Group Theory.docx

A group with only one generator x, is the cyclic group generated by x and can now be written as gp{𝑥}. To illustrate these ideas, let us once more turn to the group G(≅ 𝑆3 ) of order 6 which is exhibited in the table below

1

a

b

c

d

e

1

1

a

b

c

d

e

a

a

b

1

e

c

d

b

b

1

a d

e

c

c

c

d

e

1

a

b

d

d

e

c

b

1

a

e

e

c

d

a

b

1

table 1

we find that each of the six elements can be expressed in terms of a and c thus, 1 = 𝑐 2 = (𝑎3 ), 𝑎 = 𝑎, 𝑏 = 𝑎2 , 𝑐 = 𝑐, 𝑑 = 𝑐𝑎, 𝑒 = 𝑐𝑎2 Hence in this case we can write 𝐺 = 𝑔𝑝{𝑎, 𝑐}

(2.24)

Alternatively, it can be shown that 𝐺 = 𝑔𝑝{𝑏, 𝑑}

(2.2.5)

For a and c and hence the whole group, can be expressed in terms of b and d, namely 𝑎 = 𝑏 2 , 𝑐 = 𝑑𝑏 The generators in (2.24) or in (2.25) are certainly irredundant, because the group is non-Abelian and cannot be generated by a single element, in which case it would be cyclic and hence non-Abelian. It is important to realize that irredundant generators may nevertheless be linked by non-trivial relations. Thus, by referring to the table 1, we see that 𝑎𝑐 = 𝑐𝑎2 (2.26) Which is equivalent to (𝑎𝑐)2 = 1,

(2.26)′

Because (𝑎𝑐)2 = 𝑎𝑐𝑎𝑐 = 𝑎1𝑎2 = 𝑎3 = 1. It is impossible to solve either of these equations so as to express one of the generators in terms of the other. An equation like (2.26) or (2.26)′ is called a defining equation. It is often convenient to specify a

particular group by a set of generators and a set of defining relations. We shall return to this principle in more detail later but mention here that in the present case the equations 𝑎3 = 𝑎2 = (𝑎𝑐)2 = 1

(2.27)

Serve as a set of defining relations. Indeed, the information contained in (2.27) suffices to construct the entire multiplication table. First, we remark that the six elements 1, 𝑎 , 𝑎2 , 𝑐. 𝑐𝑎, 𝑐𝑎2

(2.28)

are certainly distinct: for example, if a where equal to 𝑐𝑎2 ,it would follow that 𝑎 −1 = 𝑐,contradicting the fact that a and c are irredundant generators. Next, we check that the system (2.28) is closed under multiplication by virtue of the relation (2.27); for example (𝑐𝑎)(𝑐𝑎2 ) = 𝑐(𝑎𝑐)𝑎2 = 𝑐𝑐𝑎2 𝑎2 = 𝑐 2 𝑎4 = 𝑎 𝑎2 𝑐 = 𝑎(𝑎𝑐) = 𝑎𝑐𝑎2 = 𝑐𝑎4 = 𝑐𝑎, and so on, a factor c being systematically moved to the left by virtue of (2.26), until the product is identical with one of the elements (2.27). In this notation the complete multiplication table is as follows;

1

a

𝑎2

c

ca

c𝑎2

1

1

a

𝑎2

c

ca

c𝑎2

a

a

𝑎2

1

c 𝑎2

c

ca

𝑎2

𝑎2

1

a

ca

c𝑎2

c

c

c

ca

c𝑎2 1

a

𝑎2

ca

ca

c𝑎2

c

𝑎2

1

a

c

ca

a

𝑎2

1

c𝑎2 c𝑎2 Table 2

and this provides yet another version of the group first presented in table 1. If A , B ,C… are subsets of f,the group generated by them is denoted by 𝑔𝑝{𝐴 , 𝐵 , 𝐶, … } And is defined as a collection of all finite products in which each factor is an element of A or B or C … or the inverse of such an element, taken in any order with or without repetition. This reduces to the previous concept of generators if we regard as generators of all the elements of A∪ 𝐵 ∪ 𝐶 ∪ … We might equally well right instead 𝑔𝑝{𝐴 ∪ 𝐵 ∪ 𝐶 ∪ … }, of course, if A is a subgroup, we have that A=gp{A}

THE DIRECT PRODUCT We shall now discuss a simple method of constructing a new group out of two given groups. Let H and K be any groups and consider the set of all pairs (u, v), Where u and v range over H and K respectively. The set of these pairs is denoted by G= H x K And is called the (exterior) direct product of H and K. The set G is turned into a group by endowing it with the composition law (u,v)(𝑢′ , 𝑣 ′ ) = (𝑢𝑢′ , 𝑣𝑣 ′ ).

