On group invariants determined by modular group algebras: even versus odd characteristic

Let $p$ be a an odd prime and let $G$ be a finite $p$-group with cyclic commutator subgroup $G'$. We prove that the exponent and the abelianization of the centralizer of $G'$ in $G$ are determined by the group algebra of $G$ over any field of characteristic $p$. If, additionally, $G$ is $2$-generated then almost all the numerical invariants determining $G$ up to isomorphism are determined by the same group algebras; as a consequence the isomorphism type of the centralizer of $G'$ is determined. These claims are known to be false for $p=2$.


Introduction
Let G be a group, R a commutative ring and let RG denote the group ring of G with coefficients in R. The problem of describing how much information about the group G is carried by the group algebra RG has a long tradition in mathematics, with applications in particular to the representation of groups and in general to group theory; cf.[Hig40, PW50, Bra63, Pas65, Seh67, Whi68, Pas77, Seh78, CR81, San85, CR87, BKRW99].The last question can be rewritten more compactly as: Which group invariants of G are algebra invariants of RG?By a group invariant of G we understand a feature of G that is shared with any group isomorphic to G while an algebra invariant is a feature that is shared with any group H with the property that RG and RH are isomorphic as R-algebras.For instance, the cardinality of G can be expressed as the R-rank of RG and is thus an algebra invariant of RG.Moreover, the group G is abelian if and only if RG is a commutative ring, i.e. the property of being abelian is an algebra invariant of RG.The ultimate version of the above question is the Isomorphism Problem which asks for the determination of pairs (G, R) for which the isomorphism type of G is an algebra invariant of RG: Isomorphism Problem for group algebras: Given a commutative ring R and two groups G and H, does RG and RH being isomorphic as R-algebras imply that the groups G and H are isomorphic?In symbols, The answer to this question is a function of the ring R: for instance, it is easily shown that any two nonisomorphic finite abelian groups of the same order have isomorphic group rings with complex coefficients.However, by a seminal result of G. Higman [Hig40], if G and H are non-isomorphic abelian groups then ZG and ZH are not isomorphic.More surprisingly, there even exist two non-isomorphic finite metabelian groups G and H such that kG and kH are isomorphic for every field k [Dad71].Nonetheless, the Isomorphism Problem has a positive solution for R = Z and G and H metabelian [Whi68].This extends Higman's result for abelian groups [Hig40] and has been followed by positive results for more families of groups, such as nilpotent groups [RS87] and supersolvable groups [Kim91].These early results yielded to strong expectations that the Isomorphism Problem for integral group rings (R = Z and G, H finite) would have a positive solution until Hertweck's construction of two non-isomorphic finite groups with isomorphic integral group rings [Her01].Among the classical variations of the Isomorphism Problem, the one that remained unanswered the longest deals with the case where R is a field of positive characteristic p and G and H are p-groups, formally: (1) C G (G ′ ) and C H (H ′ ) have the same exponent.
Our next results concern 2-generated p-groups with cyclic commutator subgroup.In order to present them we introduce some numerical invariants of these groups.Since the Modular Isomorphism Problem has a positive solution for abelian groups [Des56] we only consider non-abelian groups.To this end, let G be a 2-generated non-abelian p-group, that is, G ′ is non-trivial and G is generated by exactly 2 elements.A basis of G is then a pair (b 1 , b 2 ) of elements of G such that

Moreover, we define
where min lex refers to the minimum with respect to the lexicographic order.The following result implies that if G ′ is cyclic then so is H ′ and O(G) is an algebra invariant of kG.
Theorem B. Let k be a field of odd characteristic p and let G and H be finite p-groups with kG Observe that the hypothesis that p is odd in Theorem B is necessary because This shows again a clear contrast between odd and even characteristic.
For any p-group G that is 2-generated and for which G ′ is cyclic, the vector O(G) can essentially be extended to a vector inv( ) of numerical invariants characterizing the isomorphism class of G; cf.Section 3 and [BGLdR21].It is well known that the first four entries p, m, n 1 and n 2 of inv(G) are algebra invariants of kG.However, the sixth entry is not determined by the modular group algebra because inv(G m,n ) = (2, 2, n, m, −1, −1, 0, 0, 0, 0, 1, 1) is different from inv(H m,n ) = (2, 2, n, m, −1, 1, 0, 0, 0, 0, 1, 1).Note that, for p > 2, one always has σ 1 = σ 2 = 1 and therefore the counterexample from [GLMdR22] does not have a direct equivalent in odd characteristic.We will see that Theorem B is actually equivalent to the following.
