Group structure
A setGof elements is a group if there exists an internal composition law∗defined for all elements and satisfying the following properties:
2 the law admits an identity elemente a∗e=e∗a=a, ∀a∈G,
LetF andGbe two groups A composition law onF×Gis defined by (f 1 , g 1 )(f 2 , g 2 ) = (f 1 f 2 , g 1 g 2 ), (f i ∈F, g i ∈G, i= 1,2); the groupF×Gis called the direct product of the groupsF andG.
Examples 1 Cyclic groupC n of ordernthe elements of which are
(b, b 2 , b 3 , , b n =e) and wherebrepresents, for example, a rotation of 2π/naround an axis.
2 Dihedral groupD n of order 2ngenerated by two elementsaandbsuch that a 2 =b n = (ab) 2 =e.
One has in particularb − h a=ab h (h= 1ã ã ãn); indeed, since
(ab) −1 =b −1 a −1 =b −1 a=ab, one has b −1 (b −1 a)b=b −2 ab=b −1 ab 2 =ab 3 and thus b −2 a=ab 2 ; by proceeding similarly by recurrence, one establishes the above formula.
Finite groups of order n ≤ 8
The finite groups of ordern≤8,except the quaternion group which will be treated separately, are the following.
1 n= 1, there exists only one group (1 =e).
2 n= 2, only one group exists, the cyclic groupC 2 consisting of the elements (b, b 2 =e).
Examples (a) the group constituted by the elements (−1,1);
(b) the group having the elements (b : rotation of ±π around an axis, b 2 =e).
3 n = 3, only one group is possible: the cyclic group C 3 of elements (b, b 2 , b 3 =e) whereb, b 2 are elements of order 3.
(a) the cyclic groupC 4 constituted by the elements (b, b 2 , b 3 , b 4 =e) where the elementb 2 is of order 2, and where (b, b 3 ) are elements of order 4; (b) the Klein four-group defined by
I 2 =J 2 = (IJ) 2 = 1 or, equivalently I 2 = J 2 = K 2 = IJ K = 1 with K = IJ and the multiplication table
The Klein four-group is isomorphic to the direct product of two cyclic groupsC 2 ,
Example The group constituted by the elements (I: rotation of π around the axis Ox, J: rotation of π around the axis Oy, K = IJ: rotation ofπaround the axisOz).
5 n= 5, there exists only one group, the cyclic groupC 5 having the elements (b, b 2 , b 3 , b 4 , b 5 =e).
The dihedral group D3 is characterized by the relations a² = b³ = (ab)² = e, establishing its structure and operations The multiplication table illustrates the interactions between the elements, revealing that it is the first noncommutative group in the series The relations and operations within D3 highlight its unique properties and significance in group theory.
Example The symmetry group of the equilateral triangle (see Fig 1.1).
7 n= 7, there exists only one group, the cyclic groupC 7 of elements (b, b 2 , b 3 , b 4 , b 5 , b 6 , b 7 =e).
8 n= 8, there exist five groups, among them the quaternion group which will be treated separately.
(a) The cyclic groupC 8 of elements (b, b 2 , b 3 , b 4 , b 5 , b 6 , b 7 , b 8 =e). ab(M)
The symmetry group of the equilateral triangle includes a rotation of point M by 2π/3 around the Oz axis, demonstrating a symmetry of M in the ABC plane relative to the mediatrix CD, with M representing an arbitrary point within the triangle.
(b) The groupS 2×2×2 , direct product of the Klein four-group withC 2 ,
(1, I, J, K)×(1,−1) = (±1,±I,±J,±K) with 1 = (1,1), −1 = (1,−1), ±I = (I,±1), ±J = (J,±1), ±K (K,±1) ; the multiplication table is given by
(c) the groupS 4×2 , direct product of C 4 with C 2 and constituted by the elements
Quaternion group
The group D4, characterized by its noncommutative properties, is defined by the relations a² = b⁴ = (ab)² = e Its multiplication table illustrates the interactions between the elements b, a, and their combinations, resulting in various products that ultimately return to the identity element e Specifically, the structure reveals how the elements combine in a noncommutative manner, emphasizing the unique relationships within the group The notation also indicates the behavior of the elements under multiplication, showcasing the complexity and richness of D4's algebraic framework.
Example The symmetry group of the square (see Fig 1.2).
Figure 1.2: Symmetry group of the square;b is a rotation ofπ/2 around the axis
Oz, a a symmetry with respect to the axis Ox in the plane ABCD and M an arbitrary point of the square.
The quaternion group, represented as Q, was introduced by William Rowan Hamilton in 1843 and consists of eight elements: ±1, ±i, ±j, and ±k These elements adhere to specific relations, including i² = j² = k² = ijk = -1, and the multiplication rules ij = -ji = k, jk = -kj = i, and ki = -ik = j.
The first column element serves as the initial multiplier, with 1 acting as the identity element Notably, the element -1 has an order of 2, meaning its square equals 1, while the elements ±i, ±j, and ±k possess an order of 4 The subgroups of Q are also significant in this context.
