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 Post subject: Group?Posted: Fri, 23 Jul 2010 19:41:19 UTC
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This is actually an old nut. I'll post it here because this thread reminded me of it.

Let X be a set with an associative multiplication (which we will just write as juxtaposition in what follows). Suppose for every x in X we have a unique such that . Must X be a group?

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 Post subject: Posted: Mon, 6 Sep 2010 12:06:56 UTC
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Joined: Fri, 3 Sep 2010 09:29:45 UTC
Posts: 140
Nice question. I don't think it is true - had a quick play, but I can't find a proof (either way).

Closure and associativity are clear. Also if it is a group, then

(i.e. is the inverse.)

Appreciate a hint on where to go from here. I assume there is some simple algebraic trick using the fact but I am stumped.

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 Post subject: Posted: Mon, 6 Sep 2010 12:44:32 UTC
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I'm not going to give the answer away, but you should play with it some more.

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 Post subject: Posted: Mon, 6 Sep 2010 21:37:16 UTC
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Joined: Fri, 3 Sep 2010 09:29:45 UTC
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If we let then .

(Note that )
Similarly .
This shows the existence of an identity element (we know they are equal by associativity)

But then which shows the inverse property, hence X is a group.

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 Post subject: Posted: Mon, 6 Sep 2010 22:16:12 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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Location: Austin, TX
qwirk wrote:
If we let then .

(Note that )
Similarly .
This shows the existence of an identity element (we know they are equal by associativity)

But then which shows the inverse property, hence X is a group.

That doesn't guarantee that it fixes every element, just because it fixes x.

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 Post subject: Posted: Mon, 6 Sep 2010 22:45:00 UTC
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Joined: Fri, 3 Sep 2010 09:29:45 UTC
Posts: 140
Absolutely correct, of course (in my mind it was a group, and hence the identity is unique). Back to the drawing board

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