Sylow Theorems

n_p 1 p and n_p | m.

The idea

Lagrange's theorem says the order of a subgroup divides $|G|$, but not the converse: a divisor of $|G|$ need not be the order of any subgroup at all. The Sylow theorems supply the converse in the one case where it does hold — prime powers — and that single case turns out to be enough to control the whole group.

Definition (Sylow p-subgroup).

Let $G$ be a finite group and write $|G| = p^{a} m$ with $p$ prime and $p \nmid m$, so that $p^{a}$ is the largest power of $p$ dividing $|G|$. A subgroup of $G$ of order exactly $p^{a}$ is a Sylow $p$-subgroup.

Theorem (Sylow theorems).

Let $G$ be a finite group with $|G| = p^{a} m$, where $p$ is prime and $p \nmid m$. Then: 1. $G$ has a Sylow $p$-subgroup. 2. Any two Sylow $p$-subgroups $P, Q$ are conjugate: $Q = gPg^{-1}$ for some $g \in G$. 3. The number $n_p$ of Sylow $p$-subgroups satisfies $n_p \equiv 1 \pmod p$ and $n_p \mid m$.

The first part needs no further hypothesis on $G$. The second says that from inside $G$ the Sylow $p$-subgroups are indistinguishable. The two conditions in the third part together usually leave only a couple of candidates for $n_p$, and often only one.

The third part is what makes the theorems a classification tool: it converts a question about the structure of an unknown group into arithmetic on its order, and the conjugacy in the second part then converts the answer back into structure.

Ways to work on it

• Walkthrough. Sylow subgroup order and the congruence/divisibility rules.
• Practice. Order of a Sylow p-subgroup.
• Hardest. Prove a group of a given order must be cyclic.

Not sure where to start? Take the ten-question placement test.