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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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