I'm not sure if this simple first day Abstract Algebra exercise is correct

[jdinatale]

Prove: If $x$ has a right inverse given by $a$ and a left inverse given by $b$, then $a = b$.

The Attempt at a Solution

One thing that bothers me: how can we even talk about a left inverse or a right inverse without establishing that x is in an algebraic structure? I wrote this in my proof but I'm not sure if it's necessary to do so or even correct.

Then my last line is kind of questionable. I'm not entirely sure if xa = bx, then a = b. What if I said b*(x*a) = (b*x)*a = e? Would that justify a = b?

 

[SammyS]

Prove: If $x$ has a right inverse given by $a$ and a left inverse given by $b$, then $a = b$.

The Attempt at a Solution

One thing that bothers me: how can we even talk about a left inverse or a right inverse without establishing that x is in an algebraic structure? I wrote this in my proof but I'm not sure if it's necessary to do so or even correct.

Then my last line is kind of questionable. I'm not entirely sure if xa = bx, then a = b. What if I said b*(x*a) = (b*x)*a = e? Would that justify a = b?

How does x*a = b*x = e imply that a = b ?

I don't know about needing an algebraic structure, but it looks to me that the associative law will need to hold.

 

[Robert1986]

Yeah, I think you need some sort of algebraic structure. First, inverse doesn't even make sense without having closure and identities. Especially the identity thing since it is required for a definition of inverse. And I, too, think you have to have associativity. So, you need closure, identity, inverses, and associativity. In other words, it seems to me, you have to be dealing in a group. I'm guessing the author is implying this. What book is this from?

 

[gauss^2]

It has to be in a monoid, which is a set $G$ with an associative law of "multiplication" and an identity element $e \in G$ such that $e x = x e = x$ for all $x \in G$. However, a monoid need not have a two-sided inverse for all elements like a group would. The proof should be

$$a = e a = (bx)a = b(xa) = b e = b$$

 

[Robert1986]

Yeah, but monoids don't need inverses on any side. This proof shows that the set is a group.

 

[gauss^2]

Yeah, but monoids don't need inverses on any side. This proof shows that the set is a group.

The question as written seems to be asking about a specific x and not making a universally quantified statement, though I have no doubts the problem is meant to illustrate that inverses are two-sided in groups.

 

[jdinatale]

It has to be in a monoid, which is a set $G$ with an associative law of "multiplication" and an identity element $e \in G$ such that $e x = x e = x$ for all $x \in G$. However, a monoid need not have a two-sided inverse for all elements like a group would. The proof should be

$$(bx)a = b(xa)$$

The problem doesn't state that we have this associative property.

edit: plus the teacher has never even mentioned a monoid so I doubt that's what he has in mind.

 

[Robert1986]

Well, some algebraic structure HAS to be implied here, otherwise x could be anything, say a tree. I am willing to bet that whoever wrote this question just assumed that you could read his mind and infer that he meant something like "In a group, x has a left inverse and right inverse..." Is that possible? Otherwise, I really don't see how to do anything. Even if there is a way to prove it without associativity, you have to define inverses which means you need at least an identity element.