Elementary Abelian p-groups of rank greater than or equal to 4p − 2 are not CI-groups

DOI 10.1007/s10801-007-0059-2

Elementary Abelian p-groups of rank greater than or equal to 4p − 2 are not CI-groups

Pablo Spiga

Received: 27 July 2006 / Accepted: 17 January 2007 / Published online: 17 April 2007

© Springer Science+Business Media, LLC 2007

Abstract In this paper we prove that an elementary Abelianp-group of rank 4p−2 is not a CI⁽²⁾-group, i.e. there exists a 2-closed transitive permutation group con- taining two non-conjugate regular elementary Abelianp-subgroups of rank 4p−2, see Hirasaka and Muzychuk (J. Comb. Theory Ser. A 94(2), 339–362,2001). It was shown in Hirasaka and Muzychuk (loc cit) and Muzychuk (Discrete Math. 264(1–3), 167–185,2003) that this is related to the problem of determining whether an elemen- tary Abelianp-group of ranknis a CI-group.

As a strengthening of this result we prove that an elementary Abelianp-groupE of rank greater or equal to 4p−2 is not a CI-group, i.e. there exist two isomorphic Cayley digraphs overE whose corresponding connection sets are not conjugate in AutE.

Keywords Cayley graph·CI-group·Schur ring·2-closure

1 Introduction

Let H be a group and S a subset of H. The Cayley digraph of H with con- nection set S, denoted Cay(H, S), is the graph with vertex set H and edge set {(h₁, h₂)|h₁h⁻₂¹∈S}. Two Cayley digraphs Cay(H, S)and Cay(H, T )are said to be Cayley isomorphic if there exists an elementg∈AutHsuch thatS^g=T. A sub- setSof a groupHis said to be a CI-subset (or Cayley isomorphic subset) if for each T ⊆H the digraphs Cay(H, S)and Cay(H, T )are isomorphic if and only if they are Cayley isomorphic. Finally, a groupH is said to be a CI-group if each subset of

This research was supported by a fellowship from the Pacific Institute for the Mathematical Sciences.

P. Spiga (

Department of Mathematics and Computer Science , University of Lethbridge, 4401 University Drive, Lethbridge, AB, Canada

e-mail: [email protected]

H is a CI-subset. We refer the interested reader to the survey article [5] for details, examples and the main results on CI-groups. We remark that the classification of CI- groups is not known and at the time of this writing it seems that the classification problem strongly depends on whether an elementary Abelianp-group of ranknis a CI-group, see [5].

It was proved in [1] thatS⊆H is a CI-subset if and only if any regular subgroup of Aut(Cay(H, S))isomorphic toHis conjugate toHin Aut(Cay(H, S)). This result is the starting point of most of the results on CI-groups.

In [3] the authors proved that ifV is a regular elementary Abelianp-subgroup of rankn(n≤4,p >2) of a 2-closed permutation groupG, then any regular subgroup W ofGisomorphic toV is conjugate toV inG. In particular, by the former para- graph, as the automorphism group of a digraph is a 2-closed group we have that an elementary Abelianp-group of rank less than or equal to 4 is a CI-group. Motivated by this result the authors of [3] gave the following definition, see [3] page 341.

Definition 1 The groupHis said to be a CI⁽²⁾-group if for any 2-closed permutation groupGcontaining the right regular permutation representation ofH, we have that any two regular subgroups ofGisomorphic toH are conjugate inG.

Clearly ifH is a CI⁽²⁾-group, thenHis a CI-group. The authors of [3] asked for a complete classification of the CI⁽²⁾-elementary Abelianp-groups. At the time of this writing it is known that if an elementary Abelianp-group of ranknis a CI⁽²⁾-group, thenn <2p−1+_2p−1

p

, see [6], and, ifn≤4, then an elementary Abelianp-group of ranknis a CI⁽²⁾-group, see [3].

In Sect.3we prove the following theorem.

Theorem 1 An elementary Abelianp-group of rankn≥4p−2 is not a CI⁽²⁾-group.

We remark that a proof of Theorem1could be given using cohomological argu- ments. As a matter of consistency we present a proof closely related to the arguments in [6].

Finally, in Sect.4, we prove Theorem2, which is the main result of this paper.

Theorem 2 An elementary Abelianp-group of ranknis not a CI-group ifn≥4p−2.

We remark that although Theorem1is a corollary of Theorem2, the proof of the former is needed in order to prove the latter.

