Some examples of group algebras without filtered multiplicative basis (Q1098913)

The multiplicative basis theorem asserts that, for an algebraically closed field k, if A is a finite dimensional, representation finite k- algebra then A has a filtered (or normed) multiplicative basis [see \textit{R. Bautista}, \textit{P. Gabriel}, \textit{A. Rojter}, \textit{L. Salmerón}, Invent. Math. 81, 217-285 (1985; Zbl 0575.16012)]. In other words, there exists a k-basis B of A such that \(B\cup \{0\}\) is closed under multiplication and such that \(B\cap rad(A)\) is a k-basis of rad(A). If G is a finite group, kG is a finite dimensional algebra which has G as a multiplicative basis but, at the time of the above mentioned publication, it was not known if kG has always a filtered multiplicative basis. The present note answers this question giving examples for either possibility. The author observes first that, if char k\(=p\), a prime number, and if \(G=Z_ p\times Z_ p\), then kG has a filtered multiplicative basis even though it is of wild representation type. On the other hand, it is proved that if G is defined by 2 generators a, b with the relations \[ a^{p^{n-1}}=b^ p=a^{-1}b^{-1}a^{1+p^{n-2}}b=1 \] (n\(\geq 4)\), then kG cannot have such a basis. Finally, for the quaternion group of order 8, Q, it is shown that the existence of a filtered multiplicative basis may depend on the field k; kQ (char k\(=2)\) has a filtered multiplicative basis if and only if k contains the cubic roots of unity.