Appendix A Symmetry Groups

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Appendix A
Symmetry Groups
A mathematical group G is a set of objects (the group's elements) with a binary op-
eration denoted by "+" or by "*" defined on the elements that satisfies the following
requirements.
1. Closure: for any a, b G G, the sum (a + b) is an element of G.
2. Associativity: any a,b,c G G satisfies (a + b) + c = a + (b + c).
3. Identity: there exists e G G such that for all a G (7 (a + e) = (e + a) = a.
4. Inverses: for each a G G, there exists a unique element a~l G G such that a-fa"1 —
a"1 + a = e.
5. If the group operation is commutative, i.e., if a + b = b + a for any a,b £ G, the
group is called Abelian.
Question: What's purple and commutes?
Answer: An Abelian grape.
Examples of groups:
1. The set of all the integers with integer addition. The identity element is the integer
0. This is an infinite group.
2. The (finite) set of the integers 0, 1, 2,... ,m — 1 with modulo-m addition.
3. The integers 1, 2,. .. ,g — 1 for a prime q with modulo-g multiplication.
4. The set of all rotations in two dimensions under the operation: The sum of the two
rotations by a and (3 degrees is a rotation by a + (3 degrees.
The set (0,1, 2,3) with modulo-4 addition is a group denoted by G(4). It obeys the
448 Appendix A Symmetry Groups
addition table
+ 0 1 2 3
0 0 1 2 3
1 1 2 3 0
2 2 3 0 1
3 3 0 1 2
The order of a group (its cardinality) is the number of elements. It is denoted by
ord(G). The order of G(4) is 4.
A subgroup is a subset of the elements of a group that is closed under the group's
operation. A theorem by Lagrange states that if S is a subgroup of G, then ord(5)
divides ord(G). For example, if S is the subgroup (0,1) of G(4), then ord(5) = 2
divides ord(G(4)) = 4 and G(4) can be partitioned into the cosets S and S + 2.
This appendix deals with symmetry groups. The elements of such a group are sym-
metry operations (or transformations) on an object; they are not numbers. Figure A.la
shows four symmetric objects: a rhombus, a rectangle, a square, and a pentagon. The
term symmetric means an object that retains its shape and location under certain trans-
formations. A square, for example, is highly symmetric, because it preserves its shape
and position when rotated by a multiple of 90° or when reflected about the four axes
shown by dashed lines in Figure A.lb. A rectangle is less symmetric because a rotation
of 90° changes its shape from horizontal to vertical or vice versa.
b a b a b
c d c
(a)
(b) c)
Figure A.I: Symmetries of Rhombus, Square, and Pentagon.
For simple geometric objects, it is possible to express rotations and reflections by
listing the new position of each vertex of the object. When the square is rotated 90°
clockwise, for example, vertex a moves to b, b moves to c, and so on, which can be
expressed as the permutation
ab c dN
b c d a
Symmetry Groups 449
Reflections of the square about a vertical axis and about the main diagonal are expressed
by
a b c d \ /abed'
bade/' Vadcb
The connection between symmetry transformations and groups becomes clear when we
consider combinations of transformations.
The rectangle is transformed to itself after (1) a 0° rotation, (2) a reflection about
a central horizontal axis, (3) a reflection about a central vertical axis, and (4) a 180°
rotation. Examining the diagram, the following properties become clear:
1. Transformation 1 followed by transformation i (or i followed by 1) is equivalent to
just transformation i for any i.
2. Any of the four transformations followed by itself returns the rectangle to its original
shape, so it is identical to transformation 1.
3. Transformation 3 followed by 2 is equivalent to 4.
An analysis of all the combinations of two transformations of the rectangle yields
Table A.2a. The table can be considered the definition of a symmetry group of four ele-
ments, because it specifies the group operation for the elements. A direct check verifies
that element 1 (the null transformation) is the group's identity, that the operation is
closed, and that it is noncommutative. This symmetry group is denoted by D4 (D for
dihedral, meaning bending the arms up; anhedral means the opposite).
(A dihedral group is a group whose elements correspond to a closed set of rotations
and reflections in the plane. The dihedral group with 2n elements is denoted by either
D or D2n- The group consists of n reflections, n — 1 rotations, and the identity
n
transformation.)
