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 Post subject: Group?
PostPosted: 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 X\times X\to X (which we will just write as juxtaposition in what follows). Suppose for every x in X we have a unique x^*\in X such that xx^*x=x. Must X be a group?

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Mon, 6 Sep 2010 12:06:56 UTC 
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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

$x x^* x = x \implies x^* x = e

(i.e. $x^* is the inverse.)

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


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 Post subject:
PostPosted: 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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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject:
PostPosted: Mon, 6 Sep 2010 21:37:16 UTC 
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If we let $b = x x^* then $bx=x.

(Note that $b \in X)
Similarly $b' =  x^* x \implies x b' = x.
This shows the existence of an identity element $b=b'=e (we know they are equal by associativity)

But then $x^* x =e which shows the inverse property, hence X is a group.

Something doesn't feel right about this proof...


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PostPosted: Mon, 6 Sep 2010 22:16:12 UTC 
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qwirk wrote:
If we let $b = x x^* then $bx=x.

(Note that $b \in X)
Similarly $b' =  x^* x \implies x b' = x.
This shows the existence of an identity element $b=b'=e (we know they are equal by associativity)

But then $x^* x =e which shows the inverse property, hence X is a group.

Something doesn't feel right about this proof...


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

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PostPosted: Mon, 6 Sep 2010 22:45:00 UTC 
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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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