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Proof that a certain semigroup is also a group

  1. Aug 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Let S be a finite multiplicative semigroup in which these 2 cancellation laws hold. For all a,x,y [tex]\in[/tex] S, a*x=a*y implies x=y and for all a,x,y, [tex]\in[/tex] S x*a=y*a implies that x=y.
    Show that (S, *) is a group.

    For given a [tex]\in[/tex] S, let [tex]\lambda[/tex] a: S [tex]\rightarrow[/tex] S, s [tex]\rightarrow[/tex] a*s (This should just be a single arrow, can't get it to work!)
    a) Show that [tex]\lambda[/tex] a is 1-1 using the cancellation laws. Deduce that it is also onto. Thus show that there is an element ea=a

    b) Show that ea * ea= ea by using a*ea= a twice, associativity and then right cancellation.

    c) Show that ea acts as an identity, using these solutions for any b[tex]\in[/tex]S.
    ea*b =y
    ea * (ea *b) = ea
    (ea*ea) * b = ea *y
    b=y

    and
    b*ea=y
    (b*ea)*ea= y*ea
    b* (ea*ea) =y*ea
    b=y

    d) Now show that any b[tex]\in[/tex]S has a right inverse using the onto property of the function [tex]\lambda[/tex] a. This element, call it br ^-1, itself has a right inverse using the same logic. Call it c. By examining the product of b*br ^-1*c in two ways, show that b=c.

    e) Hence obtain that the right inverse br ^-1 is an inverse of b.

    f) Conclude logically that (S, *) is a group.


    2. Relevant equations
    The Group axioms.
    G1 Associativity
    G2 Identity element e*g=g*e= g
    G3 inverse for each element g^-1*g=e


    3. The attempt at a solution
    I haven't really got any idea how to start. I only did this before I got stuck:

    Outlined the group axioms as above and stated that:
    As (S, *) is defined as a semigroup, * is associative, therefore G1 has been satisfied.

    a) did not know where to begin as did not know how to go about showing 1-1 and onto.


    b) Show (ea*ea) =ea) given a*ea) =a
    I don't know whether to work with just the LHS or the whole thing and do the same operation to both sides. I have tried the latter.
    a *ea*ea = a*ea
    (a*ea)*ea) =a*ea
    a* ea=a
    a=a ????????????


    I am very, very lost! Please help if you can.
     
  2. jcsd
  3. Aug 20, 2010 #2

    Office_Shredder

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    Gold Member

    Ok, how to show the map in a) is 1-1:

    Suppose there are two elements s and r such that a*s=a*r. If you can show that s=r then you've proven it's 1-1. You should be able to see how to use the cancellation law to prove that s=r
     
  4. Aug 21, 2010 #3
    Thanks very much for getting me started. I have shown that the function is 1-1 and that it is onto. I still do not know how to tackle the last part of question a), as I am confused by the lambda notation.

    Thus show that there is an element ea for each a, such that lambda a(ea)=a.

    My undersatnding is that the function lambda takes ea and it will map to a. Is that equivalent to saying that a maps a..... which would mean that e is the identity element for a?

    Even if that's correct, I am unsure of how to tackle it!
    Many thanks again.
     
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