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Suppose $G$ is a semigroup in which every equation of the form $ax=b$ or $ya=b$ has a solution. Does this solution have to be unique?

SE Anarchist
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  • yes/no answer, probably with a link, can suffice. – SE Anarchist Sep 09 '13 at 20:49
  • Does this solution or do these solutions?? – Mikasa Sep 09 '13 at 20:59
  • @BabakS.: I mean if there's a unique solution for ax=b for fixed a and b. – SE Anarchist Sep 09 '13 at 21:02
  • If $G$ is finite then your answer is yes. – Mher Sep 09 '13 at 21:05
  • well I think there's an answer in the link suugested as duplicate of this question. – SE Anarchist Sep 09 '13 at 21:07
  • @CrossChris: You mean, you fixed a and b or you make them free wandering the set? Indeed, you can find a semigroup in which for just some fixed "a" and "b" ax=b has different solutions. – Mikasa Sep 09 '13 at 21:08
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    @BabakS.: The complete statement is in the duplicate suggested: http://math.stackexchange.com/a/346661/93637 – SE Anarchist Sep 09 '13 at 21:10
  • @Bill Cook: This is not a duplicate, since in the question the uniqueness is not required. – Boris Novikov Sep 09 '13 at 21:41
  • The other question is a different question, but it does have an answer to this one there. Nevertheless, I'm voting to re-open because I think someone searching for this question will not find that one. – Ben Millwood Sep 09 '13 at 22:49
  • This question does not have two key points of information: the context in which the question was encountered (what class? what book?) and, just as important, what you have already tried. If the question is reopened, it should be edited to include these things. – Carl Mummert Sep 09 '13 at 23:34
  • @CarlMummert: I wanted a yes/no answer. that piece of info does not need trying. because I'm not gonna try and read any proof. just yes/no. + cclear math question does need context. 'Hey what time is it?' 'oh first say where do you want to use that time. Is your class late? will your boss blame you? what's the context to ask this question. Have you tried to listen to a radio, radio says the time, what have you tried? In future give me more info OK, now get lost' – SE Anarchist Sep 10 '13 at 07:14

3 Answers3

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This is the counter example @Rebecca noted. In fact, we see that in some cases we don't find any solutions and this happens cause there are more than one solutions in other case.

  gap> a:= Transformation([2, 3,3]);;
       b:= Transformation([ 3,2,3]);;
       s:= Semigroup(a, b);;
       e:=Elements(s);;
       for k in [1..3] do
           for j in [1..3] do
              for i in [1..3] do
              if e[j]*e[i]=e[k] then Print("k:=", k,"   ", "j:=",j, "   ",i,"\n"); 
              fi;
              od; 
           od;
        od;

        k:=1   j:=1   2
        k:=2   j:=2   2
        k:=3   j:=1   1
        k:=3   j:=1   3
        k:=3   j:=2   1
        k:=3   j:=2   3
        k:=3   j:=3   1
        k:=3   j:=3   2
        k:=3   j:=3   3
Mikasa
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If a semigroup has a left identity and any element has a left inverse, then it is a group (Hall M.,The theory of groups).

Your semigroup has these properties. Indeed, let $ea=a$ and $ax=b$. Then $eb=eax=b$, so $e$ is a common identity.

Boris Novikov
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$x$ has to be $a^{-1}b$ and $y$ has to be $ba^{-1}$. They are unique of course.

QED
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