$G$ 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 $a^{-1}*(a^{-1})^{-1}=(a^{-1})^{-1}*a^{-1}=e=a^{-1}*a=a*a^{-1}$.

Thanks in advance!

  • $\begingroup$ Is multiplication commutative? Otherwise your proof is wrong (or needs more intermediate steps). $\endgroup$ – gammatester Aug 28 '13 at 11:02
  • $\begingroup$ 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. $\endgroup$ – user87690 Aug 28 '13 at 11:02
  • $\begingroup$ My proof has assumed that inverses commute. I know my proof is wrong, as I have made an unjustified assumption. I am looking for a proof under the conditions stated in the problem. $\endgroup$ – fierydemon Aug 28 '13 at 11:06
  • $\begingroup$ 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)... $\endgroup$ – AlexR Aug 28 '13 at 11:26
  • $\begingroup$ See also the related (but different) question math.stackexchange.com/questions/433546/… $\endgroup$ – J.-E. Pin Aug 28 '13 at 11:51

In your question a right identity must be common for all elements. Then a solution is as follows:

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 $a$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.