Algebra 449269.FixaandcinSand consider the functionfa,c:S{a, c}→S,
fa,c(b)=a∗(b∗c).Becausea∗fa,c(b)∗c=(a∗a)∗b∗(c∗c)=b, the function is one-to-one. It follows
that there are exactly two elements that are not in the image offa,c. These elements are
preciselyaandc. Indeed, ifa∗(b∗c)=a, then(a∗a)∗(b∗c)=a∗a,sob∗c=a∗a,
and thenb∗(c∗c)=(a∗a)∗c, which impliesb=c. This contradicts the fact that
a, b, care distinct. A similar argument rules out the casea∗(b∗c)=c.
Now choosea′,c′different from bothaandc. The union of the ranges offa,cand
fa′,c′, which is contained in the set under discussion, is the entire setS. The conclusion
follows.
Remark.An example of such a set is the Klein 4-group.
(R. Gelca)
270.Consider the set
U={h(x, y)|h(−x,−y)=−h(x, y)}.It is straightforward to check thatUis closed under subtraction and taking reciprocals.
Becausef (x, y)=xandg(x, y)=yare inU, the entire setSis inU. ButUdoes not
contain nonzero constant functions, so neither doesS.
(American Mathematical Monthly, 1987, proposed by I. Gessel, solution by O.P.
Lossers)
271.All three parts of the conclusion follow from appropriate substitutions in the identity
from the statement. For example,
(e∗e′)◦(e′∗e)=(e◦e′)∗(e′◦e)simplifies toe′◦e′=e∗e, which further yieldse′=e, proving (a). Then, from
(x∗e)◦(e∗y)=(x◦e)∗(e◦y),we deducex◦y=x∗y, for everyx, y∈M, showing that the two binary operations
coincide. This further yields
(e∗x)∗(y∗e)=(e∗x)◦(y∗e)=(e◦y)∗(x◦e)=(e∗y)∗(x∗e),and sox∗y=y∗x. Thus∗is commutative and (c) is proved.
(Romanian high school textbook)
272.Substitutingx =u∗vandy =v, withu, v∈S, in the given condition gives
(u∗v)∗(v∗(u∗v))=v .Butv∗(u∗v)=u, for allu, v∈S.So(u∗v)∗u=v, for all