Theorem C. Let k be a field of odd characteristic p and let G be a finite non-abelian p-group.If G is 2-generated with G ′ cyclic then all but the last 2 entries of inv(G) are algebra invariants of kG.
In other words, Theorem C ensures that, for p > 2, the first 10 entries of inv(G) are determined by the modular group algebra kG of G over any field k of characteristic p. Unfortunately we have not been able to decide whether the last two entries of inv(G) are algebra invariants of kG.The smallest groups for which Theorem C does not solve the Modular Isomorphism Problem occur for (n 2, 2, 0, 1, 1, 1), in which case u 2 = 1 and u 1 ∈ {1, . . ., p − 1}.That is, the last parameters yield p − 1 nonisomorphic 2-generated p-groups with cyclic commutator subgroup.In a paper in preparation, we develop new techniques (different from those presented in this paper) to prove that, for this special case and many others, u 1 is actually also an algebra invariant of kG.Theorem C enables us, however, to improve Theorem A in the 2-generated case by showing that the isomorphism type of C G (G ′ ) is an algebra invariant of kG: Corollary D. Let k be a field of odd characteristic p and let G be a finite 2-generated p-group with G ′ cyclic.Then the isomorphism type of C G (G ′ ) is an algebra invariant of kG.
The following corollary also follows from Theorem C (see Section 2 for the definition of the type invariants of a p-group).
Corollary E. Let k be a field of odd characteristic p and let G be a finite 2-generated p-group with G ′ cyclic.Then the type invariants of G are algebra invariants of kG.
The paper is organized as follows.In Section 2 we establish the notation, recall some known facts about the Modular Isomorphism Problem and prove a key lemma which we refer to as the Transfer Lemma (Lemma 2.6).In Section 3 we recall the classification of finite 2-generated p-groups with cyclic commutator subgroup from [BGLdR21] in the specific case where p > 2. Additionally, we prove a series of results about these groups which will be used in the next and final section to prove the main results of the paper.

Notation and preliminaries
In this section, we introduce the notation that will be used throughout this paper.We also collect some classical results on the Modular Isomorphism Problem that will be useful in the coming sections, as well as a new criterion for the transfer of ideals between modular group algebras; cf.Lemma 2.6.
Throughout the paper, p will denote a prime number, k a field of characteristic p and G and H finite p-groups.The modular group algebra of G over k is denoted by kG and the augmentation ideal of kG is denoted by I(G).It is a classical result that I(G) is also the Jacobson ideal of kG.For every normal subgroup N of G, we write I(N ; G) for the relative augmentation ideal I(N )kG.
We let ≤ lex denote the lexicographic order on tuples of integers of the same length.Then min lex and max lex stand for minimum and maximum with respect to ≤ lex .For a non-zero integer n, let v p (n) denote the p-adic valuation of n, that is, the greatest integer t such that p t divides n.Moreover, set v p (0) = +∞.For coprime integers m and n, write o m (n) for the multiplicative order of n modulo m, i.e. the smallest non-negative integer k with n k ≡ 1 mod m.Given non-zero integers s, t and n with n ≥ 0 we set The last notation allows us, in some cases, to compactly express powering of products in a group G.For instance, if g, h ∈ G and r, s, n are integers with n ≥ 0 then, writing a = [h, g] = h −1 g −1 hg, we get the following identities: (2.2) if a g = a r and a h = a s then (gh) n = g n h n a T (r,s|n) .
The next lemma describes elementary properties of the operators S and T that are collected in Lemmas 8.2 and 8.3 of [BGLdR21].
Lemma 2.1.Let p be an odd prime number and let n > 0 and s, t be integers satisfying s ≡ t ≡ 1 mod p.
Then the following hold: Lemma 2.2.Let p be an odd prime number and m and r be integers with m > 0 and r ≡ 1 mod p. Then for every integer 0 ≤ x < p m there is a unique integer 0 ≤ y < p m such that S (r | y) ≡ x mod p m .
Proof.Let x and y be integers with 0 ≤ x ≤ y.Then S (r | y) − S (r | x) = r x S (r | y − x) and hence from Lemma 2.1(1) it follows that S (r | x) ≡ S (r | y) mod p m if and only if x ≡ y mod p m .This shows that S (r | •) induces an injective, thus bijective, map Z/p m Z → Z/p m Z. Then the result follows immediately.