We refer to the precious book [9] for the main results on Schur rings and to [3]

and [6] for their connections to CI⁽²⁾-groups and for the notation. We strongly advise the reader to use [6] as a crib.

2 Preliminaries

Letpbe a prime number, letV , W be non-trivial elementary Abelianp-groups and letU=VwrWW be the wreath product ofV withW acting on=W×V in its natural imprimitive action. We recall that(x, y)^wf =(x+w, y+f (x+w))for any

(x, y)∈ andwf ∈U. We denote by B=V^W the base group ofU. We denote byξ_m(U ) thenmth term of the upper central series of U. In particular, the centre ξ₁(U )ofUconsists of the constant functions ofB. We shall often identify a constant function ofB, i.e. an element ofξ₁(U ), with its image. LetGbe a subgroup ofU containingW ξ₁(U ). Note thatW ξ₁(U )is an elementary Abelian regular subgroup of U. FurthermoreG=W LwhereL=G∩B. We denote byL0the stabilizer inGof the element(0,0)of. The groupL0defines a mapH:W→2^V = {S|S⊆V}by H (w)= {f (w)|f ∈L0}.

Lemma 1 Hsatisfies the following properties:

(i) H (w)is a subspace ofV; (ii) H (0)= {0};

(iii) H (βw)=H (w)for anyβ=0 inFp;

(iv) H (w1+w2)⊆H (w1)+H (w2)for anyw1, w2∈W.

Proof (i) L0is anFp-vector subspace ofBsoH (w)is anFp-vector subspace ofV. (ii)By definition ofL0we get thatH (0)= {0}.

(iii)Letf be inL0andwinW. Ifβ∈Fp, theng_β =f^−βw−f (βw)(here we are identifyingf (βw)∈V with the constant functionx →f (βw)) is an element in L₀(in factg_β(0)=0) andg_β(w)=f ((β+1)w)−f (βw). Thusf (w), f (2w)− f (w), . . . , f ((p−1)w)−f ((p−2)w)and−f ((p−1)w)are elements ofH (w).

Sof (βw)∈H (w)for anyβ∈Fp. In particularH (βw)⊆H (w)for anyβ∈Fp. If β=0, we haveH (w)=H (βw).

(iv) Now let f be in L₀ and w₁, w₂∈W. Consider g=f^−w2 −f (w₂) (as usual f (w₂) denotes the element of ξ₁(U ) with image f (w₂)). Now g∈L and g(0)=(f^−w2−f (w2))(0)=f (w2)−f (w2)=0, sog∈L0. Moreoverg(w1)= f (w1+w2)−f (w2), thereforef (w1+w2)=g(w1)+f (w2)∈H (w1)+H (w2).

We denote by Der(W, L)the group of derivations fromWtoLand by Inn(W, L) the group of inner derivations, see [8].

Lemma 2 Any regular elementary Abelian subgroup ofGis conjugate toW ξ₁(U )if and only if Der(W, L)=Hom(W, ξ1(U ))+Inn(W, L).

Proof Letδbe an element in Der(W, L). ThenH=X_δξ₁(U )is a regular elementary Abelian subgroup ofG, whereX_δ= {ww^δ|w∈W}. In particular ifHis conjugate to W ξ₁(U ), then there exists l∈L such thatX_δ ⊆W^lξ₁(U ). In other words, for any w∈W there exists a unique c_w ∈ξ₁(U ) such thatw^δ= [w, l] +c_w. The map c₋:w →c_wis an homomorphism ofW intoξ₁(U ). In fact

[w₁, l]^w2+ [w₂, l] +c_w1+c_w2=([w₁, l] +c_w1)^w2+ [w₂, l] +c_w2

=(w₁^δ)^w2+w^δ₂

=(w1+w2)^δ= [w1+w2, l] +cw₁+w₂

and soc_w1+w2=c_w1+c_w2. Thusδ=c₋+ [−, l] ∈Hom(W, ξ1(U ))+Inn(W, L).

As any regular elementary Abelian subgroupH ofGis of the formX_δξ₁(U ), for someδ∈Der(W, L), the other side of the implication is totally trivial.

