Im trying to prove $(a^{-1})^{-1}=a^{-1}$.But a statement is confusing(Please see the highlighted portion in the image,i tried to type in the equation but its not working) .

How can we say that the inverse*(inverse of the inverse) is equal to the identity element?

enter image description here


enter image description here

  • $\begingroup$ This is wrong, unless $a^{-1}=a.$ BTW, add $ signs before and after the equations can produce the formulae. $\endgroup$ – awllower Nov 8 '14 at 10:38
  • 1
    $\begingroup$ The highlighted row does not say that the inverse of the inverse is equal to the inverse; it says that the inverse of the inverse is the inverse of the inverse... $\endgroup$ – Mauro ALLEGRANZA Nov 8 '14 at 10:40
  • $\begingroup$ @MauroALLEGRANZA well,how does it become equal to the identity element that is my question. $\endgroup$ – techno Nov 8 '14 at 10:41
  • $\begingroup$ $(a^{-1})^{-1} \cdot a^{-1} = e$ because $a^{-1} \in G$, because the inverse of an elemet of $G$ is in $G$; thus, $a^{-1}$ has an inverse : $(a^{-1})^{-1}$ and by def of inverse, the product of an element and its inverse is $e$, i.e. $(a^{-1})^{-1} \cdot a^{-1} = e$ . $\endgroup$ – Mauro ALLEGRANZA Nov 8 '14 at 10:43
  • $\begingroup$ @MauroALLEGRANZA where did you get $(a^{−1})^{−1}=e$ $\endgroup$ – techno Nov 8 '14 at 10:43

$(a^{-1})^{-1} = a$ because $a^{-1} * a = a *a^{-1} = e$. Hence $a^{-1}*(a^{-1})^{-1} = a^{-1} * a = e$ etc.

Maybe this makes it more clear: Write $b:=a^{-1}$. Then we try to prove that $b*a = a*b = e$ so that $b^{-1} = a$.

  • $\begingroup$ Please see the question,i have posted the full proof,so you are saying its correct. $\endgroup$ – techno Nov 8 '14 at 10:49
  • $\begingroup$ Yes, eventhough it is not written very readable. $\endgroup$ – user42761 Nov 8 '14 at 10:50
  • $\begingroup$ @sorry :),screen captured $\endgroup$ – techno Nov 8 '14 at 10:51
  • $\begingroup$ You have to know that inverses are unique (which is an easy task to show) so that if $a*b=b*a= e$ then $a^{-1}=b$ and $b^{-1} = a$. $\endgroup$ – user42761 Nov 8 '14 at 10:53
  • $\begingroup$ yeah, i proved it using left cancellation $a*e1=a*e2$ so e1=e2 $\endgroup$ – techno Nov 8 '14 at 10:54

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.