212 CHAPTER 4 Abstract Algebra
Sinceτ is one-to-one, we conclude that s=s′, which proves
thatσ◦τ is also one-to-one.
σ◦τ is onto: We need to prove that for any s∈ S there exists
somes′∈Ssuch thatσ◦τ(s′) =s. However, sinceσis onto,
there must exist some element s′′ ∈S such thatσ(s′′) = s.
But sinceτ is onto there exists some elements′∈Ssuch that
τ(s′) = s′′. Therefore, it follows thatσ◦τ(s′) =σ(τ(s′)) =
σ(s′′) =s, proving thatσ◦τ is onto.Before looking further for examples, I’d like to amplify the issue of
“closure,” as it will given many additional examples of binary opera-
tions.
Definition of Closure. Let S be a set, let ∗be a binary operation
onS, and let ∅ 6=T ⊆S. We say thatT is closedunder the binary
operation ∗ if t∗t′ ∈ T whenever t, t′ ∈ T. In this case it is then
follows that∗ also defines a binary operation onT. Where the above
IB remark is misleading is that we don’t speak of a binary operation as
being closed, we speak of a subset being closed under the given binary
operation!
More examples...- LetRbe the real numbers. ThenZandQare both closed under
both addition and multiplication. - Note that the negative real numbers are not closed under multi-
plication. - Let Z[
√
5] ={a+b√
5 |a, b∈Z}. ThenZ[√
5] is easily checked
to be closed under both addition + and multiplicaton · of com-
plex numbers. (Addition is easy. For multiplication, note that if
a, b, c, d∈Z, then(a+b√
5)·(c+d√
5) = (ac+ 5bd) + (ad+bc)