Assume that the groupGis 2-closed. Every orbital ofGcorresponds to an orbit of the point stabilizerL0, i.e. every orbital ofGcorresponds to a suborbit of the form (w, v)^L0= {(w, f (w)+v)|f ∈L0} =(w, v+H (w)). The groupGcontains the regular subgroupW ξ₁(U )and so every orbital is a Cayley digraph overW×V. In particular the orbital corresponding to the suborbit(w, v)^L0 is{((w₁, v₁), (w₂, v₂))| (w₁−w₂, v₁−v₂)∈(w, v+H (w))} =Cay(W×V , (w, v+H (w))). Furthermore G=

(w,v)Aut((w,v)), where(w,v)=Cay(W×V , (w, v+H (w))).

Let us define theFp-vector space

HomH(W, V )= {ψ∈B|ψ (w1+w2)−ψ (w1)−ψ (w2)∈H (w1)∩H (w2) for anyw₁, w₂∈W, ψ (0)=0},

see [6].

Next ifδ∈Der(W, L), we defineθ (δ):W→V byθ (δ)(w)=(w^δ)(0).

Lemma 3 θis an injective homomorphism of Der(W, L)into HomH(W, V ). IfGis 2-closed, thenθis an isomorphism.

Proof Clearlyθ (δ)(0)=(0^δ)(0)=0. Letw1, w2be elements ofW. Setw^δ₁=f+g wheref∈L0andg∈ξ1(U ). We have

θ (δ)(w1+w2)−θ (δ)(w1)−θ (δ)(w2)=(w1+w2)^δ(0)−w^δ₁(0)−w^δ₂(0)

=((w₁)^δ)^w2(0)+w^δ₂(0)−w₁^δ(0)−w^δ₂(0)

=w₁^δ(−w2)−w^δ₁(0)=f (−w2)∈H (w2).

By symmetry (the groupWis commutative) we get that θ (δ)(w1+w2)−θ (δ)(w1)−θ (δ)(w2)∈H (w1).

Thereforeθ (δ)∈HomH(W, V ).

The mapθ is an homomorphism. Ifδ₁, δ₂∈Der(W, L)andw∈W, thenθ (δ₁+ δ₂)(w)=w^δ1+δ2(0)=w^δ1(0)+w^δ2(0)=(θ (δ₁)+θ (δ₂))(w). Moreoverθ (0)=0.

Finallyw^αδ=αw^δfor anyα∈Fpand soθ (αδ)=αθ (δ).

The mapθis injective.θ (δ)=0 if and only ifw^δ(0)=0 for anyw∈W. Sow^δ∈ L0for anyw∈W. In particular 0=(w+w1)^δ(0)=((w^δ)^w1+w^δ₁)(0)=w^δ(−w1) for anyw1, w. Thusw^δ=0 for anyw. Thereforeδ=0.

Assume thatGis 2-closed. Letψbe an element of HomH(W, V ). Defineη(w)= ψ (−w) for any w∈W. The function η lies in the base group B. Furthermore η lies in HomH(W, V ), in factη(0)=ψ (0)=0 andη(w₁+w₂)−η(w₁)−η(w₂)= ψ (−w₁−w₂)−ψ (−w₁)−ψ (−w₂)∈H (−w₁)∩H (−w₂)=H (w₁)∩H (w₂). We claim thatηnormalizesG, i.e.[W, η] ⊆L. Letw be inW. We have to prove that [w, η] ∈L, i.e.g=η−η^w+η(−w)∈L0. We claim that_(w,v)^g =(w,v)for every

(w, v)∈. Let ((w₁, v₁), (w₂, v₂)) be in_(w,v). In particular w₁−w₂=w and v₁−v₂=v+x for somexinH (w). We have

((w1, v1), (w2, v2))^g=((w1, v1+g(w1)), (w2, v2+g(w2))) and

v1+g(w1)−(v2+g(w2))=v+x+η(w1)−η(w1−w)−η(w2)+η(w2−w).

Now, the functionηlies in HomH(W, V ), therefore

η(w₂+w)−η(w₂)−η(w)∈H (w)

−η(w₂+w−w)+η(w₂−w)+η(w)∈H (w)

so η(w1)−η(w1−w)−η(w2)+η(w2 −w)∈H (w). Hence v1+g(w1)− (v2 +g(w2)) ∈v +H (w). This yields _(w,v)^g = (w,v) for every (w, v) ∈ . So, g ∈ G⁽²⁾ =G. Therefore [W, η] ⊆L and [−, (−η)] ∈Der(W, L). Finally θ ([−, (−η)])(w) = [w,−η](0)= (η^w − η)(0) = η(−w) − η(0) =ψ (w), so θ ([−, (−η)])=ψ. This provesθis surjective and Der(W, L)∼=HomH(W, V ).