The group theoretic notation we use is mostly standard.For an arbitrary group G, let |G| denote its order, Z(G) its center, {γ i (G)} i≥1 its lower central series, G ′ = γ 2 (G), its commutator subgroup, exp(G) its exponent and d(G) = min{|X| : X ⊆ G and G = X }, its minimum number of generators.Moreover, if g ∈ G and X ⊆ G then |g| denotes the order of g and C G (X) the centralizer of X in G.We write × both for internal and external direct products of groups.For n ≥ 1, we denote by C n the cyclic group of order n.
Let now G be a finite p-group and let p e = exp(G).For every 0 ≤ n ≤ e we define the following subgroups of G: If N is a normal subgroup of G, we also write The group G is said to be regular if for every g, h ∈ G there exist c 1 , . . ., c k ∈ g, h ′ such that (gh It is well known that if p is odd and G ′ is cyclic then G is regular, while this is not the case for p = 2; cf. [Hup67, Satz III.10.2,Satz III.10.3(a)].Moreover, if G is regular, [Hup67, Hauptsatz III.10.5, Satz III.10.7] ensure that the following hold: For every n ≥ 1 and G regular we define w n by means of we define e i = |{1 ≤ n ≤ e : ω n ≥ i}|.It follows that e = e 1 ≥ e 2 ≥ . . .≥ e ω(G) and the entries of the list (e 1 , . . ., e ω(G) ) are called the type invariants of G.
The Jennings series The Jennings series is also known as the Brauer-Jennings-Zassenhaus series or the Lazard series or the dimension subgroup series of kG (see [Pas77, Section 11.1] for details).A property of these series that we will use is that, for abelian groups, the orders of the terms completely determine the structure of the group.For more on the Jennings series, see for instance [Seh78, Section III.1], [Mar22, Section 4] and [MS22, Section 2.3].
The next proposition and lemma collect some well-known results which will be used throughout the paper.
Proposition 2.3.Let k be a field of positive characteristic p and let G be a finite p-group.Then the following statements hold: (1) If H is a finite p-group and φ : kG → kH is an isomorphism of k-algebras then (2) The following group invariants of G are algebra invariants of kG: (3) The Modular Isomorphism Problem has a positive solution in the following cases: Lemma 2.4.Let k be a field of characteristic p > 0, let G and H be finite p-groups and let L G and L H be normal subgroups of G and H, respectively.Assume that there is an isomorphism φ : kG → kH such that φ (I(L G ; G)) = I(L H ; H).Then, for each i ≥ 1, there is an isomorphism of groups Remark 2.5.Although all the statements of Proposition 2.3 are well known, some of them appear in the literature with the assumption that k = F p , the field with p elements, and the proof of others is hidden inside proofs of other statements.We add a few words so that the reader can track the results in the literature.The proof of (1) can be found inside the proof of [BK07, Theorem 2.(ii)].Statements (a) and (b) from (2) are proven in [Seh67,War61], [San85,War61] and [Kü82], respectively.The statement in (2)(c) for the Jenning series of G is proved in [Pas77, Lemma 14.2.7(i)] while the statement for G ′ is proved for k = F p in [Bag88, Lemma 2] and it can be generalized to arbitrary k using the argument from [Pas77, Lemma 14.2.7(i)].Observe that the two statements in (2)(c) also follow from Lemma 2.4 specialized to G) .In [San85, Theorem 6.11] point (2)(e) is stated for k = F p , but its proof can be easily generalized to hold for any k.
Statement (3)(a) is proven in [Des56, Theorem 2] while statements (b) and (c) of (3) are proven for k = F p in [Bag88,San96] and [BdR21], respectively.The latter proofs generalize gracefully for any k.Note that the analogue of (3)(c) for p = 2 appears in [BdR21] for k = F 2 , but the proof does not generalize to arbitrary k.
Finally, Lemma 2.4 is proven for k = F p in [San85, Lemma 6.26] and the proof can be easily generalized to hold for any field of characteristic p.
We close this section with a lemma that we will make use of to obtain new group algebra invariants from old ones.A version of this lemma, specialized for N Γ = Γ ′ and with a different proof, appears in [Sal93].
Lemma 2.6 (Transfer Lemma).Let p be a prime number and G and H finite p-groups.
) and hence we assume without loss of generality that t = 1.For brevity write L Γ = ℧ 1 (Γ, N Γ ) and we will prove that φ(I(G; L G )) = I(H; L H ).
