Lecture Notes - Week 1 : What are groups ?
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Groups and Symmetries
A symmetry is an operation which can be applied to a physical system (e.g a molecule, a
crystal ) or a mathematical system (e.g a system of equations, a geometrical construction )
which leaves the the system invariant.
Group theory is a mathematical construction, which abstracts the properties of symmetries, i.e states them in generality, forgetting the specifics of their particular realizations.
Example Time reversal and time translation as a symmetry of Newtons’ Law of gravitation. Rotations by 120 degrees as symmetries of an equilateral triangle.
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Definition of a group
A group is a collection (or set) {g1 , g2 , · · · } which is closed under an operation of group multiplication . Call this set G. Take an ordered pair (i.e. pick a pair from the set and remember
which one you picked first and which second), denote it (g1 , g2 ), the group multiplication
will give you another element of G. We can denote this element of G as
g1 g2
(1)
m(g1 , g2 )
(2)
SOmetimes we can use another notation
You can think of m as a function from the set of ordered pairs, denoted G × G, to G.
m:G×G→G
(3)
1. Closure : For any pair (g1 , g2 ), we have g1 g2 ∈ G.
2. Associativity :
(g1 g2 )g3 = g1 (g2 g3 )
(4)
3. Identity : There is a unit element I such that for any g ∈ G
gI = Ig = g
(5)
4. Inverse : For every g ∈ G, there is a g −1 ∈ G such that
g · g −1 = g −1 · g = I
(6)
Exercise : Write the above 4 defining properties using the notation m(g1 , g2 ) for the
product.
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2.1
Examples and terminology
1. The set of real numbers, denoted R, with the addition operation as group multiplication, satisfies all the axioms and forms a group. Zero is the identity element. If we use
odinary multiplication as the group multiplication operation, the set R is not a group,
since 0 is not invertible. The set R+ = R \ {0}, i.e real numbers excluding zero, is a
group.
2. The group axioms do not require commutativity : g1 g2 = g2 g1 . When this property
holds, as it does in the case of R, then we say that the group is abelian. Groups
G where not all pairs (g1 , g2 ) obey the commutativity condition, are said to be nonabelian.
3. The set of invertible 2 × 2 matrices
A11 A12
A=
A21 A22
(7)
where Aij ∈ R forms a group. This group called GL(2, R). Since matrix multiplication
is in general non-commutative, this is an example of a non-abelian group.
The group GL(N, R) is the is the group of N × N invertible matrices. The group
GL(1, R) is the set of invertible elements of R. It is denoted R+ .
4. GL(2, R) is an example of a non-abelian group. Given two matrices A, B, in general
AB 6= BA. In index notation
X
(AB)ij =
Aik Bkj
k
X
X
Akj Bik 6= (AB)ij
(8)
Bik Akj =
(BA)ij =
k
k
Exercise : Use index notation for matrices to show that (AB)C = A(BC). Review
your notes from MT3 if this is a bit rusty.
5. R or GL(2, R) are examples of continuous and infinite groups. The set of integers,
with addition as the group multiplication operation, and 0 as the identity, is an infinite
group, but not a continuous group. It is a discrete group.
Exercise : Check that the set {1, −1}, with ordinary multiplication of numbers as the
group multiplication operation, is a group.
6. This 2-element group is called Z2 . It is a finite group. The number of elements in a
finite group is called the order of the group.
7. The multiplication table for a group is a table of all the products. The cyclic group
Z3 has multiplication table :
2
e
ω
ω2
e
e
ω
ω2
ω
ω
ω2
e
ω2
ω2
e
ω
8. Generators and relations
If a subset S of elements of a group G have the property that any element g ∈ G can
be written as a product of elements of S, then the elements of the subset S are said to
form a set of generators for G. The set of complex numbers {1, e2πi/3 , e−2πi/3 } is closed
under the multiplication of complex numbers. It forms a group which we can write
as {e, ω, ω 2 } where e is the identity element for the group multiplication. This is also
the group of rotations about the origin on a plane, by angles which are multiple of 120
degrees. The group has one generator ω and one relation
ω3 = e
(9)
Just by knowing the generators and relations, you can list the group elements
{e, ω, ω 2 }
(10)
Exercise : A group has one generator ω which obeys ω n = e, where e is the identity
element. What is the order of the group ? This is called the cyclic group Zn .
The same group can be described by different sets of generators. A choice of generators
and relations is called a presentation.
9. Direct product Group
Given groups G1 , G2 , we can form the direct product group, denoted G1 × G2 , whose
elements are pairs (g1 , g2 ). The multiplication of two such pairs is given by multiplications in each group.