* 0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
* 0 1 2 3 4 5 6 7 1 1 2 3 4 0 6 7 8 9 5
0 0 1 2 3 4 5 6 7 2 2 3 4 0 1 7 8 9 5 6
1 1 2 3 0 6 7 5 4 3 3 4 0 1 2 8 9 5 6 7
2 2 3 0 1 5 4 7 6 4 4 0 1 2 3 9 5 6 7 8
* 1 2 3 4 3 3 0 1 2 7 6 4 5 5 5 9 8 7 6 0 4 3 2 1
1 1 2 3 4 4 4 7 5 6 0 2 3 1 6 6 5 9 8 7 1 0 4 3 2
2 2 1 4 3 5 5 6 4 7 2 0 1 3 7 7 6 5 9 8 2 1 0 4 3
3 3 4 1 2 6 6 4 7 5 1 3 0 2 8 8 7 6 5 9 3 2 1 0 4
4 4 3 2 1 7 7 5 6 4 3 1 2 0 9 9 8 7 6 5 4 3 2 1 0
(a) (b) (c)
Table A.2: The D4, D%, and D10 Symmetry Groups.
Similarly, the rhombus has limited symmetry. Its four symmetry transformations
are (1) the null transformation, (2) a reflection about the line bd, (3) a reflection about
450 Appendix A Symmetry Groups
line ac, and (4) a 180° rotation. An analysis of all the combinations of two of these
transformations, however, results in the same symmetry group. Thus, even though the
rhombus and rectangle are different objects and their symmetry transformations are
different, we can say that they have the same symmetries and we call them isometric.
Intuitively, a square is more symmetric than a rectangle or a rhombus. There are
more transformations that leave it unchanged. It is easy to see that these are the four
rotations by multiples of 90° and the four reflections about the vertical, horizontal, and
two diagonal axes. These eight transformations can be written as the permutations
_/abcd\ _ /abcd\ _ /ab c d\ _ / a b c d
~ \ a b c d j' " y b c d a y 5 \cdaby' ""\dabc
__/abcd\ /abcd\ /ab cd\ 7 __ /^ a ^ c d
ybadcy' ydcbay yadcby ycbad
which can immediately be used to construct the symmetry dihedral group Dg listed in
Table A.2b.
Finally, the pentagon is used to create the larger symmetry group Dio, because it
has 10 symmetry transformations. Figure A.la shows that the pentagon is transformed
to itself by any rotation through a multiple of 60°, while Figure A.lc shows that it can
be symmetrically reflected about five different axes. These ten transformations give rise
to the DIQ symmetry group of Table A.2c (identical to Table 2.14a), and it is this group
that is used by the Verhoeff check digit method of Section 2.11.
The mathematical sciences particularly exhibit order, symmetry, and
limitation; and these are the greatest forms of the beautiful.
—Aristotle, Metaphysica
Appendix B
Galois Fields
This appendix is an introduction to finite fields for those who need to brush up on
this topic. Finite fields are used in cryptography in the Rijndael (AES) algorithm and
in stream ciphers. In the field of error-control codes, they are used extensively. The
Reed-Solomon codes of Section 1.14 operate on the elements of such a field.
B.I Field Definitions and Operations
The mathematical concept of a field is based on that of a group, which has been intro-
duced at the start of Appendix A. A field F is a set with two operations—addition "+"
and multiplication ax"—that satisfies the following conditions.
1. F is an Abelian group under the 4- operation.
2. F is closed under the x operation.
3. The nonzero elements of F form an Abelian group under x.
4. The elements obey the distributive law (a -{- b)xc ~ axc-{- bxc.
Examples of fields are
1. The real numbers under the normal addition and multiplication;
2. The complex numbers; and
3. The rational numbers.
Notice that the integers do not form a field under addition and multiplication
because the multiplicative inverse (reciprocal) of an integer a is I/a, which is generally
a noninteger. Also, a finite set of real numbers is not a field under normal addition and
multiplication because these operations can create a result outside the set. In order for
a finite set of numbers to be a field, its two operations have to be defined carefully so
that they satisfy the closure requirement. Finite fields are intriguing because the finite
452 Appendix B Galois Fields
number of elements implies that the two operations could be performed by computers
exactly (with full precision). This is why much research has been devoted to the use of
finite fields in practical applications.
