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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 [1].

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?

Better to be sorry than safe
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  • What do you mean by "prognosticate" ? – DonAntonio Dec 23 '13 at 17:34
  • I can make it less magical: replace the word "prognostication" with the phrase "wax on, wax off." – Andrew Dudzik Dec 23 '13 at 20:19
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    Also: I have recently written about this elsewhere, but the technique of "wishful thinking" is one of the most widely useful techniques in mathematics. You look for things that would* make your life easier, if they were true. Then, if you are lucky, they turn out to be true. But even if you are unlucky, you may learn a good deal about what makes the problem "tick", or where the difficulty lies.
    • http://math.stackexchange.com/questions/613417/the-cauchy-schwarz-master-class-problem-1-2/613439#613439
     – Andrew Dudzik Dec 23 '13 at 20:21
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    Here, for example, we know that we want to show that every group has exactly one idempotent. Well, what would make our lives very easy? If the most obvious answer were the correct one. There is only one element that we know for sure every group has: the identity. So it is a very mundane act of wishful thinking to ask: is the identity element idempotent? We can do this without even knowing the meaning of the word "idempotent"! – Andrew Dudzik Dec 23 '13 at 20:23
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    +1 for "wishful thinking is one of the most widely useful techniques in mathematics". – Eric Stucky Dec 24 '13 at 01:33
  • @DonAntonio Please see my profile. It's supposed to mean "really educated guess that turns out right" –  Dec 24 '13 at 07:54
  • @User-33433 Thanks. I upvoted your comment. Do you want to make that an answer? I can upvote. And where did you learn about "wishful thinking"? My instructor didn't talk about this. What's more, where can you find books or resources that would teach wishful thinking and other 'widely useful techniques in mathematics'? –  Dec 24 '13 at 07:55
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    @FrankMuer For your first lesson in wishful thinking, imagine that you had an instructor that told you about it. :) I believe that I first read about it in this form in "The Art and Craft of Problem Solving" by Paul Zeitz. "Wax on, wax off," is a quote from The Karate Kid. The scene is a lesson about how we become masters through countless hours of practice on very basic (often boring) things. The main character is supposed to be doing favors (like waxing cars) in exchange for being taught the secret of Karate. Slowly he realizes that all his hard work is the secret... – Andrew Dudzik Dec 24 '13 at 09:00
  • @FrankMuer That's not what the word 'prognosticate' means. One usually talks about "finding" or "seeing" the solution (or its key idea). You might also ask how one motivates a certain approach to a problem. Regardless, there is no prophesizing involved. – Potato Dec 24 '13 at 11:48
  • @User-33433 Thanks for the explanation. What about other 'widely useful techniques in mathematics'? Any other good books or resources? I am passionate about the second third .... lessons in wishful thinking. –  Dec 25 '13 at 05:22
  • @Potato Thanks. Maybe I'll try 'find'. I don't want to use 'see' because it might mean 'understand' and I understand the proof here. But does it work if I swap 'prognosticate' with 'motivate' up above? –  Dec 25 '13 at 05:27
  • @FrankMuer Yes. – Potato Dec 25 '13 at 08:48

3 Answers3

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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.

Alex Wertheim
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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.

Tyler
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  • Thanks. I upvoted. But I still don't see why 'it has to be the identity'? –  Dec 24 '13 at 07:58
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    The group axioms specify there is an identity for every group. The identity is idempotent. Therefore, we already know that there is an idempotent element in every group. Now that you know that, you have to show that it is the only idempotent element. There is a theorem to prove the identity is unique, so if you can show that any idempotent element is the identity, you're done. – Tyler Dec 24 '13 at 15:33
  • Great. Thanks. Too bad I can't upvote twice for your comment. –  Dec 25 '13 at 04:45
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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!!!

Robert Lewis
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