# Prove that (G,*) is a group .

1. Nov 3, 2012

### Maths Lover

hi ,
let (G,*) be a semigroup with the property that for any two elements a,b belongs to G , the equations:
a*x=b , y*a=b
have solutions x,y in G , verify that (G,*) forms a group.

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my attempt

first one ,
* is associative as (G,*) is a semigroup

assume a=b ,then , a*x1=a , y1*a=a
so x is a left identity and y is the right one .
let c belongs to G then
c*x2=c , y2*c=c

if I can prove that x1 = x2 then x1 is the identity , but I have no idea about the way to do
that ! ,

if I could prove that e is an identity element in G then it's easy to proof that there is an inverse for all a belongs to G :
let b = e then
a*x=e , and x belongs to G " that is mentioned in the question " so the solotion is the inverse .

so the problem is in the identity
the solution shouldn't use any theorem related with groups because the exercise is after the defenition of group and it's examples and no theorem has explained

2. Nov 3, 2012

### superg33k

Here's my attempt, I hope it helps. I find reading this stuff is always pretty cryptic:

Let a=b therefore exists 1xa,1ya such that a*1xa=a, 1ya*a=a
Let b=1xa therefore exists -xa such that a*-xa=1xa
Let b=1ya therefore exists -ya such that -ya*a=1ya
Therefore 1xa=a*-xa=(1ya*a)*-xa=1ya*(a*-xa)=1ya*1xa
Similarly 1ya=-ya*a=-ya*(a*1xa)=(-ya*a)*1xa=1ya*1xa
Therefore 1xa=1ya (so I can call them just 1a)

So for any particular element a, it's left identity is it's right identity. Still have to show that if a*1a=a and b*1b=b then 1a=1b.