# Proof that a certain semigroup is also a group

1. Aug 20, 2010

### dndod1

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 $$\in$$ 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 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$$\in$$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$$\in$$S has a right inverse using the onto property of the function $$\lambda$$ 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. Aug 20, 2010

### Office_Shredder

Staff Emeritus
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

3. Aug 21, 2010

### 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 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.