WRITTEN ASSIGNMENT 2
MATH0043
Please attempt all problems, and show ALL working for each question.
DEADLINE: 31st March 23:59. Submit through Crowdmark
1. Find the maximum and minimum possible areas bounded by the curve y(x) in
the interval 0 ≤ x ≤ 1, subject to the constraint that
Z 1
1
and y(0) = y(1) = 1.
(y ′ )2 dx =
48
0
We are trying to find the extreme values taken by
Z 1
A[y] =
ydx,
0
subject to y(0) = y(1) = 1 and the integral constraint
Z 1
1
.
F [y] =
(y ′ )2 dx =
48
0
This is an isoperimetric problem and so we must form the composite functional
Z 1
K[y] = A[y] + λF [y] =
y + λ(y ′ )2 dx,
0
with λ as our Lagrange multiplier. the integrand H(y, y ′ ) = y + λ(y ′ )2 has no
explicit dependence on x, so we use Beltrami’s identity
∂H
− H = C,
∂y ′
or 2λ(y ′ )2 − (y + λ(y ′ )2 ) = C,
r
y′
or y ′ =
y+C
.
λ
Integrating,
Z
1
dy =
(y + C)1/2
Z
λ−1/2 dx
x
2(y + C)1/2 = √ + D
λ
2
1
x
√ + D − C.
y=
4
λ
1
The extremal of both the maximising and the minimising problem is therefore a
parabola with coefficients determined by C, D and λ. To determine these, first use
the boundary conditions to give:
D2
−C
4
2
1
1
√ +D −C
1 = y(1) =
4
λ
Subtracting the second equation from the first gives:
2
1
1
2
D − √ +D =0⇒D =− √ .
λ
2 λ
Then
1
D2
−1=
− 1.
C=
4
16λ
Hence,
2
1
1
1
y=
+1−
x−
.
4λ
2
16λ
To find λ we use the integral constraint:
2
Z 1
Z 1
1
1
1
=
(y ′ )2 dx = 2
x−
dx
48
4λ 0
2
0
"
3 # 1
1
1
x−
=
12λ2
2
0
1
1
1 1
=
+
=
12λ2 8 8
48λ2
1 = y(0) =
Hence λ = ±1 and we have two extremal curves y = f1 (x), y = f2 (x) where
2
2
1
1
1
15
1
17
and f2 (x) = −
f1 (x) =
x−
+
x−
+ .
4
2
16
4
2
16
The first of these must be the minimiser of A and the second the maximiser. The
minimum and maximum areas can be calculated by direct integration to be
23
25
Amin = A[f1 ] =
and Amax = A[f2 ] =
.
24
24
10 marks: lose a mark if min/max areas not stated. Arithmetic error/small
slips: lose a mark but do error-carried forward.
2. Let Zn denote the group of integers modulo n where the group operation is
modular addition. In this question we will explore group products of such groups.
(1) List the elements of Z2 × Z3 and construct the group table.
(2) What are the orders of the elements in Z2 × Z3 ?
(3) Hence, or otherwise, deduce whether Z2 × Z3 is cyclic.
(4) Is Z2 × Z3 ∼
= Z3 × Z2 ?
Solution:
(1) The elements of Z2 × Z3 are
{(0, 0), (1, 0), (0, 1), (1, 1), (0, 2), (1, 2)}.
The group table is
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+
(0,0) (1,0) (0,1) (1,1) (0,2) (1,2)
(0,0) (0,0) (1,0) (0,1) (1,1) (0,2) (1,2)
(1,0) (1,0) (0,0) (1,1) (0,1) (1,2) (0,2)
(0,1) (0,1) (1,1) (0,2) (1,2) (0,0) (1,0)
(1,1) (1,1) (0,1) (1,2) (0,2) (1,0) (0,0)
(0,2) (0,2) (1,2) (0,0) (1,0) (0,1) (1,1)
(1,2) (1,2) (0,2) (1,0) (0,0) (1,1) (0,1)
4 marks: 2 for elements, 2 for group table
(2) The orders are
Element (0,0) (1,0) (0,1) (1,1) (0,2) (1,2)
Order
1
2
3
6
3
6
4 marks for correct answer
(3) The element (1, 1) has order 6 (= |Z2 × Z3 |), and so is a generator for the
group. Hence the group is cyclic. Alternatively, The map ϕ : Z2 × Z3 → Z6
defined by ϕ((1, 1)) = [1] gives an isomorphism. The same argument may
be used with the element (1, 2).
6 marks for any correct answer
(4) Yes. Observe that Z2 × Z3 ∼
= C6 , by (3). Also, (1, 1) has order 6 in Z3 × Z2
and so Z3 × Z2 is also cyclic of order 6 and so
Z3 × Z2 ∼
= Z2 × Z3 .
= C6 ∼
An alternate solution is any correct isomorphism.
6 marks for any correct solution.
3. Let C6 be a cyclic group of order 6 generated by an element x, so C6 =
{e, x, x2 , x3 , x4 , x5 }, where e is the identity element and x6 = e.
(1) Let S3 be the symmetric group. Is S3 isomorphic to C6 ? Justify your
answer briefly.
(2) Show that there is no representation ρ of C6 such that
1 1
ρ(x) =
.
0 1
(3) Let ω = exp(2πi/6) be a complex 6th root of 1. You are told that there
are two representations ϕ and ψ of C6 such that
ω 0
ϕ(x) =
,
0 1
1
ψ(x) =
0
0
.
ω
Are ϕ and ψ equivalent? Justify your answer in detail.
Solution:
(1) No. “C6 is abelian/cyclic and S3 isn’t” is fine, as is observing that if
f : C6 → S3 were an isomorphism f (x) would have order 6 but S3 has no
element of order 6, as is any other correct argument. 3 marks: 1 for correct
Yes/No, 2 for a correct justification
(2) We needρ(x)6= ρ(x6 ) = ρ(e) = I2 , as ρ is a group homomorphism, but
1 6
ρ(x)6 =
. It’s OK to compute the first couple of powers of ρ(x) and
0 1
spot the pattern, no proof needed. 3 marks
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0 1
takes ϕ(x) to ψ(x), and hence does the same
1 0
for every group element. Students may observe that the matrices have the
same distinct eigenvalues. 4 marks
(3) Yes. Conjugating by
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