Download Generating sets, Cayley digraphs. Groups of permutations as

Generating sets, Cayley digraphs. Groups of permutations as
universal model. Cayley’s theorem
Sergei Silvestrov
Spring term 2011, Lecture 10
Contents of the lecture
+ Generating sets and Cayley digraphs.
+ Groups of permutations as universal model. Cayley’s theorem
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Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Directed graphs: definition
Definition 1. A directed graph (or just digraph) is a finite set of points called vertices and some
arcs (with a direction denoted by an arrowhead or without a direction) joining vertices.
For each generating set S of a finite group G, we can construct the following Cayley digraph
D . The number of vertices in D is ∣G∣. For any a ∈ S, there exist arcs of type a. An arc of type
a points from x ∈ G to y ∈ G if and only if y = xa. If a ∈ S and a2 = e, it is customary to omit the
arrowhead from the arc of type a.
Example: Cayley digraph for G = ℤ6 and S = {1}
Example 1. Let G = ℤ6 and S = {1}. The Cayley digraph has the form
@
0
5
>>
>>
>>
>
1
O
4
2
^>>
>>
>>
>
3
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1
Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Example: Cayley digraph for G = ℤ6 and S = {2, 3}
/
Example 2. Let G = ℤ6 and S = {2, 3}. Let
in ℤ6 , the arrow of type 3 must be
_ _ _
be an arrow of type 2. Because 32 = 0
. The Cayley digraph has the form
0
4
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E 3
333
33
33
33
33
3
33
>>
@
3
>>
>> 333
>
33
33
o
33
5
1
33
>
> 333
> 3
>3
o
2
2
Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
A characterisation of Cayley digraphs
Theorem 1. A digraph G is a Cayley digraph of some generating set H of a finite group G if and
only if the following four properties are satisfied.
À
Á
Â
Ã
G is connected.
At most one arc goes from vertex g to a vertex h.
Each vertex g has exactly one arc of each type starting at g, and one of each type ending at g.
If two different sequences of arc types starting from vertex g lead to the same vertex h, then
those same sequences of arc types starting from any vertex u will lead to the same vertex v.
Cayley used this theorem to construct new groups. For example, the following digraph satisfies all conditions of Theorem 1.
⋅
>
O
>
>
/
>
>
>
⋅
_ _ _
by a and
O
⋅
o
⋅
C
O
e
>
⋅
⋅
/
⋅
>
⋅ o
If we label
group of order 8:
/
C
C
C
by b, we obtain a Cayley digraph of a new
C
{
b
o
{
{
/
{
a
A
ab
O
a3 b
}
a3 o
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}
}
}
/
a2 b
A
A
A
a2
3
Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Groups of permutations as universal model. Cayley’s theorem
Permutations and symmetric group
Definition 2. A permutation of a non-empty set A is a mapping σ : A 7→ A that is both one to one
and onto.
Let A = {1, 2, . . . , n}. A rearrangement is a list, with no repetitions, of all the elements of A. A
rearrangement i1 , i2 , . . . , in of A determines a function σ : A 7→ A, namely, σ (1) = i1 , σ (2) = i2 , . . . ,
σ (n) = in . We use a two-rowed notation to denote the function corresponding to a rearrangement;
if σ ( j) is the jth item on the list, then
(
1
σ=
σ (1)
2
σ (2)
...
...
j
σ ( j)
...
...
)
n
.
σ (n)
That a list contains all the elements of A says that the corresponding function σ is onto;
that there are no repetitions on the list says that distinct points have distinct values; that is, σ is
one-to-one. Thus, each list determines a one-to-one correspondence σ : A 7→ A; that is, each rearrangement determines a permutation. Conversely, every permutation σ determines a rearrangement, namely, the list σ (1), σ (2), . . . , σ (n) displayed as the bottom row. Therefore, rearrangement
and permutation are simply different ways of describing the same thing. The advantage of viewing
permutations as functions, however, is that they can now be composed and their composite is also
a permutation.
Definition 3. The family of all the permutations of a set A, denoted by SA , is called the symmetric
group on A. When A = {1, 2, . . . , n}, SA is usually denoted by Sn , and it is called the symmetric
group on n letters.
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Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Example: S3 and D3
Example 3. Denote the elements of S3 as
(
1
ρ0 =
1
(
1
ρ1 =
2
(
1
ρ2 =
3
2
2
2
3
2
1
)
3
,
3
)
3
,
1
)
3
,
2
µ1
µ2
µ3
(
1
=
1
(
1
=
3
(
1
=
2
2
3
2
2
2
1
)
3
,
2
)
3
,
1
)
3
.
3
The multiplication table of S3 is as follows.
