Question 1172370
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<pre>

Let x and y be two elements of the multiplicative group G.


Then their commutator is the element  {{{x*y*x^(-1)*y^(-1)}}}  of the group.


Let's denote it C(x,y) = {{{x*y*x^(-1)*y^(-1)}}}.


The problem asks about the inverse of the commutator C(x,y),  which is  {{{1/C(x,y)}}}.


We have 


    {{{1/C(x,y)}}} = {{{(x*y*x^(-1)*y^(-1))^(-1)}}} = {{{y*x*y^(-1)*x^(-1)}}} = C(y,x).


Thus  {{{(C(x,y))^(-1)}}} = C(y,x).      <U>ANSWER</U>
</pre>

Answered, solved, explained and completed.