(g1 , g2 )(h1 , h2 ) = (g1 h1 , g2 h2 )
(11)
when G1 , G2 are finite, the order is the product of the orders of the individual groups.
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Subgroups, Cosets and Lagrange’s theorem
A subset of a group, which contains the identity, is closed under the group multiplication and inversion, is called a subgroup.
Let G = {g1 , · · · , gN } be a finite group. Let H = {h1 , · · · , hn } be a subgroup. Pick an
element g1 in G which is not in H. Consider the list
g1 H ≡ {g1 h1 , · · · , g1 hn }
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(12)
We can show that no element in this list belongs to H. Indeed, suppose the contrary
is true, namely
g1 hi = hj
(13)
for some i, j. Then we can show that g1 = hj h−1
which contradicts our starting point
i
that g1 is not in H. So we conclude that
H, g1 H
(14)
are disjoint subsets of G, i.e they have no element in common. Now take a g2 which is
not in H ∪ g1 H and form g2 H. Again we can show that g2 H has no element in common
with g1 H. Indeed, suppose
g1 hi = g2 hj
(15)
then g2 = g1 hi h−1
j ∈ g1 H. We conclude that
H, g1 H, g2 H
(16)
have no element in common. Proceeding in this we we can continue until there is no
element of G left. So there is an integer k such that
G = H t g1 H · · · t gk H
(17)
The symbol t denotes disjoint union, i.e a special case of union, where there is no
intersection. This is called the left coset decomposition of G with respect to H. Each
ga H is called a coset of G. The set of cosets is called the coset space and is denoted
G/H. There are Nn elements in the coset space. We can analogously define left cosets
Hg and the coset space H \ G.
Each of the subsets in this decomposition has n elements. So we deduce that if H is a
subgroup of G, with n being the order of H and N the order of G, then there is some
positive integer k such that N = nk. In other words the order of a subgroup H of
group G is a divisor of the order of G. This is called Lagrange’s theorem .
Cosets can be defined using the notion of equivalence class. Two elements ga , gb of G are
said to be in the same equivalence class (by right multiplication by H) if gb = ga h for
some h ∈ H. The set of equivalence classes is denoted G/H. To relate this description
to the above right coset decomposition, note that if gi , gj are two elements in the same
subset ga H (for some a ∈ {1, · · · , k}), i.e.
gi = ga hi for some hi
gj = ga hj for some hj
(18)
−1
then gi = ga hj (h−1
j hi ) = gj (hj hi ), so that gi and gj belong to the same equivalence
class.
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4
The symmetric group Sn - an important example
The symmetric group Sn is the group of all permutations (re-arrangments) of n objects,
which can be taken to be {1, 2, · · · , n}. We can write permutations in 2-row notation
e.g in S3 we have
1 2 3
σ1 =
1 2 3
1 2 3
1 2 3
1 2 3
σ2 =
σ3 =
σ4 =
2 1 3
1 3 2
3 2 1
1 2 3
1 2 3
σ5 =
σ6 =
(19)
3 1 2
2 3 1
Permutations in Sn are bijective (one-to-one and onto) maps from the set {1, 2, · · · , n}
to itself. This way of thinking is emphasized when we describe, for example, σ2 as
σ2 [1] = 2 σ2 [2] = 1 σ2 [3] = 3
(20)
A convenient way to write permutations is in cycle notation.
σ1 = e = (1)(2)(3) = ()
σ2 = (1, 2)(3) = (1, 2) σ3 = (1)(2, 3) = (2, 3)
σ5 = (1, 3, 2), σ6 = (1, 2, 3)
σ4 = (1, 3)(2) = (1, 3)
(21)
Notes : (1, 2, 3) and (2, 3, 1) are the same permutation. Often one drops the cycles
of length 1 to simplify notation.
The cycle structures can be described by listing the cycle lengths and the number of
times they appear in the permutation. For the above three types of permutations, the
cycle structures are
[13 ]
[2, 1]
[3]
(22)
The product of two permutations is the composition of the two re-arrangements (bijective maps).
(σi σj )[a] = σj (σi (a))
(23)
Exercise Write the multiplication table of S3 . Write σ1 · · · σ6 down the rows and along
the columns. Write σa σb for the entry in the a’th row and b’th column.
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Any element of Sn can be written as a product of simple transpositions (i, j), which
are swops of i and j which leave all other elements of {1, · · · , n} fixed. There can be
different ways of writing the same permutation as a product of transpositions, e.g
(1, 2, 3) = (1, 3)(2) · (1)(3, 2) = (1, 2)(3) · (1, 3)(2)
(24)
This written more neatly by dropping the 1-cycles
(1, 2, 3) = (1, 3) · (2, 3) = (1, 2) · (1, 3)
(25)
We are using the notation σ1 · σ2 for the product here, to avoid confusion when there
are multiple cycles in the same permutation.