Lemma 4 Assume thatGis a 2-closed group. Any two regular elementary Abelian subgroups ofGare conjugate if and only if for anyψ∈HomH(W, V )there exists ∈Hom(W, V )such that(ψ−)(w)∈H (w)for everyw∈W.

Proof Assume that any two regular elementary Abelian subgroups ofGare conjugate and letψ∈HomH(W, V ). So, by Lemmas2and3we haveψ=θ ([−, l] +)for somel∈L₀and∈Hom(W, ξ1(U )). Clearly

(θ ([−, l] +)−)(w)=w^[−,l]+(0)−(w)

= [w, l](0)=(−l)^w(0)=(−l)(−w)∈H (w) for any w∈W. Conversely, by Lemma3it is enough to prove that Der(W, L)= Hom(W, ξ1(U ))+Inn(W, L). Letδbe an element of Der(W, L). By hypothesis there exists∈Hom(W, V )such that(θ (δ)−)(w)∈H (w)for anyw∈W. Setg(w)= (θ (δ)−)(−w) for any w∈W. Clearly _(w,v)^g =_(w,v+g(w))=_(w,v) for any (w, v)∈. This yieldsg∈L. We leave the reader to check thatδ= [−, (−g)] +.

Thus the proof is complete.

Summing up so far, a 2-closed subgroupGofU contains two non-conjugate ele- mentary Abelian regular subgroups if and only if there existsψ∈HomH(W, V )such that there does not exist∈Hom(W, V )such that(ψ−)(w)∈H (w)for every w∈W, see Proposition 7 in [6]. Note that this description is based on the knowledge ofHand in particular the groupGdoes not appear.

Conversely, we show that a mapHsatisfying(i), (ii), (iii), (iv)is intimately re- lated to a 2-closed subgroup of U. For (w, v) ∈ W × V define _(w,v) = {((w₁, v₁), (w₂, v₂))| w₁−w₂ =w, v₁−v₂∈H (w)+v}. Finally define, only

for the remainder of this section, A=

(w,v)Aut((w,v)). Note that W×V act- ing via right multiplication is a regular subgroup of Aut((w,v)). In particular _(w,v)=Cay(W ×V , (w, v+H (w))).

We claim thatAis a subgroup of U. Firstly we prove that= {{w} ×V}w∈W

is a block system for A. Let g be an element of A and be {w} ×V. As- sume that ^g∩= ∅. So, (w, v1)^g=(w, v2)for some v1, v2∈V. Let(w, v)∈ . Then ((w, v), (w, v1))∈(0,v−v1) so ((w, v), (w, v1))^g =((w, v)^g, (w, v2))∈ (0,v−v1). This implies that (w, v)^g ∈ , so ^g =. Next, we prove that A acts regularly on . The group A is transitive on and so A is transitive on . Assume that g fixes {w} ×V and let w₁ ∈W. Now, (w,0)^g =(w, v) for some v ∈V and (w₁,0)^g =(w₂, v₂) for some w₂ ∈W and v₂∈ V. Now g ∈ Aut((w1−w,0)) and ((w₁,0), (w,0)) ∈ _(w1−w,0) so ((w₁,0)^g, (w,0)^g) = ((w2, v2), (w, v))∈(w₁−w,0), so,w2−w=w1−w. Thusw1=w2. This says that Aacts regularly on. Let us denote byLthe kernel of the permutation representa- tion ofAon(i.e. the stabilizer of the set{0W} ×V). It remains to prove thatLacts regularly on{0W} ×V. The groupLcontains{0W} ×V and soLacts transitively on each set of the form{w} ×V. Letgbe inL. Assume that(0,0)^g=(0,0). Letvbe inV and let us prove that(0, v)^g=(0, v). Note that((0, v), (0,0))∈_(0,v) and so ((0, v), (0,0))^g=((0, v1), (0,0))is an element of_(0,v)=Cay(W ×V , (0, v)), so, v₁=v. This proves thatAis a subgroup ofU.

3 The construction

Since the elementary Abelian 2-group of rank 6 is known not to be a CI-group, see [7], we assume thatp≥3.

We strongly use the well-known structure of the upper central series of the group U, see [4].

