Commutative Semigroup

Question

Let $S$ be a Semigroup with the two following properties, $(1):$ for all $x$ in $S$ we have $x^3=x$ $(2):$ for any $x,y$ in $S$ we have $xy^2x=yx^2y$. Then prove that this Semigroup $S$ is commutative.

I have found the following identities for any $x,y$ in $S$

• $(xy)^3=xy=x^3y^3$

• $xy^2x=y^2(xy^2x)$

• $(xy)^2=y(xy)^2y$

• $xy^2x^2x=yx^2yx^2$

Answer 1

$$yxy=yxyyxyyxy=yxy^2xy^2xy=yyx^2yy^2xy=y^2x^2yxy=$$ $$=yxy^2xxy=yxy^2x^2y=yxyxy^2x,$$ which gives $$yxy^2=yxyxy^2xy.$$ In another hand, $$xyxy=xyy^2xy=y(xy)^2y=yxyxy^2,$$ which gives $$xy=(xy)^3=yxyxy^2xy.$$ Thus, $$xy=yxy^2$$ and also, $$yx=xyx^2$$ or $$xy^2=yxy$$ and $$yx^2=xyx.$$ Now, $$x^2yx^2y=x^2yy^2x^2y=y(x^2y)^2y=yx^2yx^2yy=yx^2yx^2y^2.$$ Thus, $$x^2y=(x^2y)^3=yx^2yx^2y^2x^2y=xy^2xx^2y^2x^2y=$$ $$=xy^2xy^2x^2y=yx^2y^3x^2y=yx^2yx^2y,$$ Which gives $$xy=x(x^2y)=(xyx^2)(yx^2y)=(yx)(xy^2x)=yx^2y^2x=x^2yx.$$ Id est, $$x^2y=xyx=yx^2$$ and $$xy=x(x^2y)=xyx^2=yx.$$

Answer 2

Suppose $S$ is a semigroup such that \begin{align*} x^3&=x,\;\text{for all}\;x\in S\tag{1}\\[4pt] xy^2x&=yx^2y,\;\text{for all}\;x,y\in S\tag{2}\\[4pt] \end{align*} Our goal is to show $xy=yx$, for all $x,y\in S$.

We can recast $(2)$ as $$(xy)(yx)=(yx)(xy),\;\text{for all}\;x,y\in S\tag{3}$$ so $xy$ commutes with $yx$, for all $x,y\in S$.

Next, working on $y^2x^2y^2$, we get \begin{align*} y^2x^2y^2&=(y^2x)(xy^2)\\[4pt] &=(xy^2)(y^2x)&&\text{[by $(3)$]}\\[4pt] &=xy^4x\\[4pt] &=xy^2x&&\text{[by $(1)$]}\\[4pt] \end{align*} Thus we have $$y^2x^2y^2=xy^2x,\;\text{for all}\;x,y\in S\tag{4}$$ Next, working on $x^2y^2$, we get \begin{align*} x^2y^2&=(x^2y^2)^3\\[4pt] &=(x^2y^2)(x^2y^2)(x^2y^2)\\[4pt] &=x^2(y^2x^2y^2)x^2y^2\\[4pt] &=x^2(xy^2x)x^2y^2&&\text{[by $(4)$]}\\[4pt] &=x^3y^2x^3y^2\\[4pt] &=xy^2xy^2&&\text{[by $(1)$]}\\[4pt] &=(xy^2x)y^2\\[4pt] &=(yx^2y)y^2&&\text{[by $(2)$]}\\[4pt] &=yx^2y^3\\[4pt] &=yx^2y&&\text{[by $(1)$]}\\[4pt] &=xy^2x&&\text{[by $(2)$]}\\[4pt] \end{align*} Thus we have $$x^2y^2=yx^2y=xy^2x,\;\text{for all}\;x,y\in S$$ hence, by symmetry, we get $$x^2y^2=y^2x^2,\;\text{for all}\;x,y\in S\tag{5}$$ so $x^2$ commutes with $y^2$, for all $x,y\in S$.

Next, working on $x^2y$, we get \begin{align*} x^2y&=(x^2y)^3&&\text{[by $(1)$]}\\[4pt] &=x^2(yx^2y)x^2y\\[4pt] &=x^2(xy^2x)x^2y&&\text{[by $(2)$]}\\[4pt] &=x^3y^2x^3y\\[4pt] &=xy^2xy&&\text{[by $(1)$]}\\[4pt] &=(xy^2x)y\\[4pt] &=(yx^2y)y&&\text{[by $(2)$]}\\[4pt] &=y(x^2y^2)\\[4pt] &=y(y^2x^2)&&\text{[by $(5)$]}\\[4pt] &=y^3x^2\\[4pt] &=yx^2&&\text{[by $(1)$]}\\[4pt] \end{align*}

Thus we have $$x^2y=yx^2,\;\text{for all}\;x,y\in S\tag{6}$$ so squares commute with everything.

Finally, working on $xy$, we get \begin{align*} xy&=(xy)^3&&\text{[by $(1)$]}\\[4pt] &=x(yx)^2y\\[4pt] &=(yx)^2xy&&\text{[by $(6)$]}\\[4pt] &=yx(yx^2y)\\[4pt] &=yx(xy^2x)&&\text{[by $(2)$]}\\[4pt] &=yx^2y^2x\\[4pt] &=y(x^2y^2)x\\[4pt] &=y(y^2x^2)x&&\text{[by $(5)$]}\\[4pt] &=y^3x^3\\[4pt] &=yx&&\text{[by $(1)$]}\\[4pt] \end{align*} Thus we have $$xy=yx,\;\text{for all}\;x,y\in S$$ as was to be shown.