A given permutation can be written as a product of transpositions in more than one
way. However, for a given permutation, all these ways of writing it either involve odd
numbers of transpositions or an even numbers. If the number of transpositions appearing in a decomposition of a permutation into transpositions is even, the permutation
is said to be an even permutation. If the number of transpositions appearing in the
decomposition of a permutation into transpositions is odd, the permutation is said to
be an odd permutation. The identity is an even permutation.
Exercise : The even permutations in Sn form a subgroup ( denoted An ), while the odd
permutations do not. Why ?
Remark : There are an equal number n!/2 of even and odd permutations in Sn .
4.1
Conjugacy classes and cycle structures
Two elements of g1 , g2 ∈ G are said to be in the same conjugacy class if there is some
g3 such that
g1 = g3 g2 g3−1
(26)
We can use conjugacy to organize the elements of G into non-overlapping sets. Pick a
ga ∈ G and conjugate it with every g ∈ G to generate the conjugacy class Ca containing
ga
Ca = {gga g −1 , ∀g ∈ G}
(27)
Now pick a gb ∈
/ Ca and generate the conjugacy class Cb .
Cb = {ggb g −1 , ∀g ∈ G}
(28)
Ca and Cb do not overlap. If they did, we would have
gga g −1 = g 0 gb g 0−1
6
(29)
for some g, g 0 . But that would mean
ga = g −1 g 0 gb (g −1 g 0 )−1
(30)
which says that ga , gb are in the same conjugacy class - a contradiction.
Remark An abelian group of order n has n conjugacy class. Each group element is in
a conjugacy class by itself.
For the case of the symmetric group, two elements in the same conjugacy class have
the same cycle structure. Thus, the six permutations in S3 can be organized according
to conjugacy classes as follows
(1)(2)(3)
(1, 2)(3) , (1, 3)(2) , (1)(2, 3)
(1, 2, 3) , (1, 3, 2)
(31)
Proof : Define στ = τ στ −1 for σ, τ ∈ Sn .
στ [i] = τ στ −1 [i] = στ −1 [τ [i]]
στ [τ −1 [i]] = στ −1 [i] = τ −1 [σ[i]]
(32)
In the cycle decomposition of σ, every integer i is followed by σ[i]. In the cycle
decomposition of στ , the integer τ −1 [i] is followed by τ −1 [σ[i]]. This means that the
cycle decomposition of στ is obtained from that of σ by replacing every i with τ −1 [i].
Exercise : One of the homework problems will explore this property by examples.
Useful Fact : The conjugacy class of a permutation with pi cycles of length i is
Q
n!
pj
j pj !j
(33)
Proof can be seen in Ramond Chapter 2.
For permutation groups G which are subgroups of the symmetric group Sn , elements
of the same conjugacy class have the same cycle structure, but it is not always true
that any pair of permutations belong to the same conjugacy class of G.
As an example, if we take the alternating group A(3) of three elements, the permutations (1, 2, 3) and (1, 3, 2) are NOT in the same conjugacy class.
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Isomorphisms of groups
Two groups G and H are said to be isomorphic if there is a 1-1 and onto, i.e. bijective,
map between them, which preserves the product. Let (g1 , · · · , gN ) be the elements of
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G and (h1 , · · · , hN ) be the elements of H. An isomorphism φ : G → H, assigns to each
ga , an element φ(ga ) ∈ H such that
φ(ga )φ(gb ) = φ(ga gb )
(34)
Examples
• The group D3 of rotational and reflection symmetries of an equilateral triangle is
isomorphic to S3 .
• The group Zn of n’th roots of unity is isomorphic to the group generated by the
permutation (1, 2, · · · , n).
• Cayley’s theorem : Every group G is isomorphic to a subgroup of the symmetric
group SN , where N is the order of the group G. (see Ramond page 21 for a good
down-to-earth explanation ; Wikipedia article on Cayley’s theorem is a more
formal explanation). This is related to the “regular representation” of the group.
We will discuss representations more in Week 2.
For more examples see Wikipedia article on “Group isomorphism”
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Further Reading
The notes above have borrowed liberally from Ramond Chapter 2. You will probably
find that chapter to be a good supplement to these notes. Other sources below give
slightly different presentations - it is useful to look at slightly different presentations
of the same concepts for s deeper understanding.
• Arfken, Webber and Harris (7’th edition) : Chapter 17.
• Mathews and Walker : Chapter 16
• Ramond : Group Theory - A Physicist’s survey. – Chapter 2
• Riley Hobson and Bence - Chapter 28,29.
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