Let $G$ be a set and $*:G\times G \rightarrow G$ an operation with:

(i) $(G,*)$ is associative

(ii) There is a $f \in G$ with $f*a=a$ for all $a \in G$

(iii) For every $a \in G$ exists a $b \in G$ with $b*a=f$

First show that from $b*a=f$ concludes $a*b=f$

Then show that $(G,*)$ is a group.

  • 2
    $\begingroup$ What progress have you made? You're more likely to get help here if you show some effort on the problem. $\endgroup$ – rogerl Nov 18 '16 at 18:20
  • $\begingroup$ Since (i), (ii), and (iii) are the axioms of a group, there is nothing to show. $\endgroup$ – Björn Friedrich Nov 18 '16 at 18:25
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    $\begingroup$ @BjörnFriedrich Depends... many people assume in the axioms that the inverses and the identity are two-sided. $\endgroup$ – Jack M Nov 18 '16 at 18:26
  • $\begingroup$ @Jack M: Can you give an example where dropping this assumption from the axioms has the effect that the structure is not a group anymore? $\endgroup$ – Björn Friedrich Nov 18 '16 at 18:42
  • $\begingroup$ @BjörnFriedrich, the point of the OP is that there is no such example. However that doesn't mean that this one-sided definition is the standard one, or that there is nothing to prove. $\endgroup$ – vadim123 Nov 18 '16 at 18:43

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