1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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*ea)*ea= y*ea
    b* (ea*ea) =y*ea

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook