By Vyacheslav Futorny, Victor Kac, Iryna Kashuba, Efim Zelmanov

This quantity includes contributions from the convention on 'Algebras, Representations and functions' (Maresias, Brazil, August 26 - September 1, 2007), in honor of Ivan Shestakov's sixtieth birthday. This publication can be of curiosity to graduate scholars and researchers operating within the thought of Lie and Jordan algebras and superalgebras and their representations, Hopf algebras, Poisson algebras, Quantum teams, staff jewelry and different themes

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2). 2 is cocommutative if and only if g x=x g for all g ∈ G and x ∈ Mat(n, k). 6. 2 is cocommutative if and only if Ag = ξg U t Ag U −1 for all g ∈ G and x ∈ Mat(n, k) where ξg = ±1. A. CHUBAROV Proof. 2 Ag xA−1 g = g −1 g = U t Ag U −1 xU t A−1 g U x=x −1 for any matrix x. Hence the matrix U t A−1 Ag is a central matrix for any g ∈ G. g U t −1 t −1 Hence Ag = ξg U Ag U . But U = ±U = U and therefore ξg = ±1. 2 is a Z2 -graded algebra H ∗ = H0∗ ⊕ H1∗ , H0∗ = kG, H1∗ = Mat(n, k). The space of matrices Mat(n, k) is equipped with a symmetric bilinear form A, B = tr (A · S(B)) = tr A · U t BU .

If n > 1 then the group G is not cyclic. Proof. If G is cyclic then G∗ = G because H 2 (G, k∗ ) is trivial. Then all irreducible representations of G have dimension 1. 4 the representation Ψ is irreducible and has dimension n. Now we have the following situation. 5. 6. 1] which means that it is induced by a one-dimensional representation of some subgroup D ⊂ G∗ . Clearly D contains the derived central subgroup [G∗ , G∗ ] and therefore D is normal in G∗ . Consider an irreducible linear representation Ψ of G∗ in a space V of dimension n induced by a one-dimensional representation of a normal subgroup D, containing [G∗ , G∗ ].

T∈∆0 In a similar way, we see that er ∗ [er , er ] = αt (et er + [er , et ]) = t∈∆0 αt et er + [er , [er , er ]]. t∈∆0 Hence < er , er , er >= −1/2[er , [er , er ]] = 1/2[[er , er ], er ]. On the other hand, from the deﬁnition of Akivis superalgebra, we have that SJ(er , er , er ) = 3[[er , er ], er ] = 6A(er , er , er ). Thus < er , er , er >= A(er , er , er ). 2. r ≤ k < s and r = k if r ∈ ∆1 : ¯ < er , es , ek >= (er es ) ∗ ek − er ∗ ((−1)s¯k ek es + [es , ek ]) = ¯ ¯ (−1)s¯k er ek es + A(er , es , ek ) − (−1)s¯k A(er , ek , es )+ ¯ ¯ s¯k +er ∗ [es , ek ] − er ∗ [es , ek ] − (−1) er ek es + (−1)s¯k A(er , ek , es ) = = A(er , es , ek ).