A Galois field, abbreviated GF, is a finite field. These fields were "discovered,"
studied, and precisely defined by the young French mathematician Evariste Galois, and
today they have many applications in fields as diverse as error-control codes, cryptog-
raphy, random-number generation, VLSI testing, and digital signal processing. Galois
has proved that the size of a finite field must be a power m of a prime number q and
that there is exactly one finite field with any given size qm. This justifies talking about
the finite field with qm elements, and this field is denoted by GF(qm).
If m = 1, the size of the field GF(q) is a prime number q, its elements are the
integers 0, 1,... ,g — 1, and the two operations are integer addition and multiplication
modulo q. The simplest examples are GF(2) and GF(3).
The simple field GF(2) consists of the two elements 0 and 1 and is the smallest
finite field. Its operations are integer addition and multiplication modulo 2, which are
summarized by
+ 0 1 X 0 1
0 0 1 0 0 0
1 1 0 1 0 1
Notice that the addition is actually an XOR and the multiplication is a logical AND.
The next field is GF(3), whose elements are 0,1, and 2. Its operations are integer
addition and multiplication modulo 3, summarized by the truth tables
+ 0 1 2 X 0 1 2
0 0 1 2 0 0 0 0
1 1 2 0 1 0 1 2
2 2 0 1 2 0 2 1
The additive inverse of 1 is 2 because 1 + 2 = 2 + 1 = 0. Similarly, the multiplicative
inverse of 2 is itself because 2x2 = 1.
o Exercise B.I: Write the addition and multiplication tables of GF(5).
He is the only candidate who gave poor answers. He knows absolutely nothing. I was
told that this student has an extraordinary capacity for mathematics. This astonishes
me greatly, for, after his examination, I believed him to have but little intelligence or
that his intelligence is so well hidden that I was unable to uncover it. If he really is
what he appears to be, I doubt very much that he will make a good teacher.
—French physicist Jean Claude Eugene Peclet, one of Galois's examiners in 1829.
o Exercise B.2: Compute the addition and multiplication tables of GF(4) as if 4 were
a prime and show why these tables don't make sense.
B.I Field Definitions and Operations 453
If ra > 1, the elements of GF(qm) are polynomials of degree less than m over
GF(g) [i.e., polynomials whose coefficients are elements of GF(g)], and the operations
are special versions of polynomial addition and polynomial multiplication. Hence, if the
polynomial a ^\xrn~l + • • • + a\x + a$ is an element of GF(gm), then ao, ai,... ,a _i
m m
are elements of Galois field GF(q). The degree of the polynomial is the largest i for
which ai ^ 0.
Adding elements of GF(qm) is easy. If the polynomials a{x) and b(x) are elements
of GF(qm), then the sum c(x) = a(x) + b(x) is a polynomial with coefficients Ci =
(ai + bi) mod g. The sum is a polynomial whose degree is the greater of the degrees
of a(x) and b(x), so it is an element of GF(qm). Also, the rule for addition implies
that this operation is associative and that there is an identity (the polynomial whose
coefficients are all zeros).
o Exercise B,3: In order for GF(qm) to be a field, each element must have an additive
inverse. What is it?
Multiplying elements of GF(qm) is a bit trickier, because the normal multiplication
of two polynomials of degrees m and n results in a polynomial of degree ra + n. Multipli-
cation of polynomials in GF(qm) must therefore be defined (similar to addition) modulo
something. In analogy to addition, which is done modulo a prime integer, multiplication
is performed modulo a prime polynomial. Such a polynomial is called irreducible. Much
as a prime number is not a product of smaller integers, an irreducible polynomial is not
a product of lower-degree polynomials. The irreducible polynomials we are interested in
are irreducible in GF(g), which means that such a polynomial cannot be factored into
a product of lower-degree polynomials in GF(q). (Note, A polynomial irreducible over
GF(q) has no roots in GF(q). The opposite, however, isn't true. A polynomial with no
roots in GF(q) may be reducible over GF(<?).) Section B.2 shows how to multiply two
polynomials modulo a third polynomial.
