Sylow theorems and groups of small order

The three theorems

Let |G| = pᵏm with p ∤ m.

1. Existence: Sylow p-subgroups (order pᵏ) exist.
2. Conjugacy: all are conjugate; every p-subgroup lies in one.
3. Counting: n_p ≡ 1 (mod p) and n_p | m. Also n_p = [G : N(P)].

n_p = 1 ⇔ the Sylow p-subgroup is normal.

The standard drill

|G|	Conclusion
pq, p < q, p ∤ q−1	cyclic
pq, p | q−1	$\mathbb{Z}_{pq}$ or the non-abelian semidirect product
$p^{2}$	abelian: $\mathbb{Z}_{p^{2}}$ or $\mathbb{Z}_p \times \mathbb{Z}_p$
15 = 3·5	$n_{3} = n_{5} = 1 \Rightarrow$ cyclic
20 = 4·5	$n_{5} \equiv 1 (mod 5), n_{5}$ | $4 \Rightarrow n_{5} = 1$
$2023 = 7\cdot17^{2}$	$n_{7} = n_{17} = 1 \Rightarrow$ abelian, not simple
30 = 2·3·5	$n_{3} = 1$ or $n_{5} = 1 ($counting elements) ⇒ not simple
36, 48, 24	use the coset action: index too small ⇒ not simple

Counting-elements argument

If n_q = k > 1 with q prime, the Sylow q-subgroups contribute k(q − 1) elements of order q. Adding up over primes and exceeding |G| gives a contradiction — the standard proof that no group of order 30 is simple.

Normal Sylow ⇒ direct product

If every Sylow subgroup is normal, G is the direct product of them (a nilpotent group). Finite abelian groups and p-groups are the examples.

Key takeaways

• n_p ≡ 1 (mod p) and n_p | m — apply to each prime in turn.
• One normal Sylow subgroup kills simplicity immediately.
• All Sylow subgroups normal ⇒ G is their direct product.