Course 111: Algebra, 27th October 2006
1. Consider the unique factorisation theorem, as given in the notes. Assume existence of the factorisation and prove uniqueness.
Hint: the proof is by induction and you should begin by supposing that
a = pα1 1 pα2 2 . . . pαr r = q1β1 q2β2 . . . qsβs
Proof Suppose that
a = pα1 1 pα2 2 . . . pαr r = q1β1 q2β2 . . . qsβs
where p1 > p2 > . . . > pr and q1 > q2 > . . . > qs are prime numbers
and each αi > 0, βi > 0. Then we must prove that
1. r = s
2. p1 = q1 , p2 = q2 , . . . , pr = qr
3. α1 = β1 , α2 = β2 , . . . αr = βr
For a = 2 this is clearly true. Proceeding by induction we suppose it
to be true for all integers, u with 2 ≤ u < a. Now, since
a = pα1 1 . . . pαr r = q1β1 . . . qsβs
and since α1 > 0, p1 |a, hence p1 |q1β1 . . . qsβs . However, since p1 is a prime
number, by a previous result proved in lectures, it follows that p1 = qi
for some i.
Thus q1 ≥ qi = p1 .
Similarly, since q1 |a then q1 = pj for some j and so p1 ≥ pj = q1 .
Combining these results, p1 = q1 . Then
a = pα1 1 . . . pαr r = pβ1 1 . . . qsβs
and so α1 = β1 . Then
b=
a
= pα2 2 . . . pαr r = q2β2 . . . qsβs .
p α1
Now, if b = 1 say, then α2 = . . . = αr = 0 and β2 = . . . = βr = 0. And
in this case r = s = 1 giving the result.
Next, say b > 1 then since b < a you can use induction on b to deduce
the following
1. the number of distinct (prime) power factors on both sides is the
same so r − 1 = s − 1 ⇒ r = s.
2. α2 = β2 , . . . , αr = βr .
3. p2 = q2 , . . . pr = qr .
Combined with the results: p1 = q1 and α1 = β1 the theorem is proven.
Note: the assumption of the uniqueness of factorisation for integers
< a implied the uniqueness of factorisation for a. And so induction
completes the proof.
2. Show, writing the Cayley table, that Z4 forms a group under addition.
Is this an abelian or nonabelian group?
0
1
2
3
0
0
1
2
3
1
1
2
3
0
2
2
3
0
1
3
3
0
1
2
Z4 is abelian.
3. Recall the example of a group discussed in lectures. G = S3 the group
of all 1-1 mappings of the set x1 , x2 , x3 onto itself. Given φ : S → S
and ψ : S → S such that
φ : x1 → x 2
x2 → x 1
x3 → x 3
and
ψ : x1 → x 2
x2 → x 3
x3 → x 1
Show that (φ ◦ ψ) ◦ (ψ ◦ φ) = ψ and that the list of distinct elements
in G as a result of these mappings is given by e, φ, ψ, ψ 2 , φ ◦ ψ, ψ ◦ φ.
(φ ◦ ψ) ◦ (ψ ◦ φ) =
=
=
=
=
=
=
φ ◦ (ψ ◦ (ψ ◦ φ))
φ ◦ (ψ 2 ◦ φ)
φ ◦ (ψ −1 ◦ φ)
φ ◦ (φ ◦ ψ)
φ2 ◦ ψ
e◦ψ
ψ