(2.29)

It is easy to verify that the associative law holds in G, because multiplication is associative in H and in K. The unit element in G is the pair (1𝐻 , 1𝐾 ) Where 1𝐻 and 1𝐾 are the unit elements of H and K respectively. Also (𝑢, 𝑣)−1 = (𝑢− , 𝑣 −1 ). If H and K are finite groups of order h and k respectively, then H x K is a group of order hk. More generally, if 𝐻1 , 𝐻2 , … , 𝐻𝑟 are any groups, their direct product 𝐻1 × 𝐻2 × … × 𝐻𝑟 consists of all 𝑟 − 𝑡𝑢𝑝𝑙𝑒𝑠 (𝑢1 , 𝑢2 , … , 𝑢𝑟 ), where 𝑢𝑖 ∈ 𝐻𝑖 (𝑖 = 1 ,2 , … , 𝑟) and multiplication is carried out in each component of the 𝑟 − 𝑡𝑢𝑝𝑙𝑒. If each 𝐻𝑖 is finite, then clearly, 𝑟

|𝐻1 , 𝐻2 , … , 𝐻𝑟 | = ∏|𝐻𝑖 | 𝑖=1

it sometimes happens that a group G is isomorphic with the direct product of two of its subgroups, H and K, thus 𝐺 ≅ 𝐻 × 𝐾,

(2.30)

Or with a slight abuse of notation 𝐺 = 𝐻×𝐾

(2.30)′

This situation arises in the following circumstances: (1) The subgroups H and K commute elementwise, that is, if u and v are arbitrary elements of H and K respectively, then 𝑢𝑣 = 𝑣𝑢 (2.31)

(2) Every element 𝑥 ∈ 𝐺 can be expressed in the form 𝑥 = 𝑢𝑣, or more briefly, G= 𝐻𝐾 (2.32) (3) The intersection of H and K is the unit element, that is 𝐻∩𝐾 =1 (2.33) We remark that (2) and (3) are equivalent to the single condition:(2′ ).Every element 𝑥 ∈ 𝐺 Can be uniquely decomposed as 𝑥 = 𝑢𝑣, where 𝑢 ∈ 𝐻 𝑎𝑛𝑑 𝑣 ∈ 𝐾. For assume (2) and (3) and suppose we have two decompositions 𝑥 = 𝑢𝑣 = 𝑢1 𝑣1 . Then 𝑢1−1 𝑢 = 𝑣1 𝑣 −1 ,

(2.34)

and the element on either side of (2.34) belongs to H and to K, whence by (3), it must be equal to 1. Thus 𝑢 = 𝑢1 𝑎𝑛𝑑 𝑣 = 𝑣1 , which proves the uniqueness of the decomposition, as required in (2′ ). Conversely, assume (2′ ) and suppose that 𝑤 ∈ 𝐻 ∩ 𝐾. Then 𝑤 = 1𝑤 = 𝑤1 are two decompositions of w with factors in H and K respectively, and we deduce from the uniqueness that 𝑤 = 1. The foregoing conditions make it plain that each element 𝑥 ∈ 𝐺 uniquely determines a pair (u,v) where 𝑢 ∈ 𝐻 𝑎𝑛𝑑 𝑣 ∈ 𝐾 and that each such pair occurs , because (u,v) corresponds to the product 𝑢𝑣 = 𝑥. The correspondence 𝑥𝜃 = (𝑢𝑣)𝜃 = (𝑢, 𝑣) establishes the isomorphism (2.30), because, by (1), (𝑢𝑣)(𝑢′ , 𝑣 ′ )𝜃 = (𝑢𝑢′ , 𝑣𝑣 ′ ). Similarly, we have that 𝐺 ≅ 𝐻1 × 𝐻2 × … × 𝐻𝑟 , where 𝐻𝑖 (𝑖 = 1 ,2 , … , 𝑟) are subgroups of G, if the following conditions are satisfied: (a)any two groups 𝐻𝑖 , 𝐻𝑗 commute elementwise (b)Every element 𝑥 of G can be expressed in the form 𝑥 =, 𝑢1 , 𝑢2 , … , 𝑢𝑟 Where 𝑢𝑖 ∈ 𝐻𝑖 (c)

(2.35)

𝐻𝑖 ∩ 𝐻1 𝐻2 … 𝐻𝑖−1 𝐻𝑖+1 … 𝐻𝑟 = {1}, or alternatively, in place of (2) and (3), (2′ ) The decomposition (2.35) is unique. In particular, if 𝑢1 𝑢2 … 𝑢𝑟 = 1, it follows from (2′ ) that 𝑢1 = 𝑢2 = ⋯ = 𝑢𝑟 = 1, Because 1 = 11 … 1 is the only decomposition of 1.

