Proof that a certain semigroup is also a group

[dndod1]

Homework Statement

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

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

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

c) Show that e_a acts as an identity, using these solutions for any b\inS.
e_a*b =y
e_a * (e_a *b) = e_a
(e_a*e_a) * b = e_a *y
b=y

and
b*e_a=y
(b*e_a)*e_a= y*e_a
b* (e_a*e_a) =y*e_a
b=y

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

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

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

Homework Equations

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

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 (e_a*e_a) =e_a) given a*e_a) =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 *e_a*e_a = a*e_a
(a*e_a)*e_a) =a*e_a
a* e_a=a
a=a ??


I am very, very lost! Please help if you can.

 

[Office_Shredder]

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

 

[dndod1]

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 e_a for each a, such that lambda a(e_a)=a.

My undersatnding is that the function lambda takes e_a 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.