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On \(Q\)-algebras. (English) Zbl 0996.06011
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 \(x\neq 0\).

MSC:
06F35 BCK-algebras, BCI-algebras
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