450 Algebra
u, v∈S. Hence the existence and uniqueness of the solution to the equationa∗x=b
is equivalent to the existence and uniqueness of the solution to the equationx∗a=b.
The existence of the solution for the equationa∗x=bfollows from the fact that
x=b∗ais a solution. To prove the uniqueness, letc∈Sbe a solution. By hypothesis
we have the equalitiesa∗(b∗a)=b,b∗(c∗b)=c,c∗(a∗c)=a. Froma∗c=b
it follows thatc∗(a∗c)=c∗b=a.Soa=c∗b, and froma∗c=bit follows that
c∗(a∗c)=c∗b=a. Therefore,b∗a=b∗(c∗b)=c,which implies thatb∗a=c.
This completes the proof.
273.Substitutingy=ein the second relation, and using the first, we obtainx∗z=
(x∗e)∗z=(z∗e)∗x=z∗x, which proves the commutativity. Using it, the associativity
is proved as follows:
(x∗y)∗z=(z∗x)∗y=(y∗z)∗x=x∗(y∗z).(A. Gheorghe)274.The answer is yes. Letφbe any bijection ofFwith no fixed points. Define
x∗y=φ(x). The first property obviously holds. On the other hand,x∗(y∗z)=φ(x)
and(x∗y)∗z=φ(x∗y)=φ(φ(x)). Again sinceφ hasno fixed points, these two are
never equal, so the second property also holds.
(45th W.L. Putnam Mathematical Competition, 1984)
275.Froma∗(a∗a)=(a∗a)∗awe deduce thata∗a=a. We claim that
a∗(b∗a)=a for alla, b∈S.Indeed, we havea∗(a∗(b∗a))=(a∗a)∗(b∗a)=a∗(b∗a)and(a∗(b∗a))∗a=
(a∗b)∗(a∗a)=(a∗b)∗a. Using associativity, we obtain
a∗(a∗(b∗a))=a∗(b∗a)=(a∗b)∗a=(a∗ (b∗a))∗a.The “noncommutativity’’ condition from the statement impliesa∗(b∗a)=a, proving
the claim.
We apply this property as follows:
(a∗(b∗c))∗(a∗c)=(a∗b)∗(c∗(a∗c))=(a∗b)∗c,
(a∗c)∗(a∗(b∗c))=(a∗(c∗a))∗(b∗c)=a∗(b∗c).Since(a∗b)∗c=a∗(b∗c)(by associativity), we obtain
(a∗(b∗c))∗(a∗c)=(a∗c)∗(a∗(b∗c)).This means thata∗(b∗c)anda∗ccommute,so they must be equal, as desired.