88 2 Algebra
a∗b=b∗a=e, thenbis called the inverse ofaand is denoted bya−^1. If an element
has an inverse, the inverse is unique.
Just as a warmup, we present a problem from the 62nd W.L. Putnam Competition,
2001.
Example.Consider a setSand a binary operation∗onS. Assume that(a∗b)∗a=b
for alla, b∈S. Prove thata∗(b∗a)=bfor alla, b∈S.
Solution.Substitutingb∗afora, we obtain
((b∗a)∗b)∗(b∗a)=b.The expression in the first set of parentheses isa. Therefore,
a∗(b∗a)=b,as desired.
Often, problems about binary operations look like innocent puzzles, yet they can have
profound implications. This is the case with the following example.
Example.For three-dimensional vectorsX=(p,q,t)andY =(p′,q′,t′)define the
operations(p,q,t)∗(p′,q′,t′)=( 0 , 0 ,pq′−qp′),andX◦Y =X+Y+^12 X∗Y,
where+denotes the addition inR^3.
(a) Prove that(R^3 ,◦)is a group.
(b) Letα:(R^3 ,◦)→(R^3 ,◦)be a continuous map satisfyingα(X◦Y)=α(X)◦α(Y )
for allX, Y(which means thatαis a homomorphism). Prove that
α(X+Y)=α(X)+α(Y ) and α(X∗Y)=α(X)∗α(Y ).Solution.(a) Associativity can be verified easily, the identity element is( 0 , 0 , 0 ), and the
inverse of(p,q,t)is(−p,−q,−t).
(b) First, note thatX∗Y=−Y∗X. Therefore, ifXis a scalar multiple ofY, then
X∗Y=Y∗X=0. In general, ifX∗Y=0, thenX◦Y=X+Y=Y◦X. Hence in
this case,
α(X+Y)=α(X◦Y)=α(X)◦ α(Y )=α(X)+α(Y )+1
2
α(X)∗α(Y )on the one hand, and
α(X+Y)=α(Y◦X)=α(Y )◦α(X)=α(Y )+α(X)+1
2
α(Y )∗α(X).