# Equivalence Relation using Binary Operations.

Question: Let ∗ be a binary operation on a set A. Assume that ∗ is associative with identity. Let R be the relation on A defined on elements a,b ∈ R as follows: aRb if there exists an invertible element c ∈ A such that c∗b = a∗c. Prove that R is an equivalence relation. What happens if c is not required to be invertible?

I've figured out the first part of the problem, asking to prove that the relation is an equivalence relation. I'm confused as to how the answer will change if 'c' is not necessarily invertible. I believe that the relation will still be reflexive, but any guidance on transitivity and symmetry will be appreciated.

-
If for $\forall c\in A$ $c$ isn't invertible about $R$, $R=\emptyset$ by definition. – Math.StackExchange May 28 '14 at 2:42
@AranKomatsuzaki that is not true... by hypothesis, at least the identity is in the set, and the identity is related to itself... e*e=e, so choose a, b, and c as the identity, thus R is not null... – malxmusician212 May 28 '14 at 2:52

Reflexivity: $\forall a\in A, aRa$.

If $e$ is the identity element, then $a*e=e*a$, so this holds.

Transitivity: $\forall a,b,c \in A$[$(aRb$ and $bRc) \rightarrow aRc$]

So, let's assume that $aRb$ and that $bRc$. Then $\exists d,f \in A$ such that $a*d=d*b$ (1) and $b*f=f*c$. (2)

We multiply $f$ to both sides of equation (1): $a*d*f=d*b*f$. By equation (2), we can substitute $f*c$ for $b*f$, so $a*d*f=d*f*c$.

If $A$ is closed under $*$, then we can conclude $d*f \in A$ and thus from the above that $R$ is transitive.

Edit 23 oct 2014. I'm sure you're done with this problem for some time, but I need to keep thinking about algebra, so I came back to it. :)

Symmetry: $\forall a,b \in A~[aRb \rightarrow bRa]$

So, we assume $aRb$ and want to show that $bRa$. That is, we know that $\exists c \in A$ such that $a*c = c*b$ (3) and we want to show that $\exists ? \in A$ such that $?*a = b~*~?$.

This ought to be false, but I don't know enough about non-commutative monoids to prove it.

I'm still going to come back to this answer at some point. :)

Edit Sept 2015.

In the invertible scenario, we can right-multiply and left-multiply $c^{-1}$ to (3): $c^{-1}*a*c*c^{-1}=c^{-1}*c*b*c^{-1}$, so $c^{-1}*a=b*c^{-1}$.

In the non-invertible scenario...

-