Quaternion algebra H
Definitions
Consider the vector space of numbers called quaternions a, b, constituted by four real numbers a=a 0 +a 1 i+a 2 j+a 3 k
= (a 0 , a) = (a 0 , a) where S(a) = a 0 is the scalar part and V(a) = a = a the vectorial part This
1.4 Quaternion algebraH 9 vector space is transformed into the associative algebra of quaternions (denoted
H) via the multiplication ab= (a 0 b 0 −a 1 b 1 −a 2 b 2 −a 3 b 3 ) + (a 0 b 1 +a 1 b 0 +a 2 b 3 −a 3 b 2 )i + (a 0 b 2 +a 2 b 0 +a 3 b 1 −a 1 b 3 )j + (a 0 b 3 +a 3 b 0 +a 1 b 2 −a 2 b 1 )k and in a more condensed form ab= (a 0 b 0 − a ã b , a 0 b+b 0 a+a ì b) where a ã b= (a 1 b 1 +a 2 b 2 +a 3 b 3 ) anda ì b= (a 2 b 3 −a 3 b 2 )i+ (a 3 b 1 −a 1 b 3 )j+ (a 1 b 2 −a 2 b 1 )kare respectively the usual scalar and vector products Historically, these two products were obtained by W J Gibbs [17] by takinga 0 =b 0 = 0 and by separating the quaternion product in two parts.
The quaternion algebra constitutes a noncommutative field (without divisors of zero) containingRandCas particular cases Leta=a 0 +a 1 i+a 2 j+a 3 kbe a quaternion, the conjugate ofa, the square of its norm and its norm are respectively a c =a 0 −a 1 i−a 2 j−a 3 k,
|ab| 2 =|a| 2 |b| 2 , the last relation deriving from (ab) c ab=b c a c ab= (aa c )(bb c ) ; furthermore,
S(ab) =S(ba) thusS[a(bc)] =S[(bc)a] =S[b(ca)] =S[(ca)b] and therefore
To divide a quaternionaby the quaternionb( = 0), one simply has to resolve the equation xb=a or by=a with the respective solutions x=ab −1 =a b c
Examples Consider the quaternionsa= 2 + 4i−3j+kandb= 5−2i+j−3k;
1 the vectorial partsa,band the conjugatesa c , b c are a= 4i−3j+k, b=−2i+j−3k, a c = 2−4i+ 3j−k, b c = 5 + 2i−j+ 3k;
2 the norms are given by
4 one can realize the following operations a+b = 7 + 2i−2j−2k, a−b = −3 + 6i−4j+ 4k, ab = 24 + 24i−3j−3k, |ab|=√
Polar form
Any nonzero quaternion can be written a=a 0 +a 1 i+a 2 j+a 3 k
=r(cosθ+usinθ), 0≤θ≤2π withr=|a| a 2 0 +a 2 1 +a 2 2 +a 2 3 being the norm ofaand cosθ= a 0 r , sinθ=± a 2 1 +a 2 2 +a 2 3 r , cotθ=±a 0
|a|, tanθ=±|a| a 0 ; the unit vectoru(uu c = 1) is given by u=±(a 1 i+a 2 j+a 3 k) a 2 1 +a 2 2 +a 2 3 witha 2 1 +a 2 2 +a 2 3 = 0 Sinceu 2 =−1, one has via the De Moivre theorem a n =r n (cosnθ+usinnθ).
Example Consider the quaterniona; let us determine its polar form a = 3 +i+j+k,
Square root and n th root
The square root of a quaternion a=a 0 +a 1 i+a 2 j+a 3 k can be obtained alge- braically as follows The equationb 2 =awithb=b 0 +b 1 i+b 2 j+b 3 kleads to the following equations b 2 0 −b 2 1 −b 2 2 −b 2 3 =a 0 (1.1)
2b 0 b 1 =a 1 , sgn(b 0 b 1 ) = sgn(a 1 ), (1.2)2b 0 b 2 =a 2 , sgn(b 0 b 2 ) = sgn(a 2 ), (1.3)2b 0 b 3 =a 3 , sgn(b 0 b 3 ) = sgn(a 3 ) (1.4)
Writingt=b 2 0 the above equation (1.1) leads to t−a 2 1 +a 2 2 +a 2 3
Equation (1.6) then becomes witht=b 2 1 and using (1.5) a 2 1
1.4 Quaternion algebraH 13 henceb 1 (one proceeds similarly forb 2 , b 3 ); finally, one obtains b 0 = ε
Example Consider the quaterniona= 1 +i+j+k ; find its square rootb b 0 = ε
Then th root of a quaterniona=r(cosϕ+usinϕ), whereϕcan always be chosen within the interval [0, π] with an appropriate choice ofu, is obtained as follows [9].
1 Supposing sinϕ= 0, the equationb n =awithb=R(cosθ+esinθ),θ∈[0, π], leads to
R n =r, cosnθ= cosϕ, sinnθ= sinϕ, e=u and thus to
finally, one has b=r n 1 cos(ϕ+ 2kπ) n +usin(ϕ+ 2kπ) n
2 When sinϕ= 0, the vectore in b is arbitrary Ifa >0, one hasϕ= 0 and thusθ= 2 πm n (m= 0,1, , n−1) Forn= 2, one obtainsθ= 0, πand thus the real roots ±√ a Withn >2, certain values ofθ ( = 0 or π) give nonreal roots, the vectorebeing arbitrary Witha