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from (xy)^i = x^iy^i show that G is Abelian

Source: 4-th Hungary-Israel Binational Mathematical Competition 1993

May 26, 2007
abstract algebragroup theorysuperior algebra

Problem Statement

In the questions below: GG is a finite group; HGH \leq G a subgroup of G;G:HG; |G : H | the index of HH in G;XG; |X | the number of elements of XG;Z(G)X \subseteq G; Z (G) the center of G;GG; G' the commutator subgroup of G;NG(H)G; N_{G}(H ) the normalizer of HH in G;CG(H)G; C_{G}(H ) the centralizer of HH in GG; and SnS_{n} the nn-th symmetric group. Suppose k2k \geq 2 is an integer such that for all x,yGx, y \in G and i{k1,k,k+1}i \in \{k-1, k, k+1\} the relation (xy)i=xiyi(xy)^{i}= x^{i}y^{i} holds. Show that GG is Abelian.