When a group has been expressed as the direct product of subgroups, we speak of an interior direct product. Example The least positive residues prime to the modulus 15 is 1 , 2 ,4 ,7 , 8 ,11 , 13 ,14. (2.36) They form an Abelian group of order 8 , which, as we shall see, is isomorphic with the direct product of the cyclic groups generated by the elements 2 and 11 respectively. In fact, the residue 2 generates a cyclic group of order 4, namely 𝐶4 : 1, 2,4 ,8 (24 = 16 ≡ 1(𝑚𝑜𝑑15). Similarly, 11 generates a cyclic group of order 2, 𝐶4 : 1, 11 (114 = 121 ≡ 1(𝑚𝑜𝑑15)) Since the whole group is Abelian, we have only to check conditions (2) and (3). Taking all possible products, we obtain 1 , 2 , 4 , 8 , 11 , 22, 44 , 88, Which on reduction mod.15 becomes 1 , 2 , 4 , 8 , 11 , 7, 14 , 13. As this is the complete group, condition (2) holds, and we see at once that 𝐶4 ∩ 𝐶4 = {1}. This shows that the group is isomorphic with 𝐶4 × 𝐶4 . The following simple proposition, which is of some independent interest, will be used in the next section. PROPOSITION 6. Let G be a finite group in which all elements satisfy the equation 𝑥2 = 1

(2.37)

That is each element, other than the unit element, is of order 2. Then G is isomorphic with an Abelian group of type 𝐶2 × 𝐶2 × … 𝐶2 , And the order of G is therefore a power of 2. Proof By the fact that A group of prime order has no proper subgroups and is necessarily cyclic, the proposition is obviously true when G is the (only) group of order 2. Suppose that G is of order greater than 2 and let a and b be distinct elements other than 1. By hypothesis 𝑎2 = 𝑏 2 = 1,

So that 𝑎 = 𝑎−1 , 𝑏 = 𝑏 −1 . Next, we consider the element ab. By (2.37), (𝑎𝑏)2 = 1, whence 𝑎𝑏 = (𝑎𝑏)−1 = 𝑏 −1 𝑎−1 = 𝑏𝑎. This shows that G is Abelian. Let 𝑢1 𝑢2 … 𝑢𝑟 be a set of irredundant generators of G. Since G is Abelian, products of the generators can be collected in such a way that every element can be expressed in the “normal” form 𝑘

𝑘

𝑘

𝑢1 1 𝑢2 2 … 𝑢𝑟 𝑟 .

(2.38)

But by virtue of (2.37) the exponents in (2.28) can be restricted to take only the values of 0 and 1. Then all the products will be distinct; For an equality between two such products would lead to a relation 𝑙

𝑙

𝑙

𝑢11 𝑢22 … 𝑢𝑟𝑟 = 1, Where each 𝑙𝑖 is either zero or unity. This would enable us to express one of the generators in terms of the others, contradicting the hypothesis that the generators were irredundant. Thus, the normal form (2.38) is unique, which amounts to saying that 𝐺 = 𝑔𝑝{𝑢1 } × 𝑔𝑝{𝑢2 } × … × 𝑔𝑝{𝑢𝑟 }, and hence 𝐺 ≅ 𝐶2 × 𝐶2 × … × 𝐶2 (𝑟 𝑓𝑎𝑐𝑡𝑜𝑟𝑠).

SURVEY OF GROUPS UP TO ORDER OF 8 No successful method has yet been discovered for constructing all possible abstract groups of preassigned order, nor do we know in advance how many such groups exists, except in a few simple cases. However, the elementary means which we have so far developed, suffice to give a complete list of groups up to order 8. Since groups of prime order have already been discussed, it remains only to discuss in more detail the cases in which the order, g, is equal to 4 or 6 or 8.

There are two groups of order 4, both Abelian. For if g=4, an element, other than 1, can be of order 4 or 2 (corollary 1, p. 35). (1) If G contains an element a of order 4, this element generates G; In fact, the four elements of G are 1 , 𝑎 , 𝑎2 , 𝑎3 , (𝑎4 = 1) and we have that G= 𝐺4 , the cyclic group of order 4. Then all the elements, other than 1, are of order 2 and we deduce from proposition 6 that 𝐺 = 𝐶2 × 𝐶2 . Thus, G is generated by two elements a and b, and the four elements of G are 1 , 𝑎 , 𝑏, 𝑎𝑏,

(2.39)

Where 𝑎2 = 𝑏 2 = 1, 𝑎𝑏 = 𝑏𝑎

(2.40)

