Extract structure from divisibility of orbit sizes

Try

A finite p-group has order pⁿ for a prime p and a nonnegative integer n. If the group is nontrivial, n is positive. Its subgroup orders and orbit sizes are powers of p by Lagrange. In a conjugation action, a noncentral orbit cannot have size one, so its size must be divisible by p.

Predict what the class equation then says about the number of central elements. Identity supplies at least one, but the divisibility condition will force more.

Predict

Does a nontrivial center imply the whole group is abelian? Consider square symmetries: the half-turn commutes with every symmetry, while a quarter-turn and a reflection do not commute. The center can be proper. Predict why groups of order p² nevertheless face a stronger restriction.

Experiment

For D₄, the center is ｛e,r²｝. The remaining conjugacy classes each have size two.

Class	Size
｛e｝	1
｛r²｝	1
｛r,r³｝	2
｛s,r²s｝	2
｛rs,r³s｝	2

The class equation is 8=2+2+2+2. Every noncentral contribution is even, and the center size is also even. Check that conjugating r by s gives r⁻¹=r³, while r² remains fixed.

Understand

For a nontrivial finite p-group G, the class equation gives |G|=|Z(G)| plus a sum of multiples of p. Since p divides |G|, it divides |Z(G)|. The center contains identity, so its size is positive and therefore at least p. This proves the nontrivial-center theorem without assuming the group is abelian.

A useful lemma says that if G/Z(G) is cyclic, then G is abelian. Choose gZ(G) generating that quotient. Every x can be written gᵃz with z central, and every y as gᵇw with w central. Then xy=g^(a+b)zw=g^(b+a)wz=yx. The hypothesis uses the center specifically; a cyclic quotient by some other normal subgroup need not force G abelian.

Now let |G|=p². Its center has order p or p². If it had order p, the quotient by the center would have order p and be cyclic. The lemma would make G abelian, contradicting a proper center. Thus Z(G)=G.

To classify such groups, first ask whether an element has order p². If so, G is cyclic. Otherwise every nonidentity element has order p. Choose a nonidentity a and choose b outside ⟨a⟩. Their cyclic subgroups intersect trivially because each has prime order and they are distinct. Since G is abelian, the product map C_p×C_p→G is injective and its image has p² elements, so it is an isomorphism. The two possibilities are therefore C_(p²) and C_p×C_p.

Apply

Apply the classification to order forty-nine and distinguish the two possibilities by maximum element order. Then explain why the proof does not extend unchanged to order p³: the center could have order p while the central quotient has order p² and is noncyclic.

D₄ supplies the concrete counterexample at order eight. Use its noncommuting rotation and reflection to verify nonabelianness, and use the displayed center to show it still obeys the p-group theorem. A counterexample to a stronger statement should remain consistent with the theorem actually proved.

Continue

Sylow theory finds maximal prime-power subgroups inside groups whose orders contain several primes. Its proofs reuse the same orbit-divisibility idea, now on carefully chosen sets of cosets and subsets.