What Are the Properties of an Endomorphism f That Satisfies x*f(x)=k?

[mnb96]

Given a set A with a binary associative operation *, I would like to say that for each element a there exist an element a' such that a*a' = k.
Does this concept have a name?

I was very tempted to say that a' is a right inverse for a, but that means that k should then be the identity element, for which a*k=k*a=a holds, which might not be true.
Any ideas?

 

[HallsofIvy]

Then what is k?

 

[mnb96]

well, k is an element of the set, but not necessarily the identity-element.
I guess it is possible to define an associative operation on a set so that, for each element x there exists another element y, such that xy=k.

I do not want to include a further restriction that kx=xk=x. If I did, then I would essentially have defined the (right) inverse element.

If we drop the requirement that k must be an identity-element, is there a name for this concept?

 

[Bobhawke]

closure under the operation *?

 

[mnb96]

yes!
the set is closed under the operation *.
I guess we can assume it is a semigroup (closure, and associativity).

Also, the set A does not necessarily have a finite number of elements, but its cardinality can be also infinite

 

[Office_Shredder]

I'd just say k is divisible by all elements of the set

 

[mnb96]

...sorry for the lame question, but could you please specify the definition of divisible?

 

[Office_Shredder]

Not usually used in a group context, but in rings. a is divisible by b if there exists c such that bc = a

I would only use this term here for lack of a better way of describing it. It gets the point across though

 

[Hurkyl]

Just to give an example:

The set of all integers has this property, where the product is multiplication, and k is zero.

 

[mnb96]

Thanks a lot!
I cannot yet find another better definition.
However, before giving up, I want to know what you think about the following:

If we have a semigroup A, and we say that for each x there must be some y, such that xy=k, where k is a fixed element, this is equivalent to say that the following identity must hold:

$$xyx = kx$$

Obviously, since the identity kx=x does not necessarily hold, we haven't done much progress. However when the identity kx=x holds, we have:

$$xyx = x$$

and y is called weak inverse, while the element x is said to be regular. A semigroup where all its elements are regular is called regular semigroup (see Planetmath and Wikipedia).
Is it possible that there is not a weaker definition in which one must have:

$$xyx \in K \subseteq A$$ ?
where K is essentially the principal left ideal generated by k

 

[Hurkyl]

If we have a semigroup A, and we say that for each x there must be some y, such that xy=k, where k is a fixed element, this is equivalent to say that the following identity must hold:

$$xyx = kx$$

I don't see how that equivalence works in either direction. (-->) is nonobvious because you switched from "there exists a y" to "for all y". (<--) is nonobvious because you haven't proven a cancellation law.

 

[Office_Shredder]

I don't really know anything about this stuff but wikipedia says you got your terminology mixed up

http://en.wikipedia.org/wiki/Weak_inverse

 

[mnb96]

Hurkyl: you are completely right.
In fact, I concluded that I am not able to find a better way than saying that 'k must be "divisible" by all the elements in the set'.

However, I would like to avoid using improper terminology, and so I am trying to investigate on how the concept of divisibility in a semigroup is formally defined.

I just found that there are the so called Green's Relations (http://en.wikipedia.org/wiki/Green's_relations), which are supposed to help understanding the nature of divisibility in a semigroup (unless I misunderstood). At the moment, I still cannot figure out, how to use those relations, in order to describe $$\forall x, \exists y$$, such that $$xy=k$$, where k is a unique element in the set.
Any idea?

 

[Hurkyl]

Why should k be unique? Oh... did you mean for it merely to be a constant?

By the way, your structure can be made into a universal algebra: starting with the presentation of "semigroup", you add a constant k and a unary function ', and impose the identity

x x' = k​

This may or may not be suitable for your purposes -- note that it creates a distinguished solution to xy=k which you might find undesirable. (In particular, any homomorphism would be expected to map distinguished elements to distinguished elements)

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In the immortal words of Humpty Dumpty:

“When I use a word, it means just what I choose it to mean -- neither more nor less.”​

I suspect that it should be quite fair for you to define the word "divisibility" to mean whatever you want it to mean in this case. (And, of course, in any discussion where you plan to use the word, you should first give your explicit definition)

I suggest "x is a left divisor of y" to mean that there exists z such that xz=y.

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For a different description, a semigroup with the property you want has a minimum¹ nonempty right ideal.
Proof: For any x, k is an element of x S¹. Therefore, for any nonempty ideal I, we have $k S^1 \subseteq I$.

Of course, in any discussion on this subject, you should probably define "ideal" as well. I'm using "I is a right ideal" to mean "I is a subset such that $I S^1 = I$".


1: This ideal is not merely minimal, it really is a minimum under the partial ordering defined by $\subseteq$

 

[mnb96]

Thanks Hurkyl!
That was the kind of elegant definition I was looking for.