This group is called the four-group (F. Klein’s ‘Vierergruppe’) and is often denoted by V There being no other possibilities we conclude that any group of order 4 is isomorphic either with 𝐶4 or with V(≅ 𝐶2 × 𝐶2 ). In a different notation the multiplication tables of these groups are presented in (3)𝑎𝑛𝑑 (4)

1

a

b

c

1

a

b

c

1

1

a

b

c

1

1

a

b

c

a

a

1

c

b

a

a

b

c

1

b

b

c

1

a

b

b

c

1

a

c

c

b

a

1

c

c

1

a

b

table 3

table 4

There are two groups of order 6, one cyclic and one-Abelian. (1) If G possesses an element of order 6, then 𝐺 = 𝑔𝑝{𝑎} = 𝐶6 (2) Next, suppose there is no element of order 6. The order of every element, other than 1, is therefore 2 or 3 (corollary 1, p.35). Since the order of G is not a power of 2, not all its elements can satisfy (2.37). Hence there exists at least one element, a, of order 3, so that 1 , 𝑎 , 𝑎2 (2.41) are three distinct elements of G and 𝑎3 = 1,

(2.42)

If c is a further element of G, the six elements 1 , 𝑎 , 𝑎2 , 𝑐 , 𝑐𝑎, 𝑐𝑎2

(2.43)

are distinct, as we pointed out on P.40, in connection with the elements listed in (2.28) If the elements (2.43) are to form a group of order 6, the axiom of closure must be fulfilled. In particular, 𝐶 2 must be one of these elements. We cannot have an element of the form 𝐶 2 = 𝑐𝑎𝑖 (𝑖 = 0, 1 , 2), as this would imply that c belongs to the set (2.41). There remain only the following three possibilities: (𝛼)𝑐 2 = 1,

(𝛽)𝑐 2 = 𝑎

(𝛾)𝑐 2 = 𝑎2 .

(2.44)

Under the assumption (𝛽) 𝑎𝑛𝑑 (𝛾), The element c cannot be of order 2 and must therefore be of order 3. But on multiplying (𝛽) 𝑎𝑛𝑑 (𝛾) on the left by c we should obtain 1 = 𝑐𝑎 𝑎𝑛𝑑 1 = 𝑐𝑎2 respectively, neither of which can be true. Hence, we conclude that (𝛼) must hold, that is 𝑐2 = 1

(2.45)

Next, consider ac. It must be amongst the elements (2.43). As it cannot be equal to c or to a power of a, we are left with the alternatives 𝑎𝑐 = 𝑐𝑎,

or 𝑎𝑐 = 𝑐𝑎2 .

(2.46)

The first of these renders the group Abelian. Let us find the order of ac in this case. Thus (𝑎𝑐)2 = 𝑎2 𝑐 2 = 𝑎2 ≠ 1, (𝑎𝑐)3 = 𝑎3 𝑐 3 = 𝑐 3 = 𝑐 ≠ 1, and the element ac would have to be of order 6, contrary to our initial assumption. Hence the second equation (2.46) must hold, that is 𝑎𝑐 = 𝑐𝑎2

or equivalently, (𝑎𝑐)2 = 1,

See (2.26) and (2.26)′ . To summarize we can state that if there is a group G of order 6 other 𝐶6 , then 𝐺 = 𝑔𝑝{𝑎, 𝑐} Subject to the relations 𝑎3 = 𝑐 2 = (𝑎𝑐)2 = 1. This does not prove that such a group exists. But we happen to know that this is the case; indeed, its multiplication table is exhibited on table 2. Thus, there are precisely two groups of order 6. There are five groups of order 8, of which three are Abelian and two are non-Abelian. Three Abelian groups of order 8 are written down, namely

1) 𝐶8 = 𝑔𝑝{𝑎}, where 𝑎8 = 1(𝑡𝑎𝑏𝑙𝑒 5) 2) 𝐶4 × 𝐶2 = 𝑔𝑝{𝑎} × 𝑔𝑝{𝑏}, 𝑤ℎ𝑒𝑟𝑒 𝑎4 = 𝑏 2 = 1, 𝑎𝑏 = 𝑏𝑎 (𝑡𝑎𝑏𝑙𝑒 6) 3) 𝐶2 × 𝐶2 × 𝐶2 = 𝑔𝑝{𝑎} × 𝑔𝑝{𝑏} × 𝑔𝑝{𝑐}, 𝑤ℎ𝑒𝑟𝑒 𝑎2 = 𝑏 2 = 𝑐 2 = 1, 𝑎𝑏 = 𝑏𝑎, 𝑏𝑐 = 𝑐𝑏, 𝑐𝑎 = 𝑎𝑐(𝑡𝑎𝑏𝑙𝑒 7) From the general theory, which we shall develop later, it will follow that these are all positive Abelian groups of order 8, but we shall here derive this result from first principles. If the group contains an element of order 8, it must be the group 𝐶8 , and if all elements, other than 1, are of order 2, then it is isomorphic with the group (3). Henceforth we shall assume that each element, other than 1, is either of order 4 or of order 2 and that there is at least one element of order 4, say a, where 𝑎4 = 1, 𝑎2 ≠ 1.

