MAS114 Numbers and Groups Spring 2012–2013 Exam solutions

MAS114 Numbers and Groups
Spring 2012–2013
Exam solutions
Eugenia Cheng / Sam Marsh
Marking scheme
Format
The idea is:
ˆ 20% on online tests, of which 10% each for each semester
ˆ c.30% marks on the exam for semester one
ˆ c.50% marks on the exam for semester two with some dependence on semester one
Thus with a total of only 40% purely on semester one, it is effectively not possible to pass with
only semester one knowledge.
1
Question 1.
i)
(standard problem)
The Chinese Remainder Theorem tells us that pairs of remainders mod y and mod z are in
bijective correspondence with remainders mod y.z.
Thus if we find one solution, say, x = t then we know that all solutions are given precisely
by
x ≡ t (mod y.z).
(1)
ii)
(unseen problem)
11 and 10 are coprime
so by the above, if we find one solution, say, x = t then we know that all solutions are given
precisely by
x ≡ t (mod 110).
Note that we are solving the simultaneous equations
x
=
11i + a
x
=
10j + b
Equating these, we get 11i + a = 10j + b thus the Diophantine equation
11i − 10j = b − a
which does have solutions as (11, 10) = 1.
Clearly 11.1 − 10.1 = 1 so a particular solution is given by
11(b − a) − 10(b − a) = b − a.
so i = b − a, j = b − a.
(1)
Now we only need one solution, and this gives
x
=
11i + a
=
11(b − a) + a
=
11b − 10a.
By the CRT this gives all possible solutions:
x ≡ 11b − 10a
(mod 110).
(1)
NB we get the same answer by substituting into the other expression for x, using j.
iii)
(unseen problem)
We prove by induction that any such set of simultaneous congruences has a unique solution
(mod k1 · · · kn ).
Here the initial step is n = 2 and is given by the assumption of part (i).
For the induction step, we assume the result is true for n = m, say, and aim to deduce the
result for n = m + 1. So we consider simultaneous congruences
x ≡ a1
(mod k1 )
x ≡ a2
(mod k2 )
..
.
x ≡ am
(mod km )
x ≡ am+1
2
(mod km+1 )
where any pair ki , kj are coprime. Now by the induction step we know that the first m
congruences have a unique solution (mod k1 · · · km ), say
x ≡ b (mod k1 · · · km ).
(1)
Now we show that the product k1 · · · km is coprime to km+1
i.e., that their highest common factor is 1.
Suppose p is a prime factor that divides k1 · · · km and km+1 .
Then since p is prime, it must divide one of the ki for some 1 ≤ i ≤ m.
But then p|ki and p|km+1 which contradicts the assumption that every pair ki , kj are coprime.
(1)
Thus k1 · · · km and km+1 are coprime, so by the initial step we have a unique solution
(mod k1 · · · km .km+1 ) to the simultaneous congruences
x ≡ b (mod k1 · · · km )
x ≡ am+1
(mod km+1 ).
But by the induction hypothesis, this is equivalent to the simultaneous congruences
x ≡ a1
(mod k1 )
x ≡ a2
(mod k2 )
..
.
x ≡ am
(mod km )
x ≡ am+1
(mod km+1 )
as required. Thus we have proved the induction step.
(1)
Thus the result is true for all n ≥ 2 as required.
Presentation
(1)
3
Question 2.
i)
(bookwork and unseen)
Suppose that gf is surjective, i.e.
∀c ∈ C ∃a ∈ A s.t. gf (a) = c.
We need to show that g is surjective, i.e.
∀c ∈ C ∃b ∈ B s.t. g(b) = c.
Let c ∈ C. Since gf is surjective we have a ∈ A s.t. gf (a) = c.
i.e. g(f (a)) = c.
So put b = f (a). Then g(b) = g(f (a)) = c as required.
So g is indeed surjective.
(2)
It is not true that f needs to be surjective. For example let
A = {0}
B = {0, 1}
C = {0}
with functions
f
g
{0} −→ {0, 1} −→ {0}
given by f (0) = 0, and g must map everything to 0.
Then gf (0) = 0 so gf is surjective, but f is not surjective as nothing maps to 1.
(2)
ii) f is not injective.
For example f (0) = f (5) = 0.
f is surjective, since for any a ∈ {0, 1, 2, 3, 4} we have
f (a) = a.