Let τ Γ : kΓ → kΓ be the ring homomorphism extending the identity on Γ and the Frobenius map x → x p on k.Moreover, for a normal subgroup K of Γ, let π K : kΓ → k(Γ/K) denote the natural projection with ker π K = I(K; Γ).As Γ/N Γ is abelian, the assignment g → g p on Γ induces a ring homomorphism λ Γ : k(Γ/N Γ ) → k(℧ 1 (Γ/N Γ )).We denote by σ Γ the k-linear map extending the restriction of λ Γ to Γ/N Γ .By the definition of L Γ and the hypothesis φ(I(N G ; G)) = I(N H ; H), we have that σ Γ induces an isomorphism σΓ : As φ is a bijection, then so is φ.Moreover, each σΓ is bijective and each τ Γ is injective, thus we have as desired.

Finite 2-generated p-groups with cyclic commutator and p odd
In this section p is an odd prime number.We start by recalling the classification of non-abelian 2generated p-groups with cyclic commutator subgroup from [BGLdR21].Each such group G is showed to be uniquely determined, up to isomorphism, by an integral vector 2 ) of length 12 whose entries are determined as described below.The first four are straightforward and satisfy: where r(g) denotes the unique integer satisfying 2 ≤ r(g) ≤ p m + 1 and a g = a r(g) for each a ∈ G ′ .Define: Let now B r = B r (G) be the set consisting of all bases (b 1 , b 2 ) of G with the property that, for every a ∈ G ′ and i = 1, 2, one has We have described how the entries of inv(G) are computed directly as structural invariants of G. Conversely, for a list of non-negative integers , defining r 1 and r 2 as in (3.1), the group G I is defined as 1 The classification in [BGLdR21] is performed for all primes p and for p = 2 the fifth and sixth entries of inv(G) may also be −1.
Denoting by [G] the isomorphism class of a group G, the main result of [BGLdR21] for p odd takes the following form.
Theorem 3.1.The maps [G] → inv(G) and I → [G I ] define mutually inverse bijections between the isomorphism classes of 2-generated non-abelian groups of odd prime-power order and the set of lists of integers (3) One of the following conditions holds: 1 ≤ m ≤ n 1 and one of the following conditions hold: 6) One of the following conditions holds: , and u 1 ≡ 1 mod p; where In the remainder of the section, G denotes a finite non-abelian 2-generated p-group with cyclic commutator subgroup, with invariant vector . and associated r 1 and r 2 as in (3.1).
Thanks to Theorem 3.1(2) and Lemma 2.1(1) we have The following two lemmas are Lemma 2.2 and Lemma 4.2 from [BGLdR21].

and only if one of the following conditions holds:
(1) o(b 1 ) = 0. ( if and only if the following conditions hold: For the following result, recall from the introduction that Lemma 3.4.The following equality holds: Proof.For every g ∈ G let r(g) be the unique integer 2 ≤ r(g) ≤ p m + 1 such that a g = a r(g) for every a in G ′ .From Lemma 2.1(1) it follows that In . On the other hand, the two automorphisms of G ′ given by a → a bi and a → a b ′ i have order p oi .Since Aut(G ′ ) is cyclic, there exist integers x 1 and x 2 , both coprime to p, such that b In the remainder of the section let b Then the following hold: In particular, every element of G is of the form b x 1 b y 2 a z for some integers x, y, z.Moreover, it follows from (2.1) and (2.2) that, for every non-negative integer e, one has The next lemma describes some characteristic features of G.
Lemma 3.5.The following statements hold: Proof.(1) Let w be a non-negative integer.As v p (r i − 1) = m − o i and a bi = a ri , we have that a p w ∈ Z(G) if and only if, for each i ∈ {1, 2}, one has w . By (3.5), we have that exp(G) ≥ p e so we show that ℧ e (G) = 1.To this end, note that e ≥ m as a consequence of Theorem 3.1(4) and thus ℧ e (G ′ ) = 1.Now regularity yields that p e -th powering induces a homomorphism G/G ′ → G and so, as a consequence of (3.5), we get ℧ e (G) = 1.
(3) We work by induction on i and, as the base case i = 2 is clear, we assume that i > 2 and the claim holds for i − 1.In other words, write f We conclude by computing v p (p f min(r Since each S (r i | −) induces a bijection Z/p m Z → Z/p m Z (see Lemma 2.2) there are unique integers 1 ≤ δ 1 ≤ p o1 and 1 ≤ δ 2 ≤ p o2 satisfying the following congruences: Moreover, (3.1) and Lemma 2.1(1) yield that p does not divide δ 1 δ 2 .