LetV be the Galois fieldFpⁿ, wheren=2p−1, and letε1, . . . , ε_nbe anFp-basis of V. Also, letW be anFp-vector space of dimensionnwith basise1, . . . , e_n and dual basise^∗₁, . . . , e^∗_n. LetS be the symmetricV-algebra one^∗₁, . . . , e^∗_n and natural filtrationS= ⊕m∈NSm (recall that Sm is spanned by the monomials e_i^∗j1

1 ⊗ · · · ⊗ e^∗_i^jk

k of degreem). Let Mm be theV-subspace of Sm spanned by the monomials e^∗_i^j1

1 ⊗ · · · ⊗e^∗_i^jk

k of degreemsuch thatj₁, . . . , j_k≤p−1. Let π:S→B=V^W be the valuation map andZ_i =π(Mi). For instanceπ(e^∗_i^j1

1 ⊗ · · · ⊗e^∗_i^jk

k )(w)is the element of V defined by (e^∗_i

1(w))^j1· · ·(e_i^∗

k(w))^jk for any w∈W. It is known that B∩ξm(U )= ⊕i≤m−1Zi and thatπ_|Mi is injective. Without loss of generality we identifye_i^∗j1

1 ⊗ · · · ⊗e^∗_i^jk

k inMmwith its image underπ.

LetXbe the set{1, . . . , n}andX_i= {A⊆X| |A| =i}fori=0, . . . , p. IfAis a subset ofX, then we denote bye^∗_Athe element ofBdefined by⊗i∈Ae^∗_i. SimilarlyεA

denotes

i∈Aεi∈V. We letεandedenoteεXand

i∈Xei, respectively.

Consider

f =

A∈Xp

ε_X\Ae^∗_A∈B.

DefineL= [W, f]. Ifw₁, w₂∈W, then[w₁+w₂, f] = [w₁, f]^w2 + [w₂, f]. This shows thatG= W, L =W L. Since[G, f] = [W, f] ⊆G, we have thatf normal- izesG.

LetCbe a subset ofXof sizep−i. Define

gC=

B∈Xi,C∩B=∅

εX\(B∪C)e_B^∗.

Lemma 5 The groupL is generated by the set{gC|C∈Xi,1≤i≤p}andL0is generated by{g_C|C∈X_i,1≤i≤p−1}.

Proof Ifi∈A, then(e_A^∗)^ei(w)=

a∈Ae_a^∗(w−e_i)=e^∗_A(w)−e_A^∗_\{i}(w)and so (†) [e_i, e^∗_A] =

0 ifi /∈A, e^∗_A\{i} ifi∈A.

Using(†)we get

[e_i, f] =

B∈X_p−1,i /∈B

ε_X\(B∪{i})e^∗_B=g_{i}∈L for anyi∈X.

More generally we have

[e_i, g_C] =

0 ifi∈C, gC∪{i} ifi /∈C.

The groupLis generated by the left-normed commutators[f, e_i1, . . . , e_ik]fork≥1 and thereforeLis generated by{gC|C∈Xi,1≤i≤p}. Our claim onL0is just an

easy remark.

Corollary 1 Gcontainsξ₁(U ).

Proof By Lemma5we have thatL∩ξ1(U )is generated by{g_C|C∈X_p} = {ε_X\C| C∈Xp}. As usual we are identifying the elements ofV with the elements ofξ1(U ).

We leave the reader to show that{εX\C|C∈Xp}spansV. By Corollary1we have thatU⊇G⊇W ξ₁(U )and so we can apply Section2to G.

Letϕbe the non-degenerate bilinear symmetric form onV defined byϕ(ε_i, ε_j)= δij, whereδij is the Kronecker delta.

Lemma 6 H (e_i)=ε^⊥_i for anyi∈X.

Proof We have thatg_C(e_i)=0 for anyC∈X_p−j,j≥2. Therefore, by Lemma5, H (e_i)= g_C(e_i)|C∈X_p−1 = ε_X\(C∪{i})|i /∈C, C∈X_p−1, while it is an easy exercise in linear algebra to prove that this vector space isε^⊥_i . Lemma 7 Ifi=j, thenH (e_i+e_j)=(ε_i−ε_j)^⊥.

Proof Arguing is the same way as in Lemma 6 we have H (e_i + e_j) = gC(ei+ej)|C∈Xp−2∪Xp−1. It is routine computation in linear algebra to prove thatg_C(e_i+e_j)|C∈X_p−2∪X_p−1 = ε_{X\(C∪{i,j})}, ε_X\(C∪{i})+ε_X\(C∪{j})|C∈ X_p−2, C∈X_p−1, i, j /∈C∪C =(ε_i−ε_j)^⊥. Lemma 8 H (e)=ε^⊥.