For correct answers
For justifications
(1)
(2)
4
Question 3.
i)
(bookwork)
The sequence an is said to converge to a if
∀ε > 0 ∃N ∈ N s.t. ∀n ≥ N |an − a| < (1)
ii)
(unseen problem)
We must show:
∀a ∈ R ∃ε > 0 s.t. ∀N ∈ N ∃n ≥ N s.t.
Put ε = 21 .
|an − a| ≥ ε.
(1)
First note that ∀a ∈ R we have
|0 − a| ≥
since |0 − a| = |a| <
so in particular a <
1
2
1
2
iff − 12 < a <
1
1
or |1 − a| ≥
2
2
1
2
hence 1 − a ≥ 12 .
(2)
So, given any a ∈ R and any N ∈ N we can pick
(
10N
|a| ≥ 12
n=
10N + 1 otherwise
Then certainly n ≥ N and also |an − a| ≥
1
2
= ε as required.
(2)
iii) The sequence is bounded by 11 (for example) as we have
∀n ∈ N, |an | ≤ 11.
(1)
5
Question 4.
i)
(Unseen)
a) α(1 2 3 4)α−1 = (1 6 3 4).
b) α(1 2 3 4)α−1 = (1 2 5 4).
c) α(1 2 3 4)α−1 = (2 5 4 3).
(3)
It appears that α(1 2 3 4)α−1 is a 4-cycle consisting of the first four entries in the bottom
row of α; that is, α(1 2 3 4)α−1 = (α(1) α(2) α(3) α(4)).
(1)
!
1 2 3 4 5 6
If this pattern continues, then β =
should work (which it does, as
3 4 5 6 1 2
!
1 2 3 4 5 6
does
).
(1)
3 4 5 6 2 1
Also, given that α(1 2 3 4)α−1 is always a 4-cycle, it will have order 4.
(1)
ii)
(Unseen)
a)
A
−1
−1
= (det A)
0
−3
−1
2
!
= (4)
−1
0
4
6
2
!
=2
0
4
6
2
!
=
0
1
5
4
!
.
(1)
ABA
−1
=
=
=
2
3
1
0
2
3
1
0
4
3
2
3
!
!
0
6
6
0
2
3
0
6
!
0
1
5
4
!
!
!
.
(1)
b) As
B2 =
0
6
6
0
!
and
(ABA
−1 2
) =
4
3
2
3
!
0
6
6
0
!
4
3
2
3
=
1
0
0
1
!
=
!
1
0
0
1
!
it follows that the order of B and the order of ABA−1 are both 2.
iii)
(2)
(Unseen)
Let G be a group and let g, h ∈ G. Then, for any positive integer n,
(ghg −1 )n = (ghg −1 ) . . . (ghg −1 ) = g(h
. . h})g −1 = ghn g −1 .
| .{z
{z
}
|
n
n
(1)
Thus (ghg
−1 n
n −1
) = e ⇐⇒ gh g
n
= e ⇐⇒ h = g
−1
n
g ⇐⇒ h = e.
−1 n
(1)
Hence the least positive integer n such that (ghg ) = e is the same as the least positive
integer n for which hn = e (if one exists in either case); that is, the order of ghg −1 is the
same as the order of h.
(1)
6
Question 5.
i)
(Bookwork)
A permutation α in Sn is said to be even (respectively odd ) if it can be written as a product
of an even (respectively odd) number of transpositions.
(1)
The sign of a permutation α, written sgn(α), is defined to be +1 if α is even and −1 if α is
odd.
(1)
(Unseen, standard)
ˆ (Reflexivity) Let α ∈ Sn . Then sgn(α) sgn(α) = sgn(α)2 = (±1)2 = 1, so α ∼ α.
ˆ (Symmetry) Let α, β ∈ Sn and suppose that α ∼ β. Then sgn(α) sgn(β) = 1, so
sgn(β) sgn(α) = 1, so β ∼ α.
ˆ (Transitivity) Let α, β, γ ∈ Sn and suppose that α ∼ β and β ∼ γ. Then sgn α sgn β = 1
and sgn β sgn γ = 1, so
sgn α sgn β sgn β sgn γ = sgn α(sgn β)2 sgn γ = 1
so
sgn α sgn γ =
1
1
= =1
2
(sgn β)
1
so α ∼ γ.
