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.