Advanced book on Mathematics Olympiad

Algebra 453

and cancelingawe obtaina⊥a=e, for alla∈G. Using this fact, we obtain

a∗b=(a⊥a)⊥(a⊥b)=e⊥(a⊥b)=a⊥b,

which shows that the composition laws coincide. Becausea∗a=e, we see thata−^1 =a,

so fora, b∈G,

ab=(ab)−^1 =b−^1 a−^1 =ba,

which proves the commutativity.

(D. ̧Stefanescu) ̆

280.The fundamental theorem of arithmetic allows us to find the integersuandvsuch

thatus+vt=1. Sinceab=ba, we have

ab=(ab)us+vt=(ab)us

(

(ab)t

)v

=(ab)use=(ab)us=aus(bs)u=ause=aus.

Therefore,

br=ebr=arbr=(ab)r=ausr=(ar)us=e.

Using again the fundamental theorem of arithmetic we can findx, ysuch thatxr+ys=1.

Then

b=bxr+ys=(br)x(bs)y=e.

Applying the same argument, mutatis mutandis, we find thata=e, so the first part of
the problem is solved.
A counterexample for the case of a noncommutative group is provided by the cycles
of permutationsa =( 123 )andb =( 34567 )in the permutation groupS 7 of order 7.
Thenab=( 1234567 )anda^3 =b^5 =(ab)^7 =e.
(8th International Competition in Mathematics for University Students, 2001)
281.Setc=aba−^1 and observe thatca=aband thatcn=e. We have

a=ea=cna=cn−^1 ca=cn−^1 ab=cn−^2 (ca)b=cn−^2 ab^2 ,

and, inductively,

a=cn−kabk, 1 ≤k≤n.

Froma=abn, we obtain the desired conclusionbn=e.

(Gazeta Matematic ̆a(Mathematics Gazette, Bucharest), proposed by D. Batine ̧tu- ̆

Giurgiu)

282.Applying the identity from the statement to the elementsxandyx−^1 , we have