Proof LetCbe a set of sizep−i, where 1≤i≤p−1. Then ϕ(g_C(e), ε)=

B∈X_i,C∩B=∅

ϕ(ε_X\(B∪C), ε)=

B∈X_i,C∩B=∅

(p−1)

= − p+i−1 i

=0.

IfC∈X_p−1, then we haveg_C(e)=

i /∈Cε_X\(C∪{i})= −ε_X\C. Thusε^⊥⊇H (e)⊇ gC(e)|C∈Xp−1 = εX\C|C∈Xp−1 =ε^⊥. Lemma 9 H (e−e_i)=(ε+ε_i)^⊥for anyi∈X.

Proof Let C be a set of size p−j, where 1≤j ≤p −1. If i /∈C, B ∈X_j, i /∈B,B∩C= ∅, thenϕ(εX\(C∪B), ε+εi)=0. Ifi∈C, B∈Xj, C∩B= ∅, then ϕ(ε_X\(C∪B), ε+ε_i)=p−1. The functione^∗_A is not 0 on the elemente−e_i if and only ifi /∈A. Thus

(∗) ϕ(gC(e−ei), ε+εi)=

B∈Xj,i∈/B,C∩B=∅

ϕ(ε_X\(B∪C), ε+εi).

So, ifi /∈C, then(∗)is equal to 0. Ifi∈C, then(∗)is equal to

B∈X_j,C∩B=∅

(p−1)=(p−1) p+j−1 j

=0.

If C ∈ Xp−1 and i /∈ C, then gC(e −ei)=

j /∈C,j=iεX\(C∪{j}) = −εX\C − ε_X\(C∪{i}). IfC∈X_p−1 andi∈C, theng_C(e−e_i)=

j /∈Cε_X\(C∪{j})= −ε_X\C. Now,(ε+ε_i)^⊥⊇H (e−e_i)⊇ g_C(e−e_i)|C∈X_p−1 = ε_X\C, ε_X\C+ε_X\(C∪{i})| C, C∈X_p−1, i∈C, i /∈C =(ε+ε_i)^⊥. Lemma 10 f (ei)=0, f (ei +ej)=0, f (e)=ε, f (e−ei)=ε, for anyi, j ∈X, i=j.

Proof Clearly the first two identities are trivial. We have f (e)=

A∈Xpε_X\A = _2p−2

p i∈Xε_i =ε. Finallyf (e−e_i)=

A∈Xp,i /∈Aε_X\A=_2p−3

p j∈X,j=iε_j + _2p−2

p

ε_i=ε.

Letgbe the linear map defined by 2⁻¹

i∈X

ε_X\{i}e^∗_i.

We have g(e_i) = 2⁻¹ε_X\{i}, g(e_i + e_j) = 2⁻¹(ε_X\{i} + ε_X\{j}), g(e) = 2⁻¹

j∈Xε_X\{j}= −εandg(e−e_i)= −ε−2⁻¹ε_X\{i}.

Lemma 11 There exists no linear map:W→V such that(f −)(w)∈H (w) for everyw∈W.

Proof By the preliminary remark on the map gand by Lemmas 6,7,8,9,10we have that(f+g)(ei)∈H (ei), (f +g)(ei+ej)∈H (ei+ej), (f +g)(e)∈H (e).

Therefore it suffices to prove that there exists no linear mapx such thatx(ei)∈ε_i^⊥, x(e_i+e_j)∈(ε_i−ε_j)^⊥, x(e)∈ε^⊥andϕ(x(e−e_n)−2⁻¹ε_X\{n}, ε+ε_n)=0. By way of contradiction let us assume that we have such anx=

i∈Xa_ie_i^∗, fora_i∈V. In particular we haveϕ(a_j, ε_i)=ϕ(a_i, ε_j)for anyi=j andϕ(a_i, ε_i)=0. This yields

ϕ(x(e), εi)=

j∈X

ϕ(aj, εi)=

j∈X

ϕ(ai, εj)=ϕ(ai, ε).

Furthermore

ϕ(x(e−e_n), ε+ε_n)=ϕ(x(e), ε)+ϕ(x(e), ε_n)−ϕ(a_n, ε)−ϕ(a_n, ε_n)=0.

Soϕ(x(e−en)−2⁻¹εX\{n}, ε+εn)= −2⁻¹ϕ(εX\{n}, ε+εn)=1=0, a contradic-

tion. Thus the result is proved.

