The isomorphism problem for group algebras: A criterion

Problem.

Let G and H be finite p-groups. Does a unital algebra isomorphism 𝔽p⁢G≅𝔽p⁢H imply a group isomorphism G≅H?

Lemma 1.1.

Let G and H be finite groups. We denote by Epi⁡(G,H) the set of all epimorphisms from G to H. Let H be the set of all maximal subgroups of H. Then

Proof.

First, note that

Thus, by letting HomK(G,H)={f:G→H∣imf≤K}, it becomes

By the inclusion-exclusion principle, we have

(Note that if 𝒦=∅, then ⋂K∈𝒦K={h∈H∣h∈K(K∈𝒦)}=H.)∎

Lemma 1.2.

Let G and H be finite groups. Then G≅H if and only if |G|=|H| and

for every subgroup K of G.

Proof.

From |Hom⁡(G,K)|=|Hom⁡(H,K)| for every subgroup K of G, it follows that |Epi⁡(H,G)|=|Epi⁡(G,G)|≥1 by Lemma 1.1. Hence, we have an epimorphism from H to G. It must be an isomorphism because |G|=|H|. ∎

Lemma 1.3.

Let R be a unital commutative ring, G a group and A a unital R-algebra. Then there is a bijection

which is natural in G and A. (Namely, the group algebra functor is left adjoint to the unit group functor.)

Proof.

The restriction (f:RG→A)↦(f|G:G→A∗) gives rise to the desired bijection; see [17, pp. 204–205, 490] for details. ∎

Definition 1.4.

Set M={[G]∣G⁢is a finite group}, where the symbol [G] denotes the isomorphism class of a group G. It becomes a commutative monoid with an operation [G]+[H]=[G×H]. Let K⁢(M) denote the Grothendieck group^[1] of M. As it is a ℤ-module (abelian group), we can extend scalars and obtain a ℚ-vector space L⁢(M)=ℚ⊗ℤK⁢(M). For a finite unital commutative ring R, define the ℚ-subspace S⁢(R) of L⁢(M) by

Namely, S⁢(R) is the ℚ-subspace of L⁢(M) spanned by the isomorphism classes of unit groups of finite unital R-algebras. We call a finite group G a hereditary group over R if [K]∈S⁢(R) for every subgroup K of G.

Example 1.5.

Let Cq denote the cyclic group of order q. Then, from 𝔽5∗≅C4 and (𝔽5⁢C5)∗≅C4×(C5)4, we have

In particular, C5 is a hereditary group over 𝔽5.

Criterion 1.6.

Let G and H be finite groups, and let R be a finite unital commutative ring. Suppose G is a hereditary group over R. If R⁢G≅R⁢H, then G≅H.

Proof of Criterion 1.6.

Since a unital commutative ring R has invariant basis number (IBN) property, we have |G|=|H| from R⁢G≅R⁢H. Hence, by Lemma 1.2, it suffices to prove that |Hom⁡(G,K)|=|Hom⁡(H,K)| for every subgroup K of G.

Since G is a hereditary group over R, we have [K]∈S⁢(R). Therefore, there is a positive integer n and finite unital R-algebras A,B such that

Namely, A∗≅B∗×Kn. Hence,

Thus, by Lemma 1.3, we can obtain

We can calculate |Hom⁡(H,K)| similarly and conclude that

from R⁢G≅R⁢H. ∎

Remark 1.7.

Studying a finite group that is a “linear combination” of unit groups, precisely an element of S⁢(R), is essential because even the cyclic group of order five cannot be realized as a unit group of any unital ring. (See a theorem by Davis and Occhipinti [4, Corollary 3], for example.) This is quite different from the fact that every finite abelian p-group is a hereditary group over 𝔽p (Lemma 3.2).

It also should be noted that no examples of non-hereditary groups are hitherto found.

Definition 2.1.

Let A be an R-algebra. Define the quasi-multiplication on A by

An element x∈A is called quasi-regular if there is an element y∈A such that x∘y=0=y∘x. We denote the set of all quasi-regular elements by Q⁢(A). It forms a group under the quasi-multiplication, and we call it the quasi-regular group of A. If A=Q⁢(A), then A is called quasi-regular (or radical).

Definition 2.2.

Let A be an R-algebra. We denote the unitization of A by Aun: it is a direct product Aun=A×R as R-modules, and its multiplication is defined by

Lemma 2.3.

Let A be a quasi-regular R-algebra. Then an element (x,r)∈Aun is a unit if and only if r∈R is a unit.

Proof.

The “only if” part is trivial. Let us assume r∈R∗ to show (x,r)∈Aun∗. As r-1∈R exists and r-1⁢x∈A, there is an element y∈A such that

because A is quasi-regular. Then it can be shown that (x,r)-1=(r-1⁢y,r-1) by direct calculation. ∎

Lemma 2.4.

Let A be a quasi-regular R-algebra. Then Aun∗≅Q⁢(A)×R∗. In particular, [Q⁢(A)]∈S⁢(R) if R and A are finite.

Proof.

Note that there are homomorphisms

which are well-defined by Lemma 2.3. It is straightforward to check that these satisfy the universal property of a direct product. ∎

Remark 2.5.

For determining whether a finite group is a quasi-regular group of an R-algebra, a theorem by Sandling [18, Theorem 1.7] would be helpful. It should be noted that quasi-regular groups also have severely restricted structure as unit groups. See [19, p. 343].