📂Abstract Algebra

Sylow Theorems

Theorem¹

For a prime $p$ and some natural number $m$ satisfying $\gcd (p, m) = 1$, let $G$ be a finite group with $|G| = p^{n} m$. A $p$-subgroup of $G$ that is not contained in any other $p$-subgroup is called a Sylow $p$-subgroup.

• First Sylow Theorem: $G$ has a $p$-subgroup satisfying $|P| = p^{i}$ for $i=1, \cdots , n$.
• Second Sylow Theorem: For Sylow $p$-subgroups $P_{1}$, $P_{2}$ of $G$, there exists $g \in G$ satisfying $P_{2} = g P_{1} g^{-1}$.
• Third Sylow Theorem: The number $N_{p}$ of Sylow $p$-subgroups of $G$ leaves a remainder of $1$ when divided by $p$, and is a divisor of $|G|$.

Explanation

A Sylow $p$-subgroup $P$ is, in fact, precisely a $p$-subgroup satisfying $|P| = p^{n}$. In abstract mathematics, which makes active use of sets, expressions like 'not contained in any other' are often used to express being maximal.

In other words, a Sylow $p$-subgroup is none other than the 'largest' $p$-subgroup of $G$ (of course, being the largest in size does not guarantee that it is unique). Since our interest lies in Sylow $p$-groups, whether $p$-groups exist for $i = 1, \cdots , n-1$ hardly matters.

Therefore, it is perfectly fine to remember the First Sylow Theorem simply as '$G$ necessarily has a Sylow $p$-subgroup'. The reason it is stated in such a way is merely that, both in its statement and in its method of proof, it is a generalization of Cauchy’s theorem.

Meanwhile, the existence of $g \in G$ satisfying $P_{2} = g P_{1} g^{-1}$ is also expressed by saying that $P_{1}$, $P_{2}$ are conjugate to each other. With this, the above theorems can be neatly rewritten as follows.

• First Sylow Theorem: $G$ has a Sylow $p$-subgroup.
• Second Sylow Theorem: Sylow $p$-subgroups $P_{1}$, $P_{2}$ of $G$ are conjugate to each other.
• Third Sylow Theorem: If $N_{p}$ denotes the number of Sylow $p$-subgroups of $G$, then $$ N_{p} \equiv 1 \pmod{p} \\ N_{p} \mid |G| $$