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$G$ 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!

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  • $\begingroup$ 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
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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 $a$.

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