The polynomial x2 — 1 over the reals can be factored into (x — l)(x + 1), so it
is reducible. Its relative, the polynomial x2 + 1, is irreducible over the real numbers.
This same polynomial, however, is reducible over GF(2) because the polynomial product
(x + l)(x + l), which equals xxx + lxx+xxl + lxl, can also be written x2 + (l + l)x + lxl,
and in GF(2) this equals x2 + 1. Another example is the polynomial (x2 + x + I)2. It
is easy to verify that neither zero nor 1 are roots of this polynomial. It therefore does
not have any roots in GF(2), but it is not irreducible in GF(2) because it is obviously
a product of two lower-degree polynomials.
o Exercise B.4: Show that the polynomial x8 +1 with coefficients in GF(2) is reducible.
No doubt this style and this efficiency were due to his peasant heredity. Perhaps also
to the fact that manual work (whatever the demagogues may say) does not demand
a veritable genius, since it is more difficult to extract a square root than a gorse root.
—Marcel Pagnol, Jean de Florette
454 Appendix B Galois Fields
The simplest example of a Galois field of the form GF(qm) for m > 1 is GF(22) =
GF(4). Its elements are polynomials aix-\-ao over GF(2) (meaning that it has coefficients
that are 0 or 1). If we denote such an element by the two bits a\ao, then the four field
elements are 0 = 00 = Oxz+O, 1 = 01 = Oxx + 1, 2 = 10 = x+0, and 3 = 11 = x + 1.
2 2 2 2
If we now select the polynomial x2-\-x + l, which is irreducible over GF(4), and multiply
modulo this polynomial, then the two field operations become
0 12 3 X 0 1 2 3
0 12 3 0 0 0 0 0
10 3 2 1 0 1 2 3
2 3 0 1 2 0 2 3 1
3 2 10 3 0 3 1 2
The multiplication table shows that 2x2 = 3. In polynomial notation, element 2 is the
polynomial x and element 3 is x +1. This is why the product xxx, which over the reals
is x2, equals x + 1 in GF(4).
Notice that the multiplication table implies that 23 = (2x2)x2 = 3x2 = 1, so we
can consider element 2 the cube root of unity. Over the real numbers, this cube root is
(z\/3 — l)/2, which shows that the names 0, 1, 2, and 3 are arbitrary.
Choosing a different irreducible polynomial of degree m produces a different mul-
tiplication table, but all the tables that can be generated in this way are isomorphic;
they have the same essential structure in terms of the two operations and differ by the
names of the field's elements. However, as we already know, the names are arbitrary.
o Exercise B,5: Explain why GF(6) does not exist.
Another simple example is GF(23) = GF(8). Its elements are polynomials (I2X2 +
a\x + ao with coefficients a^ in GF(2) (i.e., bits). We denote such an element by the
three bits a2aiao> so element 6 = IIO2 is the polynomial x2 -f x. Addition is simple:
the sum of x2 + 1 and x + 1 is x2 + x + 1 + 1 = x2 +x. For multiplication, we select the
irreducible polynomial x3 + x + 1. The results are summarized in Table B.I.
+ 0 1 2 3 4 6 7 X 0 1 2 3 4 5 6 7
0 0 1 2 3 4 5 6 7 0 0 0 0 0 0 0 0 0
1 1 0 3 2 5 4 7 6 1 0 1 2 3 4 5 6 7
C 2 3 0 1 6 7 4 2 0 C 4 6 3 1 7 5
M M
3 3 C 1 0 7 6 5 4 3 0 3 6 5 7 4 1 C
M M
4 4 5 6 7 0 1 2 3 4 0 4 3 7 6 2 Ox 1
5 5 4 7 6 1 0 3 2 0 5 1 4 2 7 3 6
6 6 7 4 5 2 3 0 1 6 0 6 7 1 5 3 2 4
7 7 6 5 4 3 2 1 0 7 0 7 5 2 1 6 4 3
Table B.I: Addition and Multiplication in GF(8).
