Download Groups, Rings and Field, Lecture Notes - Mathematics - 7 and more Study notes Mathematics in PDF only on Docsity!
7. Groups in general and permutation groups in particular
In this section we embark on a study of groups, looking in particular at the group of permutations of a finite set. Recall from Section 2:
1.4 axioms for a group and an abelian group, and initial examples; 1.5 the theorem that (Sym(X); ◦) is a group.
Recall also
Linear Algebra II web notes, Section 1, on permutations of a finite set.
7.1 Groups: preliminaries Assume (G; ⋆) is a group. If the set G is finite, then we call its size, |G|, the order of G.
Cancellation rules in a group. For all a, x, y ∈ G,
a ⋆ x = a ⋆ y =⇒ x = y; x ⋆ a = y ⋆ a =⇒ x = y
(Problem sheet 4, Question 1).
As is customary, we shall henceforth usually suppress ⋆ in products of group elements and write ab for the product a ⋆ b in a generic group (G; ⋆).
Cayley Tables. You have seen several instances of binary operations being specified by a ‘multi- plication table’ (see 1.2). In the context of groups such a table is known as a Cayley table. This is a ‘brute force’ way of presenting a group, but can be useful when working with small finite groups. In a Cayley table, each element occurs once and once only in each row and each column. (This is so because the cancellation rules tell us that ra : x 7 → x ⋆ a and ℓa : x 7 → a ⋆ a are injective maps, for any fixed a; by the Pigeonhole Principle, ra and ℓa are bijections.)
Advice: use Cayley tables sparingly.
7.2 Subgroups: definitions and the Subgroup Test. With groups, as with vector spaces and rings, we can considerably enlarge our range of examples by looking at substructures.
Let (G, ⋆) be a group and let H be a non-empty subset of G. If H is a group when equipped with the restriction of ⋆ then H is said to be a subgroup of G. and we write H 6 G.
Extreme cases: In any group G (with identity denoted e) we have
- {e} is always a subgroup of G;
- G is always a subgroup of G.
We say that a subgroup H of a group G is non-trivial if H 6 = {e} and proper if H ( G.
The Subgroup Test: Let (G; ⋆) be a group and H a non-empty subset of G.
(i) (H; ⋆) is a subgroup of (G; ⋆) if and only if (SG1) h ⋆ k ∈ H for all h, k ∈ H; (SG2) h−^1 ∈ H for all h ∈ H.
Alternatively:
(ii) (H; ⋆) is a subgroup of (G; ⋆) if and only if it satisfies the single condition (SG) for all h, k ∈ H, h k−^1 ∈ H.
Proof. [web notes only] ⇐= is an immediate consequence of the definition of a subgroup.
⇐= Suppose ∅ 6 = H ⊆ G and that (SG1) and (SG2) hold. Note first that (SG1) is just the requirement (bop) that ⋆ defines a binary operation on H. Associativity is inherited from G so axiom (G1) is satisfied by (H; ⋆).
Now, because G is non-empty, we may take some fixed element c ∈ H. Apply (SG2) with h as c to get c−^1 ∈ H. Now apply (SG1) with h = c and k = c−^1 to get eG ∈ H. Thus (see the note above) axiom (G2) (identity element) is satisfied. Finally, (G3) (inverses) is just the condition (SG2). We have proved that H is a subgroup.
Finally we need to show that (SG1) and (SG2) together are equivalent to the condition (SG). Clearly, (SG1) and (SG2) taken together imply (SG). Conversely, if (SG) holds, we can take h = k = c in (SG), where c is some element of the non-empty set H. We get eG ∈ H. Then put h = eG in (SG) to show that (SG2) holds. Now, for any h, ˜h ∈ H, apply (SG) with k as (˜h−^1 to get h˜h ∈ H.
Warning: Observe how the fact that H 6 = ∅ is used in the above proof. When applying the Subgroup Test, don’t forget to check/note that your candidate subgroup is indeed a non-empty set. One way to do this is to verify that eG ∈ H.
