Groups and Symmetry Problem Set 2
Q1: Note that R (x) = x. Give expressions for Ra, (x), Ra, (x) and Ra, Rb, Ra, Rb, (x).
Check that your last answer is consistent with the formula in the notes for Ra, Rb, Ra, Rb, .
Q2: Three innite wall-paper p
Groups and Symmetry Problem Set 2 Solutions
Q1: First, we have Ra, = Ta R Ta , so Ra, (x) = a + R (x a) = a (x a) = 2a x. If
y = 2a x then x = 2a y; this shows that Ra, (y) = 2a y and thus Ra, = Ra, . Using this
Ra, Rb, Ra, Rb, (x) = 2a (2
Groups and Symmetry Problem Set 5 Solutions
Q1: Put G = Dir(X), which is a nite subgroup of SO3 . The normalisations of the midpoints of
the edges are poles of degree 2, the normalisations of the centres of the triangular faces are poles
of degree 3, and
Groups and Symmetry Problem Set 1
Please hand in problems 1 and 5 on Tuesday 21st October.
Q1: Find Symm(X) and Dir(X) when X R2 is
(a) The unit disc centred at the origin
(b) The isosceles triangle with vertices (0, 2), (1, 3) and (1, 3).
(c) The four po
Groups and Symmetry Problem Set 3
Please hand in questions 1 and 4 on Tuesday November 4th.
Q1: Let X be the wallpaper pattern shown below. We have seen that I(X) = Tu , Tv , R/3 , S0 ,
where u = (1, 0) and v = (1/2, 3/2). We also see geometrically that t
Groups and Symmetry Problem Set 3 Solutions
Q1: We have
GL,u/2 (0, 0) = u/2 + SL (0, 0) = (1/2, 0) + (0, 3/2) = v,
so the map f = Tv GL,u/2 = Tu/2v SL satises f (0, 0) = (0, 0). As L is parallel to the x-axis we
have (SL ) = S0 and (Ta ) = 1 for all a so
Groups and Symmetry Problem Set 1 Solutions
(a) Here Symm(X) = O2 and Dir(X) = SO2 . This just means that the unit disc is invariant
under any rotation about the origin, and under any reection across a line through the
origin, which is geometrically c
Groups and Symmetry Problem Set 4
Please hand in questions 1 and 4 on Tuesday 18th November.
Q1: Let g : R3 R3 be the map g(x, y, z) = (y, z, x).
(a) Prove that g O3 .
(b) Find a unit vector u with g(u) = u.
(c) Find the order of g and deduce that g SO3 .
Groups and Symmetry Problem Set 6 Solutions
Q1: The identity element xes all of X, so it has more than one xed point. The orbit counting
theorem says that the average number of xed points is the number of orbits, which is 1. As the
identity has more than
GROUPS AND SYMMETRY ABSTRACT GROUP THEORY
N. P. STRICKLAND
1. The Sylow theorems
Let G be a nite group of order n, say. Lagranges theorem says that if H is a subgroup of G
and |H| = d, then d is a divisor of n. It is natural to ask whether the converse is
Groups and Symmetry Examinable Proofs
All denitions are examinable. All results and examples in the notes and problem sheets may
provide insight that you will need to solve problems in the exam. The results listed below are the
ones for which you may be a
Groups and Symmetry Problem Set 6
Q1: Let G be a nite group, and let X be a set with an action of G. Suppose that there is
precisely one orbit, and that |X| > 1. Use the orbit counting theorem to show that there is an
element g G such that Fix(g) = .
Groups and Symmetry Problem Set 4 Solutions
(a) It is clear that g is linear, and we have
g(x, y, z)
= y 2 + z 2 + x2 = x2 + y 2 + z 2 = (x, y, z) 2 ,
so g preserves lengths. Thus g O3 .
(b) Visibly g(a, a, a) = (a, a, a) for any a. To get a unit ve
GROUPS AND SYMMETRY
N. P. STRICKLAND
1. Symmetry groups in Rn
1.1. General linear groups. We write Mn or Mn (R) for the set of n n matrices over the real
numbers. Recall that an nn matrix A is invertible if there is a matrix B such that AB = I = BA.
Groups and Symmetry Problem Set 5
Please hand in questions 1 and 4 on Tuesday 25th November.
Q1: The following diagram shows a cuboctahedron X in R3 , centred at the origin. Its faces are
squares and equilateral triangles.
Which of the standard nite subgr
f x g g Xt Xm g tgu H T P r T P m tg T P r T P `p ` `
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mtg h u tg X o H g T P r T P mtg T P r T P
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`g ` u h x g g h u tgu X o tX ` R ` p g x
vp(pvr vnIgiX11epqep5ueI"ee#X %@wUTRwRSRP
Algebraic Topology Problem Set 3
Please hand in questions 2 and 3 on Tuesday 11th March.
In this problem we give an alternative way of understanding the map q : S n RP n .
Suppose that A Mn+1 R and A2 = A. Put U = cfw_x Rn+1 | Ax = x and V = cfw_x
Algebraic Topology Problem Set 4
No questions to be handed in this week.
Q1: Put X = cfw_(x, y) R2 | y 2 = x3 x. (This is an example of an elliptic curve; such curves
are important in number theory and certain other areas of mathematics.)
Sketch the graph
Algebraic Topology Problem Set 1
Please hand in questions 1 and 3 on Tuesday 25th February.
Which of the following rules gives a well-dened, continuous function f : X Y ? Justify your
answer with either a geometric or an algebraic argument.
(a) X is t
Algebraic Topology Problem Set 7
Please hand in questions 2, 3 and 4 on Tuesday 6th May.
Q1: Let X be a metric space, and let f, g : X S 1 be continuous maps. Suppose that f is
linearly homotopic to g. What can you deduce?
X = cfw_z C | 1 < |z| <
Algebraic Topology Problem Set 6
Please hand in questions 1,2 and 3 on Tuesday April 1st.
Q1: Let X and Y be spaces such that 0 X and 0 Y are nite sets. Suppose that f : X Y and
g : Y X are continuous maps such that gf is homotopic to 1X .
(a) Show that f
GUIDANCE ON THE EXAM
This list covers most of the examples that you should know about. The exam may ask you to
prove that two spaces are (or are not) homeomorphic (or homotopy equivalent) to each other. The
spaces involved will either be tak
PMA333 EXAMINATION SPRING 2001 SOLUTIONS
N. P. STRICKLAND
(i) A metric space X is compact if for every sequence (xn ) in X there is a subsequence
(xnk ) and a point x X such that xnk x.
(ii) Let f : X Y be a continuous surjective map, and suppose
Algebraic Topology Problem Set 5
Please hand in questions 1,2 and 3 on Tuesday 25th March.
Q1: Let f : X R \ cfw_0 be a continuous map. Show that f is homotopic to a map g such that
g(x)2 = 1 for all x X.
Q2: Recall that for any space X and any x X we hav
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)5a~a m aH
z n c j dl b
)5a~a k fiaH
z n c d b dl Yb
z n c d b x dl b
h c dh dr vg u cfw_ z z y i c gh v ps ss dh v y
eoo| " C~ec w ipxgqt)iuo5)it!eti
QEeeC)oaeii` m qus qec k oesxrHeoeEWuqoq!sh