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For a finite p-group in characteristic p, the augmentation ideal of the group algebra is nilpotent

Statement

Let P be a finite p-group and let k be a field of characteristic p. Then the augmentation ideal IP⊆k[P] is nilpotent.

Facts & Assumptions

Given: A finite p-group P, a field k of characteristic p, and the augmentation ideal IP⊆k[P].

[F1]

The augmentation ideal is the kernel of the augmentation map (The augmentation map ε:R[G]→R and the augmentation ideal IG=ker⁡ε).

[L1]

The basis elements [g] multiply as [g][h]=[gh] in k[P] (The group ring R[G] is a unital R-algebra with basis G, and each g∈G is a unit of R[G]).

[L2]

The algebra k[P] has dimension ∣P∣ (If G is finite then dim⁡kk[G]=∣G∣).

[L3]

Every nontrivial finite p-group has a nontrivial central element of order p (Every nontrivial finite p-group has nontrivial center, in fact p divides ∣Z(P)∣).

Proof

technique · direct
1.1F1L1L3givenchooseinductionalgebra

We argue by induction on ∣P∣. If P=1, then IP=0 by [F1], so the claim is trivial. Assume P≠1. By [L3], choose a central element z∈Z(P) of order p. Then ([z]−1)p=[zp]−1=0 in characteristic p, using [L1] and the binomial theorem with vanishing intermediate coefficients. Hence the principal ideal generated by [z]−1 is nilpotent.

2.1F1L2step 1.1inductionalgebra∎

Because z is central, the quotient map P→P/⟨z⟩ induces an algebra surjection k[P]→k[P/⟨z⟩] whose kernel is the ideal ([z]−1)k[P]. Under this map the augmentation ideal of k[P] maps onto the augmentation ideal of k[P/⟨z⟩]. By the induction hypothesis, some power of IP lands inside ([z]−1)k[P]. Since step 1.1 makes that ideal nilpotent, a further power of IP is 0. Therefore IP is nilpotent.

Depends on

Used by

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