# Show that a group has exactly one idempotent element

1. Feb 20, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Prove that a group has exactly one idempotent element.

2. Relevant equations

3. The attempt at a solution
So we need to show that the identity element is the unique idempotent element in a group.

First, we know that by definition of a group there is at least one element, e, such that $e * e = e$.

Second, we need to show that there is at most one idempotent element. We do this by showing that if $x*x=x$ and $y*y=y$ then $x=y$... This is as far as I get. Am I on the right track?

2. Feb 20, 2017

### Staff: Mentor

What does it mean for $x\cdot x = x^2=x\,$? Any idea to get rid of one $x\,$?

3. Feb 20, 2017

### Mr Davis 97

I can use the cancellation law. But how does that logically show that e is the unique element in the group that is idempotent?

4. Feb 20, 2017

### Staff: Mentor

If $x$ is any idempotent element, i.e. $x^2=x$ then you can multiply (as in school on both sides) the whole equation with $x^{-1}$, which is probably what you meant by cancellation. You can do this, because all elements of a group have an inverse. Then - if you want to be very rigorous and pedantic - you can apply associativity and the existence and definition of $e$. Write it down and see what it says.

5. Feb 20, 2017

### FactChecker

You know more than that about e. You know x*e=x. Use that. The fact that e*e = e should tell you that you need to prove that x=e.