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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?

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  • $\begingroup$ your definition doesn't make sense. you quantify over $x$ and $y$ on the right hand side, but you do not on the left hand side. I.e. it is of the structure "two things x and y satisfy formula(x,y) iff forall x, forall y: anotherformula(x,y) holds" $\endgroup$
    – Merle
    Oct 8, 2020 at 12:35
  • $\begingroup$ @N.Beck I just checked but I didn't make a mistake typing. I guess there is one in the exercise then? $\endgroup$
    – 23408924
    Oct 8, 2020 at 12:40
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    $\begingroup$ "Linear algebra" tag was inappropriate. I have suppressed it. I have added "monoid" tag instead. $\endgroup$
    – Jean Marie
    Oct 8, 2020 at 12:48
  • $\begingroup$ @JeanMarie Oh my bad! Thought it would be appropriate since we covered it in our linear algebra class. Thanks for the edit. $\endgroup$
    – 23408924
    Oct 8, 2020 at 12:55
  • $\begingroup$ Surely this equivalence relation is meant to be $x \mathcal{E} y \iff zx = yz$ for some $z \in M$ (if $M$ were a group, this would be the conjugacy equivalence relation). Saying "for all $x, y \in M$" is a bit confusing, I think you should delete that part from your definition. $\endgroup$
    – Joppy
    Oct 8, 2020 at 13:28

1 Answer 1

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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.

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  • $\begingroup$ Wow thank you for this extensive answer! I think I got it now. $\endgroup$
    – 23408924
    Oct 8, 2020 at 15:09

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