# Group Theory/Simple groups and Sylow's theorem

Definition (Sylow p-subgroup):

Let ${\displaystyle G}$ be a group and let ${\displaystyle p\in \mathbb {N} }$ be a prime number such that ${\displaystyle p||G|}$. Then a Sylow ${\displaystyle p}$-subgroup of ${\displaystyle G}$ is a subgroup ${\displaystyle P\leq G}$ such that ${\displaystyle |P|=p^{k}}$, where ${\displaystyle k}$ is maximal such that ${\displaystyle p^{k}||G|}$.

Theorem (Cauchy's theorem):

Let ${\displaystyle G}$ be a group whose order ${\displaystyle |G|}$ is divisible by a prime number ${\displaystyle p\in \mathbb {N} }$. Then ${\displaystyle G}$ contains an element of order ${\displaystyle p}$.

Proof: ${\displaystyle G}$ acts on itself via conjugation. Let ${\displaystyle x_{1},\ldots ,x_{n}\in G}$ be a system of representatives of cojugacy classes. The class equation yields

${\displaystyle |G|=\sum _{k=1}^{n}[G:G_{x_{k}}]}$.

Either, there exists ${\displaystyle k\in \{1,\ldots ,n\}}$ such that ${\displaystyle [G:G_{x_{k}}]}$ is both not ${\displaystyle 1}$ and not divisible by ${\displaystyle p}$, in which case we may conclude by induction on the group order, noting that ${\displaystyle p}$ divides ${\displaystyle |G_{x_{k}}|}$ and ${\displaystyle |G_{x_{k}}|<|G|}$, or for all ${\displaystyle k\in \{1,\ldots ,n\}}$ the number ${\displaystyle [G:G_{x_{k}}]}$ is either ${\displaystyle 1}$ or divisible by ${\displaystyle p}$; but in this case, by taking the class equation ${\displaystyle \mod p}$, we obtain that ${\displaystyle Z(G)}$ is nontrivial and moreover that its order is divisible by ${\displaystyle p}$. Hence, it suffices to consider the case where ${\displaystyle G}$ is an abelian group. Take then any element ${\displaystyle h\in G}$. If ${\displaystyle h}$ has order divisible by ${\displaystyle p}$, raising ${\displaystyle h}$ to a sufficiently high power will produce an element of order ${\displaystyle p}$. Otherwise, the order of ${\displaystyle G/\langle h\rangle }$ is divisible by ${\displaystyle p}$, and by induction we find an element ${\displaystyle g+\langle h\rangle \in G/\langle h\rangle }$ whose order is divisible by ${\displaystyle p}$. Then the order of ${\displaystyle g}$ will also be divisible by ${\displaystyle p}$, because otherwise, passing to the quotient, ${\displaystyle (g+\langle h\rangle )^{k}=e+\langle h\rangle }$ for some ${\displaystyle k}$ not divisible by ${\displaystyle p}$. ${\displaystyle \Box }$

Theorem (Sylow's theorem):

Let ${\displaystyle G}$ be a finite group, such that ${\displaystyle |G|=p^{k}m}$ with ${\displaystyle p\nmid m}$. Then the following hold:

1. ${\displaystyle G}$ has a Sylow subgroup
2. The action of ${\displaystyle G}$ by conjugation on the Sylow subgroups is transitive
3. If ${\displaystyle n_{p}}$ is the number of Sylow ${\displaystyle p}$-subgroups, then ${\displaystyle n_{p}|m}$ and ${\displaystyle n_{p}\equiv 1\mod p}$
4. Every ${\displaystyle p}$-group of ${\displaystyle G}$ is contained within some Sylow ${\displaystyle p}$-group of ${\displaystyle G}$

Definition (simple group):

A group ${\displaystyle G}$ is a simple group if ${\displaystyle G}$ and ${\displaystyle \{e\}}$ are the only normal subgroups of ${\displaystyle G}$ (where ${\displaystyle e\in G}$ denotes the identity).