# Showing an equivalence relation on a monoid

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

$$x\mathcal Ey$$$$\exists$$ $$z$$ $$\in$$ $$M^×$$ 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?

• 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" Oct 8, 2020 at 12:35
• @N.Beck I just checked but I didn't make a mistake typing. I guess there is one in the exercise then? Oct 8, 2020 at 12:40
• "Linear algebra" tag was inappropriate. I have suppressed it. I have added "monoid" tag instead. Oct 8, 2020 at 12:48
• @JeanMarie Oh my bad! Thought it would be appropriate since we covered it in our linear algebra class. Thanks for the edit. Oct 8, 2020 at 12:55
• 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. Oct 8, 2020 at 13:28

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.

• Wow thank you for this extensive answer! I think I got it now. Oct 8, 2020 at 15:09