Proof.Let g = b x 1 b y 2 a z be an arbitrary element of G with x, y, z ∈ Z.We characterize when g ∈ Z(G) in terms of conditions on the exponents x, y, z.For this, note that (3.8) and (3.9) ensure that g = b x 1 b y 2 a z ∈ Z(G) if and only if the following congruences hold:  , a is regular and that ) , and these three elements generate the same subgroup a Then the following hold: ( (1) This is a direct consequence of Lemma 3.5(1) and Lemma 3.7.
(2) Let g = b x 1 b y 2 a z be an arbitrary element of G.
that is there exists an integer v with x = −yp o1−o2 + vp o1 .We have proven that To conclude, let L = ℧ t (G : Z(G)G ′ ).By (1), (3.15) and (3.16 (3) The group G ′ being cyclic, ℧ m−t (G ′ ) = Ω t (G ′ ) and so (2) and the regularity of For the other inclusion, it suffices to observe that In the following lemma, Soc(G ′ ) denotes the socle of G ′ .We remark that Soc(G ′ ) = a p m−1 , because G ′ is cyclic of order p m and m ≥ 1.
Lemma 3.9.Write G = G/Soc(G ′ ) and assume that m ≥ 2. Then G is a non-abelian group and one has Proof.That the first four entries of inv(G) are p, m − 1, n 1 and n 2 is obvious.Since we are dealing with two groups G and G, in this proof we distinguish B r = B r (G) and B r (G).Let g denote the natural image in G of an element g ∈ G. Then r(g) is the unique integer in the interval [2, p m−1 + 1] that is congruent to r(g) modulo p m−1 .By (3.4) we have Hence o(b i ) = max(0, o i − 1) and thus Lemma 3.2 yields that o i = max(0, o i − 1).Moreover, b = (b 1 , b 2 ) is an element of B r (G).Note that the following hold: and b To finish the proof we distinguish three cases according to Lemma 3.3 and search for some b ∈ B r (G) satisfying the corresponding conditions in the lemma.Then Lemma 3.3 will guarantee that o ′ i = o ′ i ( bi ) for i = 1, 2. In most cases b already satisfies the desired conditions and hence, in such cases, we take b = b and hence Otherwise we modify slightly b to obtain the desired b.Case 1. Suppose first that o 1 = 0.By Theorem 3.1(3) we have As o 1 = 0, the conditions in Lemma 3.3 hold for b = b and hence we have This yields the desired conclusion because in this case then Lemma 2.1 and Theorem 3.1(4) imply v p (k) ≥ n 2 ≥ m + o ′ 2 .Using (3.6) we get (b Therefore |b

Proofs of the main results
In this section we prove Theorem A, Theorem B and Theorem C. The first one will be included in Theorem 4.2, which relies on Lemma 4.1.Theorem B is proven shortly after Theorem 4.2, while Theorem C is a consequence of Lemmas 4.3 and 4.4.We conclude the section by proving Corollary D and Corollary E, here presented as Corollary 4.5 and Corollary 4.6, respectively.
Proof.The abelian case is straightforward so we assume that G ′ = 1.By a Theorem of Cheng [Che82] the group G can be expressed as a central product where each G i is a 2-generated group of nipotency class 2, the group H is 2-generated with H ′ = G ′ and A is abelian.For each i = 1, . . ., s, write G i = x i , y i and For the other inclusion take g = hk ∈ L with h ∈ H and k ∈ K and note that h ∈ H ∩ L ⊆ C H (H ′ ) by Lemma 3.8(2).This shows that C G (G ′ ) = L.
We now show that C G (G ′ ) ′ = ℧ m−t (G ′ ).For this, let g, h ∈ C G (G ′ ).Then h p t ∈ G ′ Z(G) and, as C G (G ′ ) has nilpotency class 2, we have that [g, h] p t = [g, h p t ] = 1.We have proven that Finally let N be a subgroup of G such that N ⊆ C G (G ′ ).Then N has nilpotency class 2, so that (2.3) yields D p n (N ) = ℧ n (N ) and the result follows .
The following result is a stronger version of Theorem A.
