The power of prime-order subgroups -- existence, counting, and conjugacy
Lagrange's theorem tells us what subgroup orders are possible. The Sylow theorems go further: for prime-power orders, they guarantee subgroups exist, constrain how many there can be, and reveal that they are all conjugate to each other.
If |G| = p^k * m where p does not divide m, a Sylow p-subgroup has order p^k. The three Sylow theorems say: (1) at least one exists, (2) all are conjugate, and (3) the count n_p divides m and satisfies n_p = 1 mod p.
Select a group and a prime p. All Sylow p-subgroups are highlighted on the Cayley diagram. Toggle between them to see how they sit inside the group.
Sylow 2-subgroups of order 2
Found: 3
Key insight: When there is only one Sylow p-subgroup, it must be normal (conjugation cannot move it anywhere else). This is a powerful tool for proving a group is not simple.
Verify the Sylow counting constraints for each prime dividing |G|. The number n_p of Sylow p-subgroups must divide |G|/p^k and be congruent to 1 mod p.
| Prime p | p^k | n_p | n_p | m? | n_p = 1 mod p? | All conjugate? |
|---|---|---|---|---|---|
| 2 | 2 | 3 | Yes(3 | 3) | Yes(3 mod 2 = 1) | Yes |
| 3 | 3 | 1 | Yes(1 | 2) | Yes(1 mod 3 = 1) | Yes |
Sylow 2-subgroups
3 conjugate subgroups of order 2. Not normal individually.
Sylow 3-subgroups
Unique Sylow 3-subgroup of order 3 -- it is normal in G.
Key insight: These counting constraints are often enough to determine n_p exactly. For S3 (order 6 = 2 * 3): n_3 must divide 2 and equal 1 mod 3, so n_3 = 1. The unique Sylow 3-subgroup is the rotation subgroup.
Pick two Sylow p-subgroups and find an element g that conjugates one to the other: gPg⁻¹ = Q. This is Sylow's second theorem in action -- all Sylow p-subgroups are "the same subgroup in different positions."
Key insight: Conjugacy means Sylow p-subgroups are structurally identical -- they are isomorphic copies of each other positioned differently within G. The number of them equals the index [G : N_G(P)].