(2.47)

If b is an element not contained in 𝑔𝑝{𝑎}, then the eight elements 1, 𝑎 , 𝑎2 , 𝑎3 , 𝑏, 𝑎𝑏 , 𝑎2 𝑏, 𝑎3 𝑏

(2.48)

Are distinct and therefore constitute the whole group, if such a group exists. Now 𝑏 2 must be one of these elements and in fact must be one of the first four, since b is not a power of a. The equation 𝑏 2 = 𝑎 or 𝑏 2 = 𝑎3 have to be ruled out, since they would imply that the order of b is eight. Thus, there remain the possibilities (𝛼)𝑏 2 = 1 or (𝛽)𝑏 2 = 𝑎2

(2.49)

(𝛼): Assume that 𝑏 2 = 1. The product ba must be one of the last three elements in (2.48). (𝛼, 𝑖): If ba=ab, the group is Abelian and is the group listed under (2). (𝛼, 𝑖𝑖): If ba=𝑎2 𝑏, we could deduce that 𝑏 −1 𝑎2 𝑏 = 𝑎, whence (𝑏 −1 𝑎2 𝑏)2 = 𝑏 −1 𝑎4 𝑏 = 𝑏 −1 𝑎2 = 𝑎, which is impossible. Hence, we must conclude that (𝛼, 𝑖𝑖𝑖): ab=𝑎3 𝑏 or equivalently, (𝑎𝑏)2 = 1. The group defined by the relations 𝑎2 = 𝑏 2 = (𝑎𝑏)2 = 1

(2.50)

Does in fact exist. It is denoted by 𝐷4 and is called the dihedral group of order 8 (𝑡𝑎𝑏𝑙𝑒 8) It belongs to a class of groups which will be discussed later, when the associative law will be confirmed.

(𝛽). assume that 𝑏 2 = 𝑎2 . In this case both a and b are of order 4. Again, ba must be one of the last three elements of (2.48), which we shall consider in turn: (𝛽, 𝑖): If ba=ab, the group is Abelian. The element 𝑐 = 𝑎𝑏 −1 is of order 2 because (𝑎𝑏 −1 )2 = 𝑎2 𝑏 −1 = 1, and since 𝑏 = 𝑐 −1 𝑎, the generator b may be replaced by c. The eight elements may therefore be written as in (2.48), but with c in place of b. Once again, we arrive at the group (2). (𝛽, 𝑖𝑖):The relation ba= 𝑎2 𝑏 is impossible as it would imply that ba=𝑏 2 𝑏, 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑎 = 𝑏 2 , which is inadmissible. (𝛽, 𝑖𝑖𝑖): The only remaining alternative, namely ba=𝑎3 𝑏, is feasible and, as we shall see, leads to a group defined by the relations 𝑎4 = 1, 𝑎2 = 𝑏 2 , ba=𝑎3 𝑏.

(2.51)

In order to demonstrate that such a group does not in fact exist, we construct a faithful matrix representation. Let 0 ], 𝐴 = [√−1 0 −√−1

𝐵=[

0 −1 ] 1 0

The reader will have no difficulty in verifying that these matrices satisfy the relations (2.51) with the appropriate change of notation and the eight matrices 𝐼, 𝐴 , 𝐴2 , 𝐴3 , 𝐵, 𝐴𝐵 , 𝐴2 𝐵 , 𝐴3 𝐵 are distinct and therefore constitute a multiplicative matrix group, which is isomorphic with the group envisaged under (𝛽, 𝑖𝑖𝑖). This group is known as the quaternion group (table 9). We recall that a quaternion is a hypercomplex number. 𝑎0 1 + 𝑎1 𝑖 + 𝑎2 𝑗 + 𝑎3 𝑘 Where the coefficients 𝑎0 , 𝑎1 , 𝑎2 , 𝑎3 are real numbers and the symbols 1 , 𝑖 , 𝑗 , 𝑘 satisfy the equations 𝑖 2 = 𝑗 2 = −1, 𝑖𝑗 = −𝑗𝑖 = 𝑘

or equivalently, 𝑖 4 = 1, 𝑖 2 = 𝑗 2 , 𝑗𝑖 = 𝑖 3 𝑗,

Which agrees with (2.51) apart from the notation. To summarize our discussion of groups of order 8 we append the complete multiplication tables of the five possible abstract groups of this order:

