# The Sylow Theorems

Recall from the Sylow p-Subgroups of a Group page that if $G$ is a finite group of order $n$ and if $n = p_1^{e_1}p_2^{e_2}...p_k^{e_k}$ is the prime power decomposition of $n$ then a Sylow $p_i$-subgroup of $G$ is a subgroup $H$ of $G$ with $|H| = p_i^{e_i}$.

We will now begin to examine the Sylow theorems.

Theorem 1 (The First Sylow Theorem): Let $G$ be a finite group. If $p$ is a prime, $k \geq 0$, and $p^k$ is a divisor of $|G|$ then $G$ has a subgroup of order $p^k$. |

**Proof:**Let $P(n)$ be the statement that if $G$ is a group of order $n$ and if $p$ is a prime such that $p^k \mid |G|$ where $k \geq 0$ then $G$ has a subgroup of order $p^k$.

**Base Step:**$P(1)$ says that if $G$ is a group of order $1$ and if $p$ is a prime such that $p^k \mid |G|$ where $k \geq 0$ then $G$ has a subgroup of order $p^k$. The only group of order $1$ is the trivial group. If $p^k \mid |G| = 1$ for some prime $p$ and some $k \geq 0$ then $k = 0$. Of course, $G$ has a subgroup of order $p^0 = 1$ - itself. So $P(1)$ is true.

**Inductive Step:**Suppose that for some $n > 1$ we have that $P(1)$, $P(2)$, …, $P(n-1)$ is true. We aim to show that $P(n)$ is true.

- Let $G$ be a group of order $n$ and suppose that $p$ is a prime and $k \geq 0$ be such that $p^k \mid n$.

- The Class Equation for $G$ is:

- Where the sum above runs over one representative for each nontrivial conjugacy class. We consider two cases.

**Case 1:**Suppose that each term in $\displaystyle{\sum [G:C_G(g)]}$ is divisible by $p$. Then $\displaystyle{p \mid \sum [G:C_G(g)]}$. From the class equation, we see that since $p \mid n$ and $\displaystyle{p \mid \sum [G:C_G(g)]}$ that then $p \mid |Z(G)|$.

- Since $Z(G)$ is itself a group, $p$ is a prime, and $p \mid |Z(G)|$ we have by Cauchy's Theorem for Groups that $Z(G)$ contains an element of order $p$. Let $g \in Z(G)$ be such that $\mathrm{ord}(g) = p$. Consider the group generated by $g$, $\langle g \rangle$. We see that $\langle g \rangle$ is a group of order $p$ that is a subgroup of $Z(G)$, and thus, a subgroup of $G$.

- Observe that since $\langle g \rangle \subseteq Z(G)$ we have that $hg^k = g^kh$ for all $h \in G$ and for all $k \in \mathbb{Z}$. So $hg^kh^{-1} = g^k$ for all $h \in G$ and for all $k \in \mathbb{Z}$, i.e., $h \langle g \rangle h^{-1} = \langle g \rangle$ for all $h \in G$. So $\langle g \rangle$ is a normal subgroup of $G$. So, we can consider the quotient group $G / \langle g \rangle$. Since $|G| = n$ and $|\langle g \rangle| = p$, we have that $|G/\langle g \rangle| = n/p$. And since $p^k \mid n$, we see that $p^{k-1} \mid n/p = |G/\langle g \rangle|$. By the induction hypothesis, $G/\langle g \rangle$ has a subgroup of order $p^{k-1}$. Let $M$ denote this subgroup.

- Let $q : G \to G/\langle g \rangle$ be defined for all $h \in G$ by $q(h) = h\langle g \rangle$ (this is simply the quotient homomorphism). Then $q^{-1}(M)$ is a subgroup of $G$. Since every coset of $\langle a \rangle$ has $p$ elements in it, we see that $q^{-1}(M)$ has order $p^k$.

**Case 2:**Suppose that there exists a term in $\displaystyle{\sum [G:C_G(g)]}$ that is not divisible by $p$. Say $g \not \in Z(G)$ is such that $p \nmid [G:C_G(g)]$. Since $C_G(g)$ is a subgroup of $G$, we have by Lagrange's Theorem that:

- Since $p^k \mid n$ and $p \nmid [G:C_G(g)]$ we have that $p^k \mid C_G(g)$. Note that if $C_G(g) = \{ h \in G : gh = hg \} = G$ then for every $g \in Z(G)$. But this is a contradiction since $g \not \in Z(G)$. Thus $C_G(g)$ is a proper subgroup of $G$. So $|C_G(g)| < |G| = n$. Since $p^k \mid |C_G(g)| < n$, by the induction hypothesis $C_G(g)$ has a subgroup of order $p^k$ - call it $M$. But then $M$ is a subgroup of $G$ of order $p^k$.

**Conclusion:**So for all finite groups $G$, if $p$ is a prime and $k \geq 0$ is such that $p^k \mid |G|$ then $G$ has a subgroup of order $p^k$. $\blacksquare$

*The First Sylow Theorem tells us that if $G$ is a finite group of order $n$ and if $p \mid n$ then $n_p \neq 0$.*

Theorem 2 (The Second Sylow Theorem): Let $G$ be a finite group of order $n$ and let $p \mid n$. Then all Sylow $p$-subgroups of $G$ are conjugate to one-another. That is, if $H_1$ and $H_2$ are Sylow $p$-subgroups of $G$ then there exists a $g \in G$ such that $H_1 = gH_2g^{-1}$. |

Theorem 3 (The Third Sylow Theorem): Let $G$ be a finite group of order $n$ and let $p$ be a prime such that $p \mid n$. If $n = p^km$ where $k \geq 1$ and $p \nmid m$ then:a) $n_p \equiv 1 \pmod p$, i.e., $p \mid (n_p - 1)$.b) $n_p \mid m$. |

*The Third Sylow Theorem (a) tells us that if $G$ is a finite group of order $n$ then the number of Sylow $p$-subgroups of $G$ MINUS 1 is divisible by $p$. In other words, there are either $1$, $p + 1$, $2p + 1$, … $\leq n$ Sylow $p$-subgroups of $n$.*

*Meanwhile, The Third Sylow Theorem (b) tells us that the number of Sylow $p$-subgroups of divides $m$ - i.e., divides $n/p^k$. Together, this allows us to usually determine $n_p$.*

## Example 1

Let $G$ be a group of order $120$. Observe that:

(3)The First Sylow Theorem guarantees the existence of a Sylow $2$-subgroup, a Sylow $3$-subgroup, and a Sylow $5$-subgroup, i.e., there exists subgroups of $G$ with order $8$, $3$, and $5$.

## Example 2

Let $G$ be a group of order $120$. We wish to find what the possible number of Sylow $5$-subgroups $G$ has. Let $n_5$ denote the number of Sylow $5$-subgroups of $G$. By the Third Sylow Theorem we must have that:

(4)Since $n_5 \mid 24$ we must have that $n_3 \in \{ 1, 2, 3, 4, 6, 8, 12, 24 \}$. Of this set, the only number congruent to $1$ modulo $5$ are $1$ and $6$. So $G$ has either $1$ Sylow $5$-subgroup or $6$ Sylow $5$-subgroups.