Sylow’s theorems

Three theorems giving a partial converse to Lagrange’s theorem: while Lagrange tells us only what subgroup orders are forbidden, Sylow tells us which prime-power orders are guaranteed and exactly how many subgroups of each kind there are.

Setup

Let $G$ be a finite group and $p$ a prime. Write $|G|=p^{r}t$ with $p\nmid t$. A Sylow $p$-subgroup of $G$ is any subgroup of order $p^r$ - i.e. the largest possible $p$-subgroup.

The Three Theorems

First Sylow Theorem

$G$ has at least one Sylow $p$-subgroup.

Second Sylow Theorem

Any two Sylow $p$-subgroups are conjugate in $G$. Equivalently, every $p$-subgroup of $G$ is contained in some Sylow $p$-subgroup.

Third Sylow Theorem

Let $a(p)$ be the number of Sylow $p$-subgroups. Then:

1. $a(p)\mid t$;
2. $a(p)\equiv 1(\mathrm{mod}p)$.

Why They Matter

• Existence (First): every prime divisor of $|G|$ guarantees a maximal $p$-subgroup.
• Uniqueness up to conjugation (Second): Sylow $p$-subgroups are structurally interchangeable.
• Counting (Third): $a(p)$ is constrained to a finite list of values; often pins it to $a(p)=1$.

When $a(p)=1$, the unique Sylow $p$-subgroup is normal (no other conjugates). This is the most common entry point for classification arguments.

Standard Application: Order $pq$ Groups Are Often Cyclic

If $|G|=pq$ with primes $p<q$ and $p\nmid q-1$, then $a(q)=1$ and $a(p)=1$, both Sylow subgroups are normal, their intersection is trivial (Lagrange), their product is $G$, so $G$ is the Internal direct product $C_p\times C_q\cong C_{pq}$. Examples: orders $15,33,35,51,65,77,85,\dots$.

When $p\mid q-1$ a non-cyclic example may exist (e.g. order $6$ allows $S_3$).

Proof Strategy

• First Sylow: act $G$ on the set of $p^r$-subsets by left multiplication; use Orbit-Stabiliser and a binomial-coefficient $p$-adic argument.
• Second / Third: act $G$ on cosets / on the set of all Sylow $p$-subgroups by conjugation.

See lecture 17 for full statements and worked examples.

Generalises Cauchy’s theorem

Cauchy: $p\mid |G|\Rightarrow$ subgroup of order $p$. Sylow First: subgroup of order $p^r$ (the maximum power of $p$ in $|G|$).