# Proving $(a^{-1})^{-1}=a$ in a set with only a one-sided identity and one-sided inverse.

$G$ is a set which is closed under mutiplication, associative, and every element has a right identity and right inverse. I have to prove that $(a^{-1})^{-1}=a$.

Is proving this at all possibe without assuming $a^{-1}$ has a two sided inverse?

The proof I thought of was $a^{-1}*(a^{-1})^{-1}=(a^{-1})^{-1}*a^{-1}=e=a^{-1}*a=a*a^{-1}$.

• How do you know they commute? You just know that $a * a' = e$ and $a' * (a')' = e$ (where $x'$ is inverse). It would work if the binary operation was commutative, but in that case all one-sided things are actually both-sided. – user87690 Aug 28 '13 at 11:02
• I can only come up with one if at least $e$ is in the center of $G$ (i.e. $e$ is a two-sided identity)... – AlexR Aug 28 '13 at 11:26
Let $x$ is a right inverse for $a^{-1}$. Then $a^{-1}a=a^{-1}ae=a^{-1}aa^{-1}x=a^{-1}ex=a^{-1}x=e$, so $a^{-1}$ is a left inverse for $a$.