Theorem 1 An elementary Abelianp-group of rankn≥4p−2 is not a CI⁽²⁾-group.

Proof By construction the elementf normalizes the 2-closureG⁽²⁾ ofG. We have G⁽²⁾=W L⁽²⁾and clearlyGandG⁽²⁾ induce the same mapH. Therefore[−, f] ∈ Der(W, L) and θ ([−, f])∈HomH(W, V ). We have θ ([−, f])(w)= [w, f](0)=

−f (−w). Thusf ∈HomH(W, V ). The groupW×V has rank 4p−2. Therefore, by Lemma4and Lemma11, the groupW×V is not a CI⁽²⁾-group. It is an easy ap- plication of the first paragraph of Sect. 5.1 in [5] to see that any elementary Abelian p-group of rankn≥4p−2 is not a CI⁽²⁾-group.

Based on some computer evidence we present the following conjecture.

Conjecture The groupGis 2-closed.

4 Proof of Theorem2

As in Section3, we assume thatp >2. In this section we shall repetitively use Propo- sition 22.1 (Schur-Wielandt principle), Propositions 22.4,23.5 and Theorem 23.9

in [9]. For the computations inside the group algebraQ[W×V]we shall stick to the notation and to the terminology of Sect. 2 and 4 in [6].

In the proof of Theorem1we have shown thatf ∈HomH(W, V ). Moreover, it was proved in [6] (Proposition 5 page 173) that the linear span,AH, of the simple quantities{(w, H (w)+v)}w∈W,v∈V is a Schur ring.

LetEbe{e_i}i∈X∪ {e_i+e_j}i=j∪ {e, e−e_n}. Note that in the proof of Lemma11 we proved that there exists no linear functionsuch that(f−)(w)∈H (w)for everyw∈E.

Proposition 1 IfSis anAH-subset such that

(w, H (w))∈ S for everyw∈E, (1) thenS is not a CI-subset ofW×V. In particularW×V is not a CI-group (S denotes the Schur ring generated byS).

Proof The following argument mimics Proposition 4 in [6]. Assume S is a CI- subset. Since f ∈ HomH(W, V ), it is easy to check that Cay(W ×V , T )^f = Cay(W×V , T^f)for any simple quantityT inAH(see Proposition 6 in [6]). So, since Sis anAH-subset, we have Cay(W×V , S^f)=Cay(W×V , S)^f∼=Cay(W×V , S).

Therefore, sinceSis a CI-subset, we getS^f =S^gfor somegin Aut(W×V ). Then, Cay(W×V , S)^f =Cay(W×V , S^f)=Cay(W×V , S^g)=Cay(W×V , S)^g. There- foref g⁻¹is an automorphism of Cay(W×V , S).

Now, using Equation (1), the reader can verify (see Theorem 2.4 in [6]) that f g⁻¹ is an automorphism of Cay(W ×V , (w, H (w))) for every w ∈ E. Thus (w, H (w))^f =(w, H (w))^g for everyw∈E. Hence (w, H (w))is a CI-subset for everyw∈Eand

[(w, H (w))]w∈E∼=Cay[(w, H (w))^f]w∈E.

Now, Proposition 7 in [6] yields that there exists ∈Hom(W, V ) such that(f − )(w)∈H (w)for everyw∈E, a contradiction.

This proves thatS is not a CI-subset of W×V. ThereforeW×V is not a CI-

group.

Proof of Theorem2 Like in Theorem1, it is enough to prove thatW×V is not a CI-group (see the first paragraph of Sect. 5.1 in [5]). LetSbe theAH-subset

i∈X

(0, εi)

i,j∈X,i=j

(e_i+e_j, H (e_i+e_j)) (e_n, H (e_n)) (e−e_n, H (e−e_n))

1≤j≤p−1

(j e_2j, H (e_2j)) (e, H (e))

1≤j≤p−1

(j e_2j−1, H (e_2j−1)+ {±ε})

.

By Proposition 1, it remains to prove that Equation(1)holds for the setS. Let us denote byAthe Schur ringS.

