92 2 Algebra
Substitutingxbyxa, we obtaina^2 xa=xa,orax^3 =xa, and sincex^3 =x, it follows
thatacommutes with all elements inG. We can therefore write
a^2 x=a(ax)=a(xa)=(xa)a=xa^2 ,whencexa^2 =a^2 x=x. This shows thata^2 is the identity element of the multiplicative
operation; we denote it bye. The relation from the statement impliesx^3 =axa=xa^2 =
xe; cancellingx, we obtainx^2 =e; hence for allx∈G,x−^1 =x. It follows thatGis a
group. It is Abelian by the well-known computation
xy=(xy)−^1 =y−^1 x−^1 =yx. Here are more examples of the kind.278.Prove that in order for a setGendowed with an associative operation to be a group,
it suffices for it to have a left identity, and for each element to have a left inverse.
This means that there should existe∈Gsuch thatex=xfor allx∈G, and for
eachx ∈G, there should existx′∈Gsuch thatx′x=e. The same conclusion
holds if “left’’ is replaced by “right.’’
279.Let(G,⊥)and(G,∗)be two group structures defined on the same setG. Assume
that the two groups have the same identity element and that their binary operations
satisfy
a∗b=(a⊥a)⊥(a⊥b),for alla, b∈G. Prove that the binary operations coincide and the group they define
is Abelian.280.Letr,s,tbe positive integers that are pairwise relatively prime. If the elementsa
andbof an Abelian group with identity elementesatisfyar=bs=(ab)t=e,
prove thata=b=e. Does the same conclusion hold ifaandbare elements of an
arbitrary nonAbelian group?
281.Assume thataandbare elements of a group with identity elementesatisfying
(aba−^1 )n=efor some positive integern. Prove thatbn=e.
282.LetGbe a group with the following properties:
(i)Ghas no element of order 2,
(ii)(xy)^2 =(yx)^2 , for allx, y∈G.
Prove thatGis Abelian.
283.A multiplicative operation on a set M satisfies (i)a^2 = b^2 , (ii)ab^2 = a,
(iii)a^2 (bc)=cb, (iv)(ac)(bc)=ab, for alla, b, c∈M. Define onMthe
operation