Sylow’s Theorems

Cauchy’s Theorem and Sylow I

The Sylow Theorems apply results on group actions to establish a partial converse to Lagrange’s Theorem.

Theorem: (Cauchy’s Theorem) Let $G$ be a finite group. If a prime $p$ divides the order of $G$, then $G$ contains an element of order $p$.

A group $G$ is called a $p$-group, where $p$ is a prime, if every element of $G$ has order a power of $p$.

Corollary: A finite group is a $p$-group if and only if its order is a power of $p$.

Theorem: (Sylow I) If $G$ is a finite group and $p$ is a prime such that $p^{r}$ divides the order of $G$, then $G$ has subgroup of order $p^{r}$.

Sylow II

See this video and these notes

Definition: A subgroup $H$ of a finite group $H$ is called a Sylow $p$-subgroup if it is a $p$-group of maximal order.

Theorem: (Sylow II) All Sylow $p$-subgroups of a finite group $G$ are conjugate to one another.

Corollary (of the proof): If $H$ is a subgroup of $G$ of prime power order $p^{s}$, then $H$ is contained in a Sylow $p$-subgroup.

Sylow III

See this video and these notes

Theorem: (Sylow III) The number $n_p$ of Sylow $p$-subgroups of a finite group $G$ is a divisor of the order of $G$ and is congruent to $1$ mod $p$.

Some applications

Theorem: Let $G$ be a group of order $pq$ where $p$ and $q$ are primes and $p<q$. Then $G$ has a unique subgroup of order $q$ (necessarily normal). If $q\not\equiv 1\pmod{p}$ then $G$ is cyclic.

Therefore all groups of order $15$, $33$, $51$ are all cyclic.

Recall that a group is called simple if it has no proper normal subgroups.

We know that groups of prime order are simple, but they are abelian.
Groups with $2$ prime factors are not simple by the theorem above.
Groups of prime power order are not simple because they have non-trivial center.
If $p$ is the largest prime divisor of the order $n$ of $G$, and $n$ has no divisors that are congruent to $1$ mod $p$, then $G$ has only one Sylow-$p$ subgroup which must be normal so $G$ is not simple.

The only possible orders for non-abelian simple groups less than $60$ are therefore $12,24,30,36,48,56$. The book shows that any group of order $48$ or $56$ is not simple.

With some work one can rule out the remaining possibilities as well and arrive at the following. For example, you can see the proof that a finite group of order $12$ is not simple in Conrad’s notes.

Proposition: A nonabelian finite simple group has order greater than or equal to $60$.

We know that $A_5$, which has $60$ elements, is simple, so it is the smallest example possible.