- Semigroup $S$ is finitely presented if $S \simeq A^{+}/R^{\sharp}$ for some finite set $A$ and for some relation $R$, which is also a finite subset of $A^{+} \times A^{+}$.
suppose that $A = \{ a,b\}$ and $R = \{ (ab,ba) , (bba , a) , (bbb,b) , (aaa,aa) \}$ and i have choosen the class $x_1=[a] ,x_2= [b] , x_3=[aa] , x_4= [bb] $ of $R^{\sharp}$.
Now $x_3^2=[aa] . [aa] = [aaaa] = [aaa] = [aa] = x_3$ , because $ (aaa, aa) \in R$ , so $(aaaa, aaa) \in R^c \subseteq R^{\sharp}$
Similarly $x_4^2=[bb]^2 = [bb] =x_4$ , $x_1^2 = [a]^2 = [aa] = x_3$ , $x_2^2 =[b]^2 = [bb] = x_4$
I want to show that $[a].[b] = [a]$ and there is no class of $R^{\sharp}$ other than thses four classes.
I would be thankful if someone help me.
The given presentation does not define $\mathbb{Z}_4$, but a semigroup with six elements, $S = \{a, b, a^2, ab, b^2, a^2b\}$. This semigroup contains two idempotents, $a^2$ and $b^2$, each one being the identity of a group of order $2$: $\{a^2, a^2b\}$ and $\{b, b^2\}$: indeed $bb = b^2$ and $(a^2b)^2 = aabaab = aabbaa = aa(bba)aa = aa(a)aa = aa$.
Since $\mathbb{Z}_4$ is a cyclic group, it suffices to take a presentation with one generator, for instance $A = \{a\}$ and $R = \{a^5 = a\}$, where the identity of the group will be $a^4$. If you allow a monoid presentation, you could take $R = \{a^4 = 1\}$.