Can a Semigroup Be a Group Under This Specific Condition?

[TheFurryGoat]

Homework Statement

Prove, that a senigruop S is a group if and only if for every x in S there exists a unique x' in S so that xx'x=x.

Homework Equations

The Attempt at a Solution

I am trying to prove that xx'=1_S for all x in S. But all I can show is that xx' is in E_S={x | x^2=x} for every x in S. I am having a lot of troubles proving xx'=yy' and I can't even prove that 1_S si in S. Am I going in the right direction?

 

[mighty2000]

Thank you for your post. It seems like you are on the right track, but there are a few key steps missing in your proof. Here is a suggested approach to proving this statement:

1. First, let's assume that S is a group. This means that S satisfies the four group axioms: closure, associativity, identity, and inverse. We can use these axioms to show that for every x in S, there exists a unique x' in S such that xx'x=x.

2. Start by choosing an arbitrary element x in S. Since S is a group, we know that there exists an identity element 1S in S. This means that xx' = 1S for some x' in S.

3. Now, we need to show that this x' is unique. To do this, assume that there exists another element y' in S such that xy'x=x. Then, we have xx' = xy'x = x(y'x). Since S is associative, we can rearrange this equation to get (xx')y = (xy')x. Using the fact that xx' = 1S, we get y = (xy')x. But we also know that xx' = 1S = (yx')x, so y = (yx')x. Since S is a group, we can cancel out the x on both sides to get y = yx'. This means that y' = x', and thus x' is unique.

4. Now, let's assume that for every x in S, there exists a unique x' in S such that xx'x=x. We need to show that S satisfies the four group axioms. To do this, we will use the fact that for every x in S, there exists a unique x' in S such that xx' = 1S.

5. Closure: Let x and y be arbitrary elements in S. Since xx' = 1S and yy' = 1S, we know that (xx')(yy') is also in S. But (xx')(yy') = (xy)(x'y') = (xy)(xy')x. Using the fact that xx' = 1S, we can simplify this to (xy)x = xy. Thus, the product of any two elements in S is also in S, and S satisfies closure.

6. Associativity: Let x,