ρ0
ρ1
ρ2
µ1
µ2
µ3
ρ0
ρ1
ρ2
µ1
µ2
µ3
ρ0
ρ1
ρ2
µ1
µ2
µ3
ρ1
ρ2
ρ0
µ2
µ3
µ1
ρ2
ρ0
ρ1
µ3
µ1
µ2
µ1
µ3
µ2
ρ0
ρ2
ρ1
µ2
µ1
µ3
ρ1
ρ0
ρ2
µ3
µ2
µ1
ρ2
ρ1
ρ0
Consider an equilateral triangle with vertices labelled as 1, 2, and 3. Let ρ j be a rotation by
2π j/3 clockwise, and let µ j be mirror imaging in the bisector of angle j. Then our table is also the
multiplication table of the group D3 of symmetries of an equilateral triangle!
Definition 4. The nth dihedral group Dn is the group of symmetries of the regular n-gon. The
group D4 is called the octic group.
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Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Cayley’s Theorem
The next theorem shows that the symmetric groups are incredibly rich and complex.
Theorem 2. (Cayley’s Theorem, Theorem 7.20, Sec. 7.4, p. 194)
Let G be a group. Then G is isomorphic to a subgroup of SG .
Sketch of proof.
defined by
1. Let f : G 7→ SG be the function which sends a ∈ G to the function fa : G 7→ G
fa (g) = ag,
g ∈ G.
For each given a, fa is a one-to-one correspondence between G and itself.
2.
f is a homomorphism, i.e., fab = fa fb .
3.
f is one-to-one and thus G is isomorphic to the image f [G] ⊂ SG .
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Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Orbits
Let σ ∈ SA . Define the following relation on A:
a ∼ b ⇔ ∃ n ∈ ℤ : b = σ n (a).
∼ is an equivalence relation.
Definition 5. Let σ ∈ SA . The equivalence classes in A determined by the equivalence relation ∼
are the orbits of σ .
The algorithm of finding the orbits of σ ∈ Sn is as follows.
Step 1 Pick the smallest element of {1, 2, . . . , n} which has not yet appeared — call it a (if you are
just starting, a = 1); write {a If no such element exist, stop.
Step 2 Read off σ (a) from the given description of σ — call it b. If b = a, close the set with a right
parenthesis } (without writing b down); this completes an orbit — return to Step 1. If b ∕= a,
write b next to a in this orbit: {a, b
Step 3 Read off σ (b) from the given description of σ — call it c. If c = a, close the set with a right
parenthesis } to complete the orbit — return to Step 1. If c ∕= a, write c next to b in this orbit:
{a, b, c Repeat this step using the number c as the new value for b until the orbit closes.
Example 4. Let σ ∈ S13 be as follows.
(
σ=
1
12
2
13
3
3
4
1
5
11
6
9
7
5
8
10
9
6
10
4
11
7
12
8
13
2
)
.
The orbits are
{1, 12, 8, 10, 4},
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{2, 13},
{3},
{5, 11, 7},
{6, 9}.
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Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Cycles
Definition 6. A permutation σ ∈ Sn is a cycle if it has at most one orbit containing more than one
element. The length of a cycle is the number of elements in its largest orbit.
In cyclic notation the cycle is written as σ = (a1 , a2 , . . . , am ), where σ (a j ) = a j+m 1 , and if ak
is not among a1 ,, a2 , . . . , am , then σ (ak ) = ak .
Example 4 shows that
σ = (1, 12, 8, 10, 4)(2, 13)(3)(5, 11, 7)(6, 9),
i.e., σ can be written as a product of disjoint cycles. This is a general fact.
The cycle decomposition theorem
Theorem 3. Every σ ∈ Sn is a product of disjoint cycles.
Proof. Let O1 , O2 , . . . , Or be the orbits of σ . For 1 ≤ j ≤ r, define the disjoint cycle µ j as
µj =
{
σ (x), x ∈ O j ,
x,
x∈
/ O j.
Clearly σ = µ1 µ2 . . . µr .
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Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Transpositions
Definition 7. A transposition is a cycle of length 2.
Clearly
(a1 , a2 , . . . am ) = (a1 , an )(a1 , an−1 ) . . . (a1 , a3 )(a1 , a2 ),
i.e., any cycle is a product of transpositions. It follows that if 2 ≤ ∣A∣ < ∞, then any σ ∈ SA is a
product of transpositions.
Theorem 4. No permutation in Sn can be expressed both as a product of an even number of
transposition and as a product of an odd number of transposition.
Proof. Let σ ∈ Sn be a product of j permutations and a product of k permutations. We have to
prove that j − k is even.
Let In be the n × n identity matrix. Because each transposition of rows multiplies the determinant of a matrix by −1, the determinant of σ In is equal to both (−1) j and (−1)k . It follows that
(−1) j = (−1)k , i.e. j − k = 2m.
The alternating group
Definition 8. A permutation σ ∈ Sn is even (odd) if it can be express as a product of an even (odd)
number of permutations.
Theorem 5. If n ≥ 2, then the set of all even permutations of the set {1, 2, . . . , n} form a subgroup An
of order n!/2 of the symmetric group Sn .
Definition 9. The group An is the alternating group on n letters.
OBS! The alternating group is a large subgroup of symmetric group.
Important Mind puzzle for you before the next lecture!
Can you joint to An some element from Sn but not from An , so that the subgroup generated
by this element and An is not the whole Sn ?
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