Advanced book on Mathematics Olympiad

454 Algebra

xy^2 x−^1 =x(yx−^1 )x(yx−^1 )=(yx−^1 )x(yx−^1 )x=y^2.

Thus for anyx, y, we havexy^2 =y^2 x. This means that squares commute with everything.

Using this fact, we rewrite the identity from the statement as

xyxyx−^1 y−^1 x−^1 y−^1 =e

and proceed as follows:

e=xyxyx−^1 y−^1 x−^1 y−^1 =xyxyx−^2 xy−^2 yx−^2 xy−^2 y

=xyxyy−^2 x−^2 xyxyy−^2 x−^2 =(xyxyy−^2 x−^2 )^2.

Because there are no elements of order 2, it follows thatxyxyy−^2 x−^2 =eand hence
xyxy=x^2 y^2. Cancel anxand ayto obtainyx=xy. This proves that the group is
Abelian, and we are done.
(K.S. Williams, K. Hardy,The Red Book of Mathematical Problems, Dover, Mineola,
NY, 1996)
283.The first axiom shows that the squares of all elements inMare the same; denote
the common value bye. Thene^2 =e, and from (ii),ae=afor alla ∈M. Also,
a∗b=a(eb)for alla, b∈M. Let us verify the associativity of∗. Using (iii) in its new
forme(bc)=cb, we obtain

a∗(b∗c)=a[e(b(ec))]=a[(ec)b].

Continue using (iv) as follows:

a[(ec)b]=[a(eb)][((ec)b)(eb)]=[a(eb)][(ec)e]=[a(eb)](ec)=(a∗b)∗c.

Here we used the fact thatde=d, for the cased=ec. Thus associativity is proved.

The elementeis a right identity by the following argument:

a∗e=a(e^2 e)=a(ee)=ae^2 =ae=a.

The right inverse ofaisae, since

a∗(ea)=a[e(ea)]=a(ae)=a^2 =e.

So there exists a right identity, and every element has a right inverse, which then implies

that(M,∗)is a group.

(M. Becheanu, C. Vraciu,Probleme de Teoria Grupurilor(Problems in Group The-

ory), University of Bucharest, 1982)

284.How can we make the sumMinteract with the multiplicative structure of? The

idea is to squareMand use the distributivity of multiplication with respect to the sum of

matrices. IfG 1 ,G 2 ,...,Gkare the elements of, then