# Is $G$ a group under those conditions? [duplicate]

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.

• What progress have you made? You're more likely to get help here if you show some effort on the problem. – rogerl Nov 18 '16 at 18:20
• Since (i), (ii), and (iii) are the axioms of a group, there is nothing to show. – Björn Friedrich Nov 18 '16 at 18:25
• @BjörnFriedrich Depends... many people assume in the axioms that the inverses and the identity are two-sided. – Jack M Nov 18 '16 at 18:26
• @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? – Björn Friedrich Nov 18 '16 at 18:42
• @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. – vadim123 Nov 18 '16 at 18:43