Thus, ∼ is reflexive, symmetric and transitive and is therefore an equivalence relation. (2)
(Part marks awarded for semi convincing arguments)
(It’s acceptable to observe that α ∼ β ⇐⇒ sgn α = sgn β)
ii)
(Unseen, standard)
H = hr3 i =
{r3 , r32 , r33 , r34
= e} = {e, r1 , r2 , r3 }.
(1)
There are 8/4 = 2 distinct left-cosets of H in D4 , by Lagrange’s Theorem.
(1)
Since the distinct left-cosets of H partition D4 , it follows that s1 H = {s1 , s2 , s3 , s4 }. As
s1 ∈ s1 H but s1 s1 = e 6∈ s1 H, SG2 fails and s1 H is not a subgroup of D4 .
(1)
iii)
(Bookwork)
Theorem 0.0.1 (Lagrange’s Theorem). Let G be a finite group and let H be a subgroup of
G. Then the order of G is a multiple of the order of H. (More precisely, |G| = m|H| where
m is the number of distinct left cosets of H in G.)
(1)
(Unseen)
Since S has size 3 and Z10 has size 10, it follows from Lagrange’s Theorem that S cannot be
a subgroup of Z10 .
(1)
Since 8 ∈ S but 8 + 8 = 6 6∈ S, S fails SG2.
iv)
(1)
(Unseen)
Denote the image of f by H. Then H = {(g, g) ∈ G × G : g ∈ G}.
SG1: As (e, e) = f (e) ∈ H, it follows that H 6= ∅.
SG2: Let (g, g), (h, h) ∈ H. Then (g, g)(h, h) = (gh, gh) = f (gh) ∈ H, so SG2 holds.
SG3: Let (g, g) ∈ H. Then (g, g)−1 = (g −1 , g −1 ) = f (g −1 ) ∈ H, so SG3 holds.
It follows that H is a subgroup of G × G by the subgroup criterion.
7
(3)
Question 6.
i)
(Bookwork)
A non-empty set G is a group under if the following four axioms hold.
G1 (Closure): a b ∈ G for all a, b ∈ G.
G2 (Associativity): (a b) c = a (b c) for all a, b, c ∈ G.
G3 (Neutral element): There is an element e ∈ G such that, for all g ∈ G, e g = g = g e.
G4 (Inverses): For each element g ∈ G there is an element h ∈ G such that gh = e = hg.
(4)
(Part marks awarded for faulty definitions, as appropriate)
ii)
(Unseen)
Suppose S contains just one element. Then S = {a} for some element a, and
ˆ S is closed under , as a a = a ∈ S;
ˆ is associative since (a a) a = a = a (a a);
ˆ a is a neutral element, as a a = a = a a;
ˆ a has inverse a, as a a = a = a s.
Thus S satisfies G1–G4 and is a group under .
(2)
Conversely, suppose S is a group and let e ∈ S be the neutral element. Then, for all a ∈ S,
a = a e since e is neutral. On the other hand, a e = e by the rule for . It follows that
a = e for all a ∈ S; that is, S = {e} contains just one element.
(2)
iii)
(Unseen, standard)
The tile has group of symmetries K = {e, r, s, t}, where r is rotation through π, and s and t
are the reflections in the lines shown.
t
s
We use the orbit counting theorem to count the number of orbits, n, for the action of K on
the colourings of the tile with 4 green and 5 blue squares.
ˆ Every colouring is fixed by e, so | fix(e)| = 95 = 126 (choose 5 squares to colour blue
from the 9 available).
ˆ The colourings fixed by r have the form
A B C
D
E
D
C B A
Thus | fix(r)| =
well).
4
2
= 6 (choose 2 of the pairs A–D to colour blue, and colour E blue as
ˆ The colourings fixed by s have the form
8
A B C
D
E
D
A B C
Thus | fix(s)| =
well).
4
2
= 6 (choose 2 of the pairs A–D to colour blue, and colour E blue as
ˆ The colourings fixed by t have the form
A C A
D
E
F
B G B
Thus | fix(t)| = 5 + 2 × 53 + 1 = 26 (select A and B, and choose 1 of C–G to colour blue
OR to select one of A or B, and choose 3 of C–G to colour blue OR colour C–G blue).
(4)
Thus, by the orbit-counting theorem,
n=
1
1 X
| fix(g)| = (126 + 6 + 6 + 26) = 41.
|K|
4
g∈K
(1)
9