Question

IF G is a finite group,  g ∈ G, how to prove for any g there is an X exists for g^-1=g^x (x∈Z)

Answer 1

this proof is done with help of contradiction ....

Proof: We statement: If G is a finite group then theme exists an integer kez fom chicha 7 '= gk and this will holds for all element of a group. this stalement in a contradietony way det if possible Thene does not exists any kez fon which The melaton holds That means fon all KER for all kez , 8879kg (post multiplying & &" =qk where & ¢G and le & tgk for all Ket e KAI in a Let ktl= Then theme no such ne z exists for which 8"=e That means onden of element g is infinite both side a gnoup product of an element with its invenie produce identity element of that groups, heme element of G ezidentity 0 (8) = infinile 7. 6) = inglimite [ at contains an elementy chich is ofinfinite omden shich implies G meest be of infite onderl which is conbsadiction because Given that G is finite group. So oun assumption is and wrong Given statement is true, "(proved CS Scanned with CamScanner