Case p≥5. The element(S+S)◦Slies inAand it can be written as

(a,b)∈S

λ(a, b)(a, b),

whereλ(a, b)is the number of solutions of the equation (‡) (u, x)+(v, y)=(a, b) with(u, x), (v, y)∈S. It can be easily shown that

(S+S)◦S=0T12pⁿ⁻²T22(n−2)T3(2(n−2)2pⁿ⁻²)T4

(2(n−2)4pⁿ⁻²)T52(n−1)T6(2(n−1)6pⁿ⁻²)T7

where

T1=

i∈X

(0, εi)(e−en, H (e−en)), T2=(e, H (e)),

T3=

i<j,{i,j}{1,2,n}

(e_i+e_j, H (e_i+e_j)), T₄=(e₂+e_n, H (e₂+e_n)),

T₅=(e₁+e₂, H (e₁+e₂))(e₁+e_n, H (e₁+e_n)),

T6=

2≤j≤p−1

(j e2j, H (e2j))

2≤j≤p−1

(j e2j−1, H (e2j−1)+ {±ε}), T7=(e1, H (e1)+ {±ε})(e2, H (e2))(e_n, H (e_n)).

For instance, ifa=0, then(‡)has no solution and soλ(0, b)=0. Also ifa=e_n, then (‡)has solutions withu=0, en, e_2p−2+e_n,−e_2p−2, e_2p−3+e_n,−e_2p−3. Studying all this possibilities we getλ(a, b)=2(n−1)6pⁿ⁻². All the other computations are similar.

Note that 0,2pⁿ⁻²,2(n−2), (2(n−2)+2pⁿ⁻²),(2(n−2)+4pⁿ⁻²), 2(n−1), (2(n−1)+6pⁿ⁻²)are all distinct. So, by Schur-Wielandt principle,T1, . . . , T7lie in A.

Ifi, j=n, thenH (e−e_n)+ε_i=H (e−e_n)+ε_j. Thus

i∈X

(0, εi)

+(e−e_n, H (e−e_n))=

i∈X

(e−e_n, H (e−e_n)+ε_i)

=(n−1)(e−e_n, H (e−e_n)+ε₁) (e−en, H (e−en)+εn).

Therefore, we have T₁+T₁=

i∈X

(0,2εi)2

i<j

(0, εi+ε_j)2(n−1)(e−e_n, H (e−e_n)+ε₁) 2(e−e_n, H (e−e_n)+ε_n)pⁿ⁻¹(2(e−e_n), H (e−e_n)).

By Schur-Wielandt principlet=(2(e−e_n), H (e−e_n))is an element ofAand so, by Theorem 23.9 in [9],(e−en, H (e−en))=(2⁻¹)t∈A.

Note that(e, H (e))+(−e+en, H (e−en))=(en, V )is a set ofA, so(en, V )◦S= (en, H (en))∈A.

Similarly, (e2+en, H (e2+en))+(−en, H (en))=(e2, V ) is a set of A, so (e2, H (e2))=(e2, V )◦S∈A.

We have

T6+T6=2pⁿ⁻²U18pⁿ⁻²U24pⁿ⁻²U32pⁿ⁻¹U4pⁿ⁻¹U5

where

U1=

2≤j1<j2≤p−1

(j1e2j₁+j2e2j₂, V ),

U₂=

2≤j₁<j₂≤p−1

(j₁e_2j1−1+j₂e_2j2−1, V ),

U₃=

2≤j2≤j1≤p−1

(j₁e_2j1+j₂e_2j2−1, V ),

U4=

1≤j≤p−1

(2j e2j−1, H (e_2j−1)),

U5=

2≤j≤p−1

(2j e2j, H (e2j))

2≤j≤p−1

(2j e2j−1, H (e2j−1)+ {±2ε}).

The elements 2pⁿ⁻²,8pⁿ⁻²,4pⁿ⁻²,2pⁿ⁻¹, pⁿ⁻¹are all distinct and so using Schur- Wielandt principle we have thatU1, . . . , U5lie inA.

We have

Fl=(e2l−1+e2l, H (e2l−1+e2l))=(l⁻¹)(U3◦(l)T3)∈A, for 2≤l≤p−1.

We have

(U4+(−2l)Fl)◦((−2)S)=pⁿ⁻²(−2le2l, H (e2l))∈A, for every 2≤l≤p−1. So,(e2l, H (e2l))∈Afor 2≤l≤p−1.

Now (e2l−1+e2l, H (e2l +e2l−1))+(−e2l, H (e2l))=(e2l−1, V ) is a set of A for 2≤l≤p−1. Therefore(e2l−1, H (e2l−1))=(e2l−1, V )◦(((2l)⁻¹)U4)∈Afor 2≤l≤p−1.