I'd like to clarify something:

why did you introduce in your proof $$xS^1$$ at all?
Given a semigroup $$A$$, and a right ideal $$I$$, for each $$y\in I$$, and $$a\in A$$, we have $$ya\in I$$. However, we know that for each $$y$$ there exists, by hypothesis, another $$y'\in A$$ such that $$yy'=k$$. From this follows that $$k$$ must be contained in any right-ideal.
As a consequence, inside $$I$$ we must also have all the elements of the form $$ka$$ (with $$a \in A$$), and these elements are nothing but the definition of the principal right-ideal $$kA$$ generated by $$k$$, which has in fact, minimal cardinality.
Proving that it is unique is rather trivial: we must be able to find a right ideal $$I$$ with same cardinality of $$kA$$, which does not contain $$kA$$; but this is not possible because we just proved that any right-ideal contains $$kA$$.

Also, it seems to me we have only proved that:
if our relation $$xy=k$$ holds (for every x, some y, and unique k) $$\Rightarrow$$ the semigroup has a unique minimal right-ideal.
One would need to prove the $$\Leftarrow$$ in order to say that the two statements are equivalent.

 

[Hurkyl]

why did you introduce in your proof $$xS^1$$ at all?

I just translated the problem into ideals and subsets at the first opportunity, using the equivalence $x \in I \Longleftrightarrow xS^1 \subseteq I$.

This point of view did clue me into an obvious fact I had not yet noticed; when I emphasized the ideals, I realized that k (generally) isn't too special -- we can replace it with any element of kS¹.

 

[mnb96]

...I realized that k (generally) isn't too special -- we can replace it with any element of kS¹.

- By this, you mean that if xy=k for a constant k, there must be necessarily other elements $$k_1,k_2,\ldots$$ for which $$xy=k_i$$ always holds?

- Did you agree on the fact that we have proved only one the $$\Rightarrow$$ and we should prove also the other way?

 

[Hurkyl]

- By this, you mean that if xy=k for a constant k, there must be necessarily other elements $$k_1,k_2,\ldots$$ for which $$xy=k_i$$ always holds?

Right. Well, except for the case where k is a (one-sided) null element -- i.e. when you have the identity kx=k -- then nothing else has that property.

Did you agree on the fact that we have proved only one the $$\Rightarrow$$ and we should prove also the other way?

Sure. Or rearrange the proof so all the steps are if-and-only-if's. e.g. something resembling

kA¹ is a minimal ideal iff kA¹ is contained in every nonempty ideal iff kA¹ is contained in every principal ideal iff k is contained in every principal ideal iff xy=k has a solution for each x.

 

[mnb96]

Maybe I am getting confused, but I think we now shifted the original problem.
Let's make things clear, and please point out if I say something incorrect.

We have proved that:
if we have $$\forall x, \exists y$$ such that $$xy=k$$, where k is a fixed unique element,
then the semigroup A has a unique minimal right-ideal (which is $$kA$$).

However, the converse is not true (in order for it to be true, $$kA$$ must have a single element, hence k should be a one sided null-element).
In any case, this simply means that the two statements are not equivalent, and does not say much on the nature of k.

I am now interested to know whether it is possible or not to have $$xy=k$$, where k is a unique element (and should not be either the identity, nor the null element).
You mentioned earlier that there are several $$k_i$$ for which $$xy=k_i$$ can hold, but I don't understand how you proved that.

 

[Hurkyl]

You mentioned earlier that there are several $$k_i$$ for which $$xy=k_i$$ can hold, but I don't understand how you proved that.

My proof method equates "For every x, xy=k has a solution" with "kS¹ is a minimum nonempty right ideal". If k' is any element of kS¹, then k'S¹=kS¹...

Once you know it's true, it's very easy to see it in terms of elements. For any particular x, if xy=k has a solution, can you find a solution to xy=kz?

 

[mnb96]

ok! now everything should be clear.
So if my right, for a fixed element k, if $$xy=k$$ holds for each x and some y:

1) if $$kA^{1}$$ is a singleton $$\Rightarrow$$ k must be a (one-sided) null-element (and k is unique)
2) if $$kA^1\subset A$$ $$\Rightarrow$$ $$xy=k$$ holds for each $$k\in kA^1$$ (and k cannot be unique)
3) if $$kA^1=A$$ $$\Rightarrow$$ k was an (one-sided) identity-element, and $$xy=k$$ always holds for any k $$\Rightarrow$$ A has a one-sided inverse

I hope there are no mistakes!

 

[mnb96]

we have essentially discussed the problem of what can be said when, given a semigroup $$(A,*)$$, for every $$x\in A$$, $$\exists x'$$, such that $$x*x'=k$$ for a fixed k.
Now, is it possible to make other observations about an endomorphism $$f:A\rightarrow A$$, which must satisfy $$x*f(x)=k$$?
Essentially I want to find the properties the endomorphism f should have, so that $$f(x)$$ always "captures" the important element $$x'$$.