𝐶8 = 𝑔𝑝{𝑎}, 𝑎8 = 1 1

a

𝑎2

𝑎3

𝑎4

𝑎5

𝑎6

𝑎7

1

1

a

𝑎2

𝑎3

𝑎4

𝑎5

𝑎6

𝑎7

a

a

𝑎2

𝑎3

𝑎4

𝑎5

𝑎6

𝑎7

1

𝑎2

𝑎2

𝑎3

𝑎4

𝑎5

𝑎6

𝑎7

1

𝑎3

𝑎3

𝑎4

𝑎5

𝑎6

𝑎7

1

𝑎4

𝑎4

𝑎5

𝑎6

𝑎7

𝑎5

𝑎5

𝑎6

𝑎7

1

a

𝑎2

𝑎3

𝑎4

𝑎6

𝑎6

𝑎7

1

a

𝑎2

𝑎3

𝑎4

𝑎5

𝑎7

𝑎7

1

a

𝑎2

𝑎3

𝑎4

𝑎5

𝑎6

1

a

a

a

𝑎2

𝑎2

𝑎3

Table 5

𝐶4 × 𝐶2 = 𝑔𝑝{𝑎} × 𝑔𝑝{𝑏}, 𝑎4 = 𝑏 2 = 1.

𝐶8 = 𝑔𝑝{𝑎}, 𝑎8 = 1 1

a

𝑎2

𝑎3

b

ab

𝑎2 𝑏

𝑎3 𝑏

1

1

a

𝑎2

𝑎3

b

ab

𝑎2 𝑏

𝑎3 𝑏

a

a

𝑎2

𝑎3

1

ab

𝑎2 𝑏

𝑎3 𝑏

b

𝑎2

𝑎2

𝑎3

1

a

𝑎2 𝑏

𝑎3 𝑏

b

ab

𝑎3

𝑎3

1

a

𝑎2

𝑎3 𝑏

b

ab

𝑎2 𝑏

b

b

ab

𝑎2 𝑏

1

a

𝑎2

𝑎3

ab

ab

𝑎2 𝑏

a

𝑎2

𝑎2 𝑏 𝑎2 𝑏

𝑎3 𝑏

𝑎3 𝑏 𝑎3 𝑏

b

Table 6

𝑎3 𝑏

𝑎3 𝑏 b ab

𝑎2

𝑎3

ab 𝑎2 𝑏

𝑎3

1

b

𝑎3 1 a

1 a 𝑎2

𝐶2 × 𝐶2 × 𝐶2 = 𝑔𝑝{𝑎} × 𝑔𝑝{𝑏} × 𝑔𝑝{𝑐}, 𝑤ℎ𝑒𝑟𝑒 𝑎2 = 𝑏 2 = 𝑐 2 = 1

1

a

b

c

ab

ac

bc

abc

1

1

a

b

c

ab

ac

bc

abc

a

a

1

ab

ac

b

c

abc

bc

b

b

ab

1

bc

a

abc

c

ac

c

c

ac

bc

1

abc

a

b

ab

ab ab

b

a

abc

1

bc

ac

c

ac ac

c

abc

a

bc

1

ab

b

𝑏𝑐 bc

abc

c

b

ac

ab

1

a

c

b

a

1

abc abc bc ac 𝑎𝑏 Table 7

Dihedral group: 𝑎2 = 𝑏 2 = (𝑎𝑏)2 = 1

1

a

𝑎2

𝑎3

b

ab

𝑎2 𝑏

𝑎3 𝑏

1

1

a

𝑎2

𝑎3

b

ab

𝑎2 𝑏

𝑎3 𝑏

a

a

𝑎2

𝑎3

1

ab

𝑎2 𝑏

𝑎3 𝑏

b

𝑎2

𝑎2

𝑎3

1

a

𝑎2 𝑏

𝑎3 𝑏

b

𝑎3

𝑎3

1

a

𝑎2

𝑎3 𝑏

b

b

b

𝑎3 𝑏

ab

ab

b

𝑎3 𝑏

ab

b

𝑎2 𝑏 𝑎2 𝑏

𝑎2 𝑏 ab

𝑎3 𝑏 𝑎3 𝑏 𝑎2 𝑏 ab Table 8

ab

ab 𝑎2 𝑏

1

𝑎3

𝑎2

𝑎2 𝑏

a

1

𝑎3

𝑎2

𝑎3 𝑏

𝑎2

a

1

𝑎3

b

𝑎3

𝑎2

a

1

a

Quaternion group: 𝑎4 = 1, 𝑎2 = 𝑏 2 , ba=𝑎3 𝑏

1

a

𝑎2

𝑎3

b

ab

𝑎2 𝑏

𝑎3 𝑏

1

1

a

𝑎2

𝑎3

b

ab

𝑎2 𝑏

𝑎3 𝑏

a

a

𝑎2

𝑎3

1

ab

𝑎2 𝑏

𝑎3 𝑏

b

𝑎2

𝑎2

𝑎3

1

a

𝑎2 𝑏

𝑎3 𝑏

b

ab

𝑎3

𝑎3

1

a

𝑎2

𝑎3 𝑏

b

ab

𝑎2 𝑏

b

b

𝑎3 𝑏 𝑎2 𝑏

ab

𝑎2

a

1

𝑎3

ab

ab

b

𝑎3 𝑏

𝑎2 𝑏

𝑎3

𝑎2

a

1

ab

b

𝑎3 𝑏

1

𝑎3

𝑎2

a

