# How to prove a semigroup with properties $x^3=x$ and $x^2y^2=y^2x^2$ is commutative

If $$S$$ is a semigroup such that $$(\forall x,y\in S)\quad x^3=x\quad\text{and}\quad x^2y^2=y^2x^2,$$prove that $$(\forall x,y\in S)\quad xy=yx.$$

All I did is prove $$x^2y^2=(x^2y^2)^2\quad\text{and}\quad xy^2x=(xy^2x)^2.$$

• Proof of $$x^2y^2=(x^2y^2)^2$$:

From $$x^2y^2=y^2x^2,$$ we deduce $$x^2x^2y^2y^2=x^2y^2x^2y^2=(x^2y^2)^2.$$

Taking into account $$x^4=x^2,$$ the result follows.

• Proof of $$xy^2x=(xy^2x)^2$$:

$$xy=x^3y^3=xx^2y^2y=xy^2x^2y=(xy^2x)(xy)$$ hence $$xyyx=(xy^2x)(xy)(yx) =(xy^2x) ^2,$$ which was the claim.

• I think you already gave "some context". What @C-RAM meant is perhaps your question would be better "received" if you present at least a proof of the 2 properties you say you already got. You could also tell where this problem comes from, why you think it is true. Feb 7, 2023 at 8:32
• Do we agree that a semigroup operation is only associative ? (no neutral element.) Feb 7, 2023 at 10:06
• (same question as Anne Bauval) What is the origin of this exercise ? Feb 7, 2023 at 10:12
• Ah, yes we have only $(xy)^4=I$. I spoke too fast... I let the comment, it might serve as a warning.
– zwim
Feb 7, 2023 at 12:11
• @JeanMarie I "agree that a semigroup operation is only associative (no neutral element)" but this theorem about semigroups is equivalent to its restriction to monoids (by adjunction of an identity element). Feb 7, 2023 at 18:10

Let $$S$$ be a semigroup satisfying the two identities \begin{align} (1) \quad &x^3 = x \\ (2) \quad &x^2y^2 = y^2x^2 \end{align} First observe that, for every $$x \in S$$, $$x^2$$ is idempotent. Indeed, $$x^2x^2 = x^3x = xx = x^2.$$ It follows by (2) that idempotents commute in $$S$$. On the other hand, (1) shows that $$S$$ is a completely regular semigroup, that is, every element of $$S$$ belongs to some subgroup of $$S$$. Indeed, every element $$x$$ belongs to the subgroup $$H(x) = \{x, x^2\}$$ of $$S$$. It follows that $$S$$ is the union of these subgroups and every $$\mathcal{H}$$-class of $$S$$ is a (maximal) subgroup of $$S$$. Moreover, (1) shows that these maximal groups have exponent $$2$$ and hence are commutative (see for instance this answer for a proof).
The structure of completely regular semigroups with commuting idempotents is described in Theorem 3 of [1]. It says that $$S$$ is a semilattice of groups, in the following sense. The idempotents of $$S$$ form a semilattice $$E$$ and $$S$$ is the disjoint union of groups $$G_e$$, $$e \in E$$. If $$e > f$$, there is a morphism $$\varphi_{e,f}:G_e \to G_f$$ such that, if $$e > f > k$$, then $$\varphi_{f,k} \circ \varphi_{e,f} = \varphi_{e,k}$$. Now, the product on $$S$$ is defined as follows. If $$g_e \in G_e$$ and $$h_f \in G_f$$ are two elements of $$S$$ and if $$k = ef$$, then the product $$g_eh_f$$ is in $$G_k$$ and is computed as follows: $$g_eh_f = \varphi_{e,k}(g_e) \varphi_{f,k}(h_f)$$ Now, in your case, since $$ef = fe$$ and since $$G_k$$ is a commutative group, this product is commutative.