MathDB

6

Part of 1993 Hungary-Israel Binational

Problems(1)

ab^2 = b^3a and ba^2 = a^3b => a=b=1

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

5/27/2007
In the questions below: GG is a finite group; H≤GH \leq G a subgroup of G;∣G:H∣G; |G : H | the index of HH in G;∣X∣G; |X | the number of elements of X⊆G;Z(G)X \subseteq G; Z (G) the center of G;G′G; 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. Let a,b∈G.a, b \in G. Suppose that ab2=b3aab^{2}= b^{3}a and ba2=a3b.ba^{2}= a^{3}b. Prove that a=b=1.a = b = 1.
group theoryabstract algebrasuperior algebrasuperior algebra unsolved