b

a

1

𝑎3

𝑎2

𝑎2 𝑏 𝑎2 𝑏

𝑎3 𝑏 𝑎3 𝑏 𝑎2 𝑏 ab Table 9

THE PRODUCT THEOREM At the beginning of this chapter, we defined the product of two subsets. We shall now examine the case in which both these subsets are subgroups of a group. It will appear that the product of two subgroups is not always a subgroup, but that explicit information can be obtained in the finite case about the number of elements in the product. THEOREM 5 (Product theorem). (i) Let A and B be subgroups. Then the subsets AB is a group if and only if AB=BA

(2.52)

(ii) In the case of finite subgroups, let |𝐴| = 𝑎, |𝐵| = 𝑏, |𝐴 ∩ 𝐵| = 𝑑.Then, irrespective of whether (2.52) holds; |𝐴𝐵| = |𝐵𝐴| = 𝑎𝑏/𝑑. Proof. (i) Since A and B are groups, we have that 𝐴2 = 𝐴 𝑎𝑛𝑑 𝐵 2 = 𝐵. First, suppose that (2.52) is satisfied and put H=AB. Then 𝐻 2 = 𝐴𝐵𝐴𝐵 = 𝐴2 𝐵 2 = 𝐴𝐵 = 𝐻, Which proves the closure of H. Obviously 1∈ 𝐻, since 1∈ 𝐴 and 1∈ 𝐵. Finally if a and b are arbitrary elements of A and B respectively, then 𝑏 −1 𝑎−1 ∈ 𝐵𝐴; and hence, by (2.52) 𝑏 −1 𝑎−1 ∈ 𝐴𝐵 = 𝐻;that is (𝑎𝑏)−1 ∈ 𝐻,which completes the proof that H is a group.

Conversely, suppose that H=AB is a group. Hence if a and b are arbitrary elements of A and B respectively, 𝑎𝑏 ∈ 𝐻, 𝑎−1 𝑏 −1 ∈ 𝐻, and also (𝑎−1 𝑏 −1 )−1 ∈ 𝐻, that is 𝑏𝑎 ∈ 𝐻. This means 𝐵𝐴 ⊂ 𝐴𝐵. In particular, 𝑎−1 𝑏 −1 = 𝑎1 𝑏1, where 𝑎1 and 𝑏1 are certain elements of A and B respectively. Hence (𝑏 −1 𝑎−1 )−1 = 𝑎𝑏 = 𝑏1−1 𝑎1−1 , that is 𝐴𝐵 ⊏ 𝐵𝐴 Hence, we conclude that AB=BA. (ii) Let D=𝐴 ∩ 𝐵. Since D is a subgroup of B, we may decompose B into cosets with respect to D, say 𝐵 = 𝐷𝑡1 ∪ 𝐷𝑡2 ∪ … ∪ 𝐷𝑡𝑛,

(2.53)

Where 𝐷𝑡𝑖 ≠ 𝐷𝑡𝑖 𝑖𝑓 𝑖 ≠ 𝑗

(2.54)

And n=b/d

(2.55)

multiplying (2.53) on the left by A and observing that AD=A because D⊂ 𝐴, we obtain that AB= 𝐴𝑡1 ∪ 𝐴𝑡2 ∪ … ∪ 𝐴𝑡𝑛 .

