= Solution
Use <group axioms from two-sided local inverse identities>. Choose any $a$ in the nonempty <semigroup> and put $e_R=a_L^{-1}a$, $e_L=aa_R^{-1}$. The given identities say $be_R=b$ and $e_Lb=b$ for every $b$. Therefore $e_Le_R=e_L$ by the right-identity property, and $e_Le_R=e_R$ by the left-identity property. Thus $e_L=e_R=e$ is a two-sided <identity element>.
For an arbitrary $a$, set $b=a_L^{-1}$ and $c=a_R^{-1}$. Inserting $e$ into the original identities gives $ba=e$ and $ac=e$. By <associativity>,
$$
b=be=b(ac)=(ba)c=ec=c.
$$
Thus every element has a two-sided <inverse element>. Closure and <associativity> were assumed, so \b[all the group axioms hold]. No cancellation or identity was used before it had been established.
Back to article page
