Course 111: Algebra, Problem Set 2, October 2006
1. Lemma
Given a mapping σ : S → T . Prove this is a bijection iff ∃ a map
µ : T → S such that σ ◦ µ and µ ◦ σ are identity mappings on S and T
respectively.
Proof
σ is a bijection so given any t ∈ T there exists and element s ∈ S such
that t = sσ by the onto property of σ. By the one-to-one property of
σ this s is unique.
Define the mapping σ −1 : T → S by s = tσ −1 iff t = sσ. The map σ −1
is the inverse map of σ.
Now compute, σ ◦ σ −1 which maps S to itself. Given s ∈ S let t = sσ
and by definition s = tσ −1 . Thus s(σ ◦ σ −1 ) = (sσ)σ −1 = tσ −1 = s.
We have shown that σ ◦ σ −1 is the identity mapping of S onto itself.
A similar computation shows that σ −1 ◦ σ is the identity mapping of T
onto itself.
Conversely, if σ : S → T is such that there exists a µ : T → S with
the property that σ ◦ µ and µ ◦ σ are the identity mapson S and T
respectively then σ is a bijection from S → T .
First, note that σ is onto since given t ∈ T, t = t(µ ◦ σ) = (tµ)σ (since
µ ◦ σ is the identity on T ) and so t is the image under σ of the element
tµ in S. Next, note that σ is one-to-one , for s1 σ = s2 σ using that
σ ◦ µ is the identity on S, we have s1 = s1 (σ ◦ µ) = (s1 σ)µ = (s2 σ)µ =
s2 (σ ◦ µ) = s2 .
This proves the lemma.
2. Lemma
Given a non-empty set S and A(S) the set of all one-to-one mappings
of S onto itself. Show that if S has more than 2 elements it is possible
to find maps σ and τ ∈ A(S) such that σ ◦ τ 6= τ ◦ σ.
Proof
Suppose S has 3 distinct elements, s1 , s2 s3 . Define the map σ : S → S
by s1 σ = s2 , s2 σ = s3 , s3 σ = s1 and sσ = s for all s ∈ S different from
s1 , s2 , s3 . Define the map τ : S → S by s2 τ = s3 , s3 τ = s2 and sτ = s
for all s ∈ S different from s2 , s3 . Clearly both σ and τ are in A(S).
Then, s1 (σ ◦ τ ) = x3 but s1 (tau ◦ σ) = s2 6= s3 . This proves the result.
3. Determine gcd(1975, 1185) and determine the coefficients m, n if this is
expressed as m1975 + n1185.
x
y
z = (x mod y)
1975 1185
790
1185 790
395
790 395
0
395
0