Showing an equivalence relation on a monoid

Question

I have difficulties solving this exercise.

Let (M, $\star$) be a monoid. Let $M^×$ $\subset M$ be the subset of invertible
elements in $M$. Define a relation $\mathcal E$ on $M$ by declaring that

$x$$\mathcal E$$y$ ⇔ $\exists$ $z$ $\in$ $M^×$ such that $z \star x$ = $y \star z$
for all $x, y \in M$.
Show that $\mathcal E$ is an equivalence relation on $M$.

I know that in order for a relation to be an equivalence relation, it has to be reflexive, symmetric and transitive. However, I don't know how to apply the theorems of my script to this exercise. How do I proceed to solve this?

Answer 1

I'll use the following slightly reworded version of your posted definition:

Let (M,*) be a monoid and let $M^{\times}$ be the set of invertible elements of $M$.

Define a relation $\mathcal E$ on $M$ by declaring that $a{\;\mathcal E\,}b$ if $u{\,*\,}a = b{\,*\,}u$ for some $u \in M^{\times}$.

The goal is to show that $\mathcal E$ is an equivalence relation on $M$.

For convenience of notation, for $a,b\in M$ we'll write $ab\;$to mean $a{\,*\,}b$.

Let the identity element of $M$ be denoted by $1$.

First we show that $\mathcal E$ is reflexive.

Let $a\in M$.

Then $ua=au$ holds using $u=1$, hence $a{\;\mathcal E\,}a$.

It follows that $\mathcal E$ is reflexive.

Next we show that $\mathcal E$ is symmetric.

Let $a,b\in M$ and suppose $a{\;\mathcal E\,}b$.

Let $u\in M^\times$ be such that $ua=bu$.

Let $u'\in M^{\times}$ be such that $uu'=1=u'u$.

Then we get $$ u'b = (u'b)(uu') = u'(bu)u' = u'(ua)u' = (u'u)(au') = au' $$ hence $b{\;\mathcal E\,}a$.

It follows that $\mathcal E$ is symmetric.

Finally we show that $\mathcal E$ is transitive.

Let $a,b,c\in M$ and suppose $a{\;\mathcal E\,}b$ and $b{\;\mathcal E\,}c$.

Let $u,v\in M^\times$ be such that $ua=bu$ and $vb=cv$

Let $u',v'\in M^{\times}$ be such that $uu'=1=u'u$ and $vv'=1=v'v$.

Let $w=vu$ and let $w'=u'v'$.

Then $w\in M^{\times}$ since $$ \left\lbrace \begin{align*} ww'&=(vu)(u'v')=v(uu')v'=vv'=1 \\[4pt] w'w&=(u'v')(vu)=u'(v'v)u=u'u=1 \\[4pt] \end{align*} \right. $$ Then we get $$ wa = (vu)a = v(ua) = v(bu) = (vb)u = (cv)u = c(vu) = cw $$ hence $a{\;\mathcal E\,}c$.

It follows that $\mathcal E$ is transitive.

Therefore $\mathcal E$ is an equivalence relation.