Constrain prime-power subgroups and prove the counting mechanism

Try

Write a finite group order as pᵃm, where p is prime and p does not divide m. A Sylow p-subgroup has order pᵃ, the largest power of p dividing the group order. Lagrange only permits this size; Sylow’s existence theorem guarantees it occurs.

For a group of order twenty-eight, the target sizes are four and seven. Do not call every subgroup of prime-power order a Sylow subgroup: maximal size matters.

Predict

If all Sylow p-subgroups are conjugate, their number should be an index of a stabilizer under conjugation. Predict why a unique Sylow subgroup is normal. Then test the proposed restrictions that its count divides m and is congruent to one modulo p.

Experiment

For order twenty-eight, apply both restrictions together.

Prime	Divisors allowed by nₚ dividing m	Congruence	Remaining counts
7	1,2,4	n₇≡1 mod 7	1
2	1,7	n₂≡1 mod 2	1 or 7

The order-seven subgroup is normal. The arithmetic alone does not decide the number of order-four Sylow subgroups. Report the remaining possibilities rather than selecting one because it looks simpler.

Understand

Here is an existence proof. Let G act by left translation on the subsets of G having pᵃ elements. Their number is the binomial coefficient choosing pᵃ from pᵃm, congruent to m modulo p. Indeed, modulo p the polynomial (1+x)^(pᵃm) equals (1+x^(pᵃ))^m; comparing the coefficient of x^(pᵃ) proves the congruence. Thus some orbit has size not divisible by p.

Let H stabilize a subset in that orbit. Orbit-stabilizer implies the full p-power pᵃ divides |H|. On the other hand, H acts freely by left translation on that subset: hx=x forces h=e. Its H-orbits therefore all have size |H|, so |H| divides the subset size pᵃ. Hence |H|=pᵃ, proving existence.

Fix a Sylow subgroup P and let any p-subgroup Q act on G/P. This coset set has m elements. Every Q-orbit has p-power size, so some orbit is a singleton. A fixed coset gP means g⁻¹Qg⊆P. Thus every p-subgroup is contained in a conjugate of P; equal orders give conjugacy of Sylow subgroups.

Their number nₚ is [G:N_G(P)], where the normalizer N_G(P) is the subgroup preserving P by conjugation. Since it contains P, this index divides m. For the congruence, let P act on the set of Sylow subgroups by conjugation. Its only fixed point is P: if P normalizes Q, then PQ is a p-subgroup because Q is normal in PQ; maximal Sylow order forces P=Q. All other orbits have size divisible by p, so nₚ≡1 modulo p.

Apply

Prove that a group of order fifteen is cyclic. Both Sylow counts are forced to one. The resulting normal subgroups of orders three and five have trivial intersection; their commutators lie in the intersection, so they commute. Their product has fifteen elements and is C₃×C₅≅C₁₅.

Identify the ingredients used: existence, conjugacy, the two counting restrictions, and a direct-product argument. Do not compress “unique Sylow” into “cyclic group”; uniqueness gives normality, while cyclicity requires additional reasoning.

Continue

Next combine these tools to classify small groups. A complete classification must construct the possibilities, distinguish them, and prove that every group of the stated order fits one of them.