Let (A,\bullet) be a monoid sush that for every x in A, x\bullet x= where e is the identity element. Show that (A, ) is an abelian group.

To show that (A, \bullet) is an abelian group, we need to prove the following properties:

1. Closure: For all a, b in A, the result of the operation a \bullet b is also in A.

2. Associativity: For all a, b, c in A, (a \bullet b) \bullet c = a \bullet (b \bullet c).

3. Identity element: There exists an element e in A such that for every element a in A, the equations a \bullet e = e \bullet a = a hold.

4. Inverse element: For each element a in A, there exists an element b in A such that a \bullet b = b \bullet a = e.

5. Commutativity: For all a, b in A, a \bullet b = b \bullet a.

Given that (A, \bullet) is a monoid, we already know that properties 1, 2, and 3 hold. We are also given that for every x in A, x \bullet x = e.

To prove property 4 (Inverse element), we can use the given condition. For any x in A, we know that x \bullet x = e. So, for any x in A, x is its own inverse.

To prove property 5 (Commutativity), consider any two elements a, b in A. We have:

a \bullet (a \bullet b) = a \bullet e = a (since e is the identity)

But also, a \bullet (a \bullet b) = (a \bullet a) \bullet b = e \bullet b = b.

So, a = b, which means that for all a, b in A, a \bullet b = b \bullet a.

Therefore, (A, \bullet) is an abelian group.