a permutation of G SolutionBy definition a permutation of G

a permutation of G .

Solution

By definition, a permutation of G is a bijection : G G. Thus, we have to show that _g is bijective. Let _g(x) = _g(y). This means gx = gy. On multiplying both sides by g^-1, we have g^-1gx= g^-1 gy, or x = y. Thus Therefore, _g is injective. Let p be an arbitrary element of G. Since g G, therefore g^-1 G. Hence g^-1 p G. Then, _g( g^-1 p ) = g g^-1 p = p. This means that every element of G has a pre-image in G under _g .Thus _g is surjective. Now, since _g is both injective and surjective, hence _g is bijective, and therefore it is a permutation of G.