Neggers, Joseph; Ahn, Sun Shin; Kim, Hee Sik On \(Q\)-algebras. (English) Zbl 0996.06011 Int. J. Math. Math. Sci. 27, No. 12, 749-757 (2001). The authors define the new notion of a \({\mathcal Q}\)-algebra and prove some fundamental results:1. Every BCH-algebra \(X\) is a \({\mathcal Q}\)-algebra and conversely every \({\mathcal Q}\)-algebra with the condition (VI): \(x*y= y*x= e\) implies \(x= y\) is a BCH-algebra. Thus the class of BCH-algebras coincides with the class of \({\mathcal Q}\)-algebras with (VI);2. Every \({\mathcal Q}\)-algebra \((X;*,0)\) satisfying the associative law is a group under the operation \(*\);3. For every \({\mathcal Q}\)-algebra of order \(3\), we have \(G(X)\neq X\), where \(G(X)= \{x\in X\mid 0*x= x\}\).Moreover, they define a quadratic \({\mathcal Q}\)-algebra and prove that every quadratic \({\mathcal Q}\)-algebra \((X;*,e)\) has the form \(x*y= x-y+e\). This is a useful representation theorem of quadratic \({\mathcal Q}\)-algebras.Unfortunately, Proposition 3.4: For every \({\mathcal Q}\)-algebra \(X\), \(x\in G(X)\) if and oly if \(0* x\in G(X)\), is not true. There is a counterexample. Let \(X\) be a nontrivial BCK-algebra. Of course it is a \({\mathcal Q}\)-algebra. For this \(X\), we have \(G(X)= \{0\}\) and \(x\in G(X)\) implies \(0* x\in G(X)\). On the other hand, we have \(0*X= 0\in G(x)\) but \(x\not\in G(X)\) for \(x\neq 0\). Reviewer: Michiro Kondo (Matsue) Cited in 11 Documents MSC: 06F35 BCK-algebras, BCI-algebras Keywords:\({\mathcal Q}\)-algebra; BCH-algebra; group; quadratic \({\mathcal Q}\)-algebra PDFBibTeX XMLCite \textit{J. Neggers} et al., Int. J. Math. Math. Sci. 27, No. 12, 749--757 (2001; Zbl 0996.06011) Full Text: DOI EuDML