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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?

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EDIT:

enter image description here

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  • $\begingroup$ 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
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    $\begingroup$ 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
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$(a^{-1})^{-1} = a$ 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 $b^{-1} = a$.

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  • $\begingroup$ 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

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