(2.56)

We claim that no two of the cosets on the right of (2.56) have an element in common; for if not, we should have an equation of the form 𝑢1 𝑡𝑖 = 𝑢2 𝑡𝑗 Where 𝑢1 , 𝑢1 ∈ 𝐴 𝑎𝑛𝑑 𝑖 ≠ 𝑗. Thus 𝑡𝑖 𝑡𝑗 −1 = 𝑢1 −1 𝑢2. Now the element on the left belongs to B by (2.53), and the element on the right lies in A. Hence either side denotes an element of D. But 𝑡𝑖 𝑡𝑗 −1 ∈ 𝐷 implies that 𝐷𝑡𝑖 = 𝐷𝑡𝑖 , which contradicts (2.54). Thus, the cosets in (2.56) are disjoint, and since each consists of a elements, we have that |𝐴𝐵| = 𝑎𝑛 = 𝑎𝑏/𝑑. The argument is clearly symmetric in A and B, so that also |𝐵𝐴| = 𝑎𝑏/𝑑.

DOUBLE COSET We saw earlier that the decomposition of a group into cosets relative to a subgroup maybe regarded as an instance of dividing a set into equivalent classes with respect to a suitably defined equivalence relation. Following Frobenius we shall now discuss a different equivalence relation, which involves two subgroups. Let A and B be subgroups of G, which need not be distinct, and term two elements 𝑥, 𝑦 ∈ 𝐺 equivalent, written x∼ 𝑦, if there exist elements 𝑢 ∈ 𝐴 and 𝑣 ∈ 𝐵 such that 𝑦 = 𝑢𝑥𝑣

(2.57)

It is easy to check that this is an equivalence relation on G. For (i) x∼ 𝑦, since we may take u=1 , 𝑣 = 1, (ii) if 𝑥 ∼ 𝑦,then 𝑦 ∼ 𝑥, because (2.57) implies that 𝑥 = 𝑢−1 𝑦𝑣 −1 , (iii) if 𝑥 ∼ 𝑦, 𝑦 ∼ 𝑧,that is 𝑦 = 𝑢𝑥𝑣, 𝑧 = 𝑢′ 𝑦𝑣 ′ , where 𝑢′ ∈ 𝐴, 𝑣 ′ ∈ 𝐵, then 𝑧 = (𝑢′ 𝑢)𝑥(𝑣𝑣 ′ ), so that 𝑥 ∼ 𝑧. Thus, the set G may be divided into the equivalence classes which stem from this definition of equivalence. The equivalence class containing 𝑥 is the complex A 𝑥𝐵, which is called a double coset of G with respect to A and B. We choose a representative from each class and obtain the decomposition 𝐺 = ⋃ 𝐴𝑡𝑖 𝑖∈𝐼

(2.58)

Where 𝛪 is a, possibly infinite, index set in one-to one correspondence with the set of double cosets. It is clear that (2.58) is a generalization of the left or right coset decomposition, as is seen by taking A or B to be the trivial group {1}. In contrast to single coset decompositions, the double cosets in (2.58) are not, in general, of the same cardinal. We will pursue the matter further when G is a finite group. Let |𝐺| = 𝑔, |𝐴| = 𝑎, |𝐵| = 𝑏. First, we observe that the subsets 𝐴𝑡𝑖 𝐵 and (𝑡𝑖−1 𝐴𝑡𝑖 )𝐵 have the same cardinal, because their elements may be put into one-to-one correspondence by matching 𝑢𝑡𝑖 𝑣 with 𝑡𝑖−1 (𝑢𝑡𝑖 𝑣). Thus |𝐴𝑡𝑖 𝐵 | = |(𝑡𝑖−1 𝐴𝑡𝑖 )𝐵 |.

Now 𝑡𝑖−1 𝐴𝑡𝑖 is a subgroup and |𝑡𝑖−1 𝐴𝑡𝑖 | = 𝐴 = 𝑎 On applying theorem 5 to the subgroups 𝑡𝑖−1 𝐴𝑡𝑖 and B we find that |𝐴𝑡𝑖 𝐵| = 𝑎𝑏/𝑑𝑖 , Where 𝑑𝑖 = |𝑡𝑖−1 𝐴𝑡𝑖 ∩ 𝐵|. Collecting these results, we obtain the following theorem. THEOREM 6 (Frobenius). Let G be a finite group of order g and let A and B be subgroups of order a and b respectively. Then there exist elements 𝑡1 , 𝑡2 , … , 𝑡𝑟 such that G is the disjoint union of double cosets, namely G= 𝐴𝑡1 𝐵 ∪ 𝐴𝑡2 𝐵 ∪ … 𝐴𝑡𝑟 𝐵. The number of elements in 𝐴𝑡𝑖 𝐵 is ab/𝑑𝑖 , Where 𝑑𝑖 = |𝑡𝑖−1 𝐴𝑡𝑖 ∩ 𝐵|, And consequently 𝑟

𝑔 = 𝑎𝑏 ∑ 𝑑𝑖−1 𝑖=1

(2.59)