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.