Re: abstract algebra~~)

From: Marc Olschok (sa796ol_at_l1-hrz.uni-duisburg.de)
Date: 11/04/04 

Date: Thu, 4 Nov 2004 14:11:45 +0000 (UTC)

mina_world <mina_world@hanmail.net> wrote:
> hello......doctor~
>
> A semigroup (G,*) (i.e. * is associative), which satisfies the axiom:
>
> 1. There is the element e in G such that ae = a for all a in G.
>
> 2. There is the element a' in G such that aa' = e for all a in G.
>
> show that G is group.
> ---------------------------------------------------------
> is this possible problem ??

Yes. These are in fact the 'one sided' versions of the
usual axioms in the definition of a group. It is a popular
exercise to show, that the these are sufficient.

>
> if "the element" => "an element"(uniqueness),
>
> then, i can deduce to left axiom from this right axiom.
>
> can we deduce the uniqueness ?

I am not sure what you mean. Once you established the both-sided
versions, the uniqueness is done as in the usual group theoretic case.
But you do not need to show the left-sided versions, in order to arrive
at the both-sided versions.

Given an element a in G, applying (2) twice gives a' and a" such
that aa' = e and a'a" = e.
Now calculate a'aa'a" in order to show a'a=e.

Now calculate aa'a in order to show ea=a

Marc

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