Trick to proving a group has exactly one idempotent element - Fraleigh p. 48 4.31

If $*$ is a binary operation on a set $S$, an element $x \in S$ is an idempotent for $*$ if $x * x = x$.

Let $\langle G, *\rangle$ be a group and let $x\in G$ such that $x*x = x.$
Then $x*x = x*e$, and by left cancellation,
$x = e$, so $e$ is the only idempotent element in a group.

The trick here looks like writing $x$ as $x*e$. How can you prognosticate (please see profile) this? I didn't see it. It also looks like you have to prognosticate the 'one idempotent element' to be the identity element. Is this right? Can someone make this less magical and psychic?

Solution 1:

This does not look so much like prognostication to me. The hint is trying to make explicitly clear something that is happening. You could instead start by right multiplying both sides by $x^{-1}$. Then you get $xxx^{-1} = xx^{-1} \implies x(e) = e \implies x = e$. By the uniqueness of the identity, there is only one such element.

Solution 2:

There are two things I want to address here:

1) There aren't that many things we can actually do. We know group elements are closed under an operation, the operation is associative, and that every element has an inverse element, and that there is an identity for the group. What could the single, unique idempotent element be? It has to be the identity.

2) We multiply things by identities all the time. Think about rationalizing denominators, or finding GCDs to add fractions. You'll probably even do it again in regards to groups, since $gg^{-1} = 1$ could be helpful at some point.

Solution 3:

I don't know Fraleigh's book, and I'm not near a library that has a copy right now so I can't check out his approach to elementary group theory, what he says, etc., but it looks like he wants to use "left cancellation", the property that $ab = ac$ implies $b = c$, to prove the only idempotent in a group is the identity element. Left (and also right) cancellation of course hold in any group by virtue of the existence of inverses. But is also possible to have cancellation in an algebraic structure without identity, for example the set $n \Bbb Z$ for integer $n > 1$, considering it equipped with only multiplication. So perhaps Fraleigh is trying to show how this argument fits into a more general pattern. Of course, if $G$ is finite and has an identity, then cancellation implies the existence of inverses, since in that case the map $g \to ag$ is injective, whence finiteness forces it to be surjective as well, so there must be some $b \in G$ with $ab = e$. Then we could argue that $x^2 = x$ forces $x =e$, as shown by AWertheim in his answer, by simply multiplying by $x^{-1}$: $x = x^{-1}x^2 = x^{-1}x = e$. In any event, if one wants to proceed via cancellation, the equation $x = xe$ is needed so that something is "left over" after one cancels out $x$! It's not so much prognostication as it is experience with such maneuvers. But it can seem a bit mysterious the first time you see it. These things being said, it seems easier, clearer and cleaner to me to simply write

$x^2 = x \Rightarrow x^{-1}x^2 = x^{-1}x = e \Rightarrow x =e. \tag{1}$

Hope this helps! Merry Christmas to One and All,

and as always,

Fiat Lux!!!