Theorem 4.2.Let k be a field of odd characteristic p and let G be a finite p-group with cyclic commutator subgroup.If H is another group with kG and kH isomorphic as k-algebras then (a) For every algebra isomorphism φ : kG → kH preserving augmentation one has that by Lemma 2.4.This implies that the lists of orders of the terms of the Jennings series of C G (G ′ ) and C H (H ′ ) are equal and, by Lemma 4.1, these groups have the same exponent.Finally, observe that since In the remainder of the section we prove Theorem B, Theorem C, Corollary D and Corollary E. For that we fix a field k of odd characteristic p, a finite 2-generated p-group G with cyclic commutator subgroup and a group H such that kG ∼ = kH.Then, by Proposition 2.3(2)(d), the group H is also 2-generated with a cyclic commutator subgroup.Moreover, Proposition 2.3(2)(a) yields that, if one of the two groups is abelian, then G ∼ = H.We assume hence without loss of generality that G and H are non-abelian.Now, by Proposition 2.3(2)(a), the first six entries of inv(G) and inv(H) coincide.Thus we set particular, if b ∈ B r and i = 1, 2, then o(b i ) = o i .Thus the first two entries of O(G) are p o1 and p o2 .To deal with the remaining two entries, fix two bases b = (b 1 , b 2 ) and b 14) In particular, the elements b p m 1 , b p m 2 and c = b δ2p m−o 2 12) we have b p m 1 , b p m 2 , d = b p m 1 , b p m 2 , c ⊆ Z(G).If o 1 = 0 and g = b x 1 b y 2 a z ∈ Z(G) then it follows from (3.1) that (3.13) is equivalent to y ≡ 0 mod p m and hence (3.14) is equivalent to x ≡ zp m−o2 mod p m .Thus g ∈ b p m 1 , b p m 2 , d and hence Z(G) = b p m 1 , b p m 2 , d , as desired.Suppose otherwise that o 1 > 0 and define B = b p m−o 1 2 , N = b p min(m,n 2 ) 2 and f : Z(G) → B/N by and hence b y 2 ∈ B. We claim that ker f = b p m 1 , b p m 2 , a p o 1 .The inclusion from right to left is clear.Assume g = b x 1 b y 2 a z ∈ ker f .If m ≤ n 2 this implies that p m divides y.Then from (3.13) and Lemma 2.1(1) we have that v p (z) ≥ o 1 and hence from (3.14) we deduce that v p (x) ≥ m − o 2 + o 1 > m.This shows that g ∈ b p m 1 , b p m 2 , a p o 1 , as desired.Suppose now that m > n 2 .Then g = b x 1 a z for some integers x and z and hence again from (3.13) we deduce that v p (z) ≥ o 1 and from (3.14) we conclude that v p (x) ≥ m − o 2 + o 1 > m.Therefore, again g ∈ b p m 1 , b p m 2 , a p o 1 and the claim is proven.We finally show that Z(G) = b p m 1 , b p m 2 , c .To this end, observe that B/N is generated by f (c), as p does not divide δ 1 .This together with the claim and the fact that f is a group homomorphism implies that Z(G) = c, ker f = b p m 1 , b p m 2 , a p o 1 , c .To complete the proof we show that a p o 1 ∈ b p m 1 , b p m 2 , c .To this end, observe that the group H = b p m−o 2 1 , b p m−o 1 2 1 and hence b does not satisfy the hypotheses of Lemma 3.3.Then we take b = (b 1 b p o 2 2 , b 2 ), which belongs to B r (G) because [b 2 p o 2 , a] = 1.Using (3.6), the regularity of G and m ≤ n 1 we compute bp b). Therefore Lemma 3.3 yields o ′ (b) = o ′ .In this case we take the basis b = (b 1 , b p o 1 1 b 2 ), which again belongs to the set B r (G) because [b
[Seh78]e a group such that kG ∼ = kH and let Γ ∈ {G, H}.It is well known that there exists an isomorphism kG → kH preserving augmentation (see e.g. the remark on page 63 of[Seh78]).Then H ′ is also cyclic as a consequence of Proposition 2.3(2)(d).Moreover, the number |Z(G) ∩ G ′ | = p t is an algebra invariant of kG by Proposition 2.3(2)(e).By Lemma 4.1 and Proposition 2.3(1), the hypotheses of Lemma 2.6 hold for L This and the fact that the groups C G (G ′ )/G ′ and C H (H ′ )/H ′ are both abelian yield, by using the argument in [Pas77, Lemma 14.2.7(ii)], that they are isomorphic.Writingp m = |G ′ |, Lemma 4.1 ensures that I(C G (G ′ ) ′ ; G) = I(G ′ ; G) p m−t and so φ induces another isomorphism φ : k(G/C G (G ′ ) ′ ) → k(H/C H (H ′ ) ′ ), and the same argument yields that C