Some examples of subgroups: (SG) can easily be verified for the following non-empty subsets of the groups specified.
- { x ∈ R | x > 0 } is a subgroup of R r { 0 } under multiplication.
- S^1 := { z ∈ C | |z| = 1 } is a subgroup of C r { 0 } under multiplication.
7.6 Permutation groups. We shall henceforth denote by Sn, rather than Sym(n), the group of permutations of the finite set { 1 , 2 ,... , n}. Recall from LAII that we have |Sn| = n!.
We can view Sn as sitting inside Sn+1 as the subgroup of permutations of { 1 ,... , n + 1} having n + 1 as a fixed point. The group S 3 is non-abelian (see detailed discussion of S 3 below). Hence Sn is non-abelian for n > 3 too.
Cycle decomposition: Every element of Sn is expressible in an essentially unique way as a product of disjoint cycles (LAII, Theorem 1.3 and accompanying discussion, and GRF 6.6), and has an associated cycle-type (see LAII notes, p. 3), specifying the lengths of the constituent cycles. When talking about permutations and cycle-types we shall suppress 1-cycles.
Parity: An element σ of Sn is classified as
even if σ is expressible as a product of an even number of transpositions (i.e. 2-cycles), odd if σ is expressible as a product of an odd number of transpositions
(see LAII, Theorems 1.5 and 1.6, and Definition 1.7). Here the transpositions are not required to be disjoint. Recall that an m-cycle (a 1 a 2... am) is even iff m is odd.
7.7 Small permutation groups. For n−1,and n = 2, the group Sn has, respectively, 1-element and 2 elements. The cases n = 3, 4 are more interesting.
The group S 3 : We can list the six elements of S 3 as
id, (1 2), (2 3), (3 1), (1 2 3), (1 3 2).
The Cayley table is ◦ id (1 2) (2 3) (3 1) (1 2 3) (1 3 2)
id (1 2) (2 3) (3 1) (1 2 3) (1 3 2)
id (1 2) (2 3) (3 1) (1 2 3) (1 3 2)
(1 2) id (1 2 3) (1 3 2) (2 3) (3 1)
(2 3) (1 3 2) id (1 2 3) (3 1) (1 2)
(3 1) (1 2 3) (1 3 2) id (1 2) (2 3)
(1 2 3) (3 1) (1 2) (1 2 3) (1 3 2) id
(1 3 2) (2 3) (3 1) (1 2) id (1 2 3)
Note in particular that
= (1 2 3) whereas (1 3)(1 2) =
So S 3 is non-abelian.
The three transpositions are odd permutations. The other elements, id, (1 2 3) = (1 2)(1 3) and (1 3 2) = (1 3)(1 2), are even permutations.
The following are the proper non-trivial subgroups of S 3 :
A 3 :={id, (1 2 3), (1 3 2)} (note (1 3 2) = (1 2 3)^2 = (1 2 3)−^1 ), {id, (1 2)}, {id, (2 3)}, {id, (3 1)}.
Note that A 3 is cyclic,
The group S 4 : This has 24 elements. You were asked on LAII problem sheet 1 (Question 2) to classify the elements according to their cycle-types. We have
1 identity even 6 2-cycles odd 8 3-cycles even 6 4-cycles odd 3 double transpositions even
The following are important subgroups of S 4 :
V 4 := { id, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3) }, A 4 := { all even permutations in S 4 }.
We note that V 4 is a subgroup of A 4 and that V 4 is not a cyclic group (why?).
Advice: get to know S 3 and S 4 well.
7.8 Alternating groups. Generalising what we noted above for n = 3, 4, we define, for any n,
An := { σ ∈ Sn | σ is even };
this forms a subgroup of Sn. We have |An| = n!/2.
[Lecture: why study permutation groups?—informal remarks]
