $p$-subgroups of a finite group must divide the order of the group. $\blacksquare$

### Try These:

Based on the questions from Theorem 1, find the possible number of Sylow 3-subgroups of groups of the following orders:
(a) $12$
Sylow subgroups of $(a)$:$2,3,4$
Sylow $3$-subgroups of $(a)$:$1,4$
(b) $24$
Sylow subgroups of $(b)$:$2,3,4,8$
Sylow $3$-subgroups of $(b)$:$1,4$
(c) $30$
Sylow subgroups of $(c)$:$2,3,5$
Sylow $3$-subgroups of $(c)$:$1,10$
(d) $36$
Sylow subgroups of $(d)$:$3,4,9$
Sylow $3$-subgroups of $(d)$:$1,4$
(e) $40$
Sylow subgroups of $(e)$:$2,4,5,8$
Sylow $3$-subgroups of $(e)$: $0$
(f) $42$
Sylow subgroups of $(f)$:$2,3,7$
Sylow $3$-subgroups of $(f)$:$1,7$
(g) $45$
Sylow subgroups of $(g)$:$3,5,9$
Sylow $3$-subgroups of $(g)$:$1$
(h) $48$
Sylow subgroups of $(h)$:$2,3,4,8,16$
Sylow $3$-subgroups of $(h)$:$1,4$
(i) $56$
Sylow subgroups of $(i)$:$2,4,7,8$
Sylow $3$-subgroups of $(i)$: $0$
(j) $60$
Sylow subgroups of $(j)$:$2,3,4,5$
Sylow $3$-subgroups of $(j)$:$1,4,10$

KammÄuller, F., & Paulson, L. C. (2000, September 7). A Formal Proof of Sylow's Theorem. Retrieved April 14, 2010, from http://www.cl.cam.ac.uk/~lp15/papers/Kammueller/sylow.pdf

The Sylow Theorems. (n.d.). In T. W. Judson (Author), Abstract Algebra (pp. 220-224).

page revision: 26, last edited: 11 May 2010 06:11