📂Abstract Algebra

Shilov's theorem

Theorem ¹

Let’s say we have a finite group that satisfies prime numbers $p$ and $\gcd (p, m) = 1$ for some natural number $m$ where $G$ is $|G| = p^{n} m$. A Sylow $p$-subgroup of $G$ is a $p$-subgroup that is not contained in any other $p$-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 of Sylow $p$-subgroups $N_{p}$ of $G$ is a remainder of $p$ dividing into $1$ and is a divisor of $|G|$.

Explanation

The Sylow $p$-subgroup $P$ effectively refers to a $p$-subgroup that exactly satisfies $|P| = p^{n}$. In abstract mathematics, which makes active use of sets, expressions like “not contained in another” often describe Maximal.

Thus, a Sylow $p$-subgroup is, in essence, the ’largest’ $p$-subgroup of $G$ (although being the largest doesn’t guarantee it’s the only one). Since our focus is on Sylow $p$-groups, whether or not other $p$-groups exist up to $i = 1, \cdots , n-1$ doesn’t matter much.

Therefore, the First Sylow Theorem essentially states that ‘$G$ must have a Sylow $p$-subgroup’. This expression was used because of its generality to Cauchy’s theorem, both in terms of expression and proof method.

Furthermore, the existence of $g \in G$ satisfying $P_{2} = g P_{1} g^{-1}$ can also be described as $P_{1}$, $P_{2}$ being Conjugate to each other. This can clean up the above theorems 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 the number of Sylow $p$-subgroups of $G$ is $N_{p}$, then