As an example, the GF(8) multiplication table indicates that 5x3 = 4, or in binary
101 x 011 = 100, or in polynomial notation (x2 + l)(x + 1) = (x3 + x2 + x + 1) =
B.I Field Definitions and Operations 455
x2 mod (x3 + x + 1). The modulo operation results in the remainder of the polynomial
division (x3 + x2 + x + \)/{x3 + x + 1).
o Exercise B,6: Choose some of the elements of the GF(4) and GF(8) multiplication
tables and show how they are computed.
o Exercise B,7: List the additive and multiplicative inverses of the eight elements of
GF(8).
The existence of the additive and multiplicative inverses makes it possible to sub-
tract and divide field elements. To subtract a — 6, just add a to the additive inverse of b
(since the additive inverse of 6 is 6 itself, subtraction in GF(8) is identical to addition).
To divide a/6, multiply a by the multiplicative inverse of b.
The particular definition of multiplication in GF(qm) satisfies the requirements for
a field. The product of two field elements is a polynomial of degree m — 1 or less, so
it is an element of the field. The multiplication is associative and there is an identity
element, namely, the polynomial 1. In order to figure out the inverse of element p(x), we
denote by m(x) the particular irreducible polynomial that we use for the multiplication
and apply the extended Euclidean algorithm. This algorithm (next paragraph) finds
two polynomials a(x) and b(x) such that p(x)a(x) + m(x)b(x) = 1. This implies that
a(x)p(x) mod m(x) = 1 or p~l{x) = a(x) mod m(x).
The extended Euclidean algorithm solves the following problem. Given two integers
ro and ri, find two other integers s and t such that s-ro + t-ri = gcd(ro, r\). This employs
Euclid's algorithm, where in each iteration the current remainder Ti is expressed in the
form Ti = SiTQ+tiTi. The signal for the last iteration is r = gcd(ro, n) = s ro+£ ri =
m m m
s-To +t-T\. This algorithm can be expressed recursively as
s = 1, t = 0,
0 Q
=0, *i = 1,
Sl
repeat
Si = Si-2 ~ Qi-lSi-li ti = ti-2 — Qi-lU-l,
for i = 2,3,....
As an example, we compute the extended Euclidean algorithm for ro = 126 and r\ = 23:
126 = 5-23 + 11, to = 0,
23-2-11 + 1, ti = 1,
11 - 11-1 + 0, t -0-5-1 = -5,
2
t = 1 — 2-(—5) = 11.
3
The Exponential Representation of Galois Fields. We start with the simple
field GF(g) and define the order of a field element. Let ft be an element in GF(q). The
order of j3 is denoted by ord(/3) and is defined as the smallest positive integer m such
that Pm = l.
It can be shown that if t is the order of ftf or some j3 in GF(g), then t divides
456 Appendix B Galois Fields
In those days, my head was full of the romantic prose of E.T. Bell's Men of Math-
ematics, a collection of biographies of the great mathematicians. This is a splendid
book for a young boy to read (unfortunately, there is not much in it to inspire a
girl, with Sonya Kovalevsky allotted only half a chapter), and it has awakened many
people of my generation to the beauties of mathematics. The most memorable chap-
ter is called "Genius and Stupidity" and describes the life and death of the French
mathematician Galois, who was killed in a duel at the age of twenty. ... "All night
long he had spent the fleeting hours feverishly dashing off his scientific last will and
testament, writing against time to glean a few of the great things in his teeming mind
before the death he saw could overtake him. Time after time he broke off to scribble
in the margin CI have not time; I have not time,' and passed on to the next frantically
scrawled outline. What he wrote in those last desperate hours before the dawn will
keep generations of mathematicians busy for hundreds of years. He had found, once
and for all, the true solution of a riddle which had tormented mathematicians for
centuries: under what conditions can an equation be solved?"
—Freeman Dyson, Disturbing the Universe (1979)
An element with order (q — 1) in GF(g) is called a primitive element in GF(q).
Every field GF(q) contains at least one primitive element a. The elements of GF(q)
can be represented as zero followed by the (q — 1) consecutive powers of any primitive
element a:
0, a, a2, a3,..., aq~2, aq~\ aq = a,....
This is the exponential representation of GF(q). Notice that we don't have to know the
value of any particular root a. All we need is this particular sequence of powers of a.
A simple example is element 2 of GF(3). The multiplication table of GF(3) shows
that the smallest n for which 2n = lisn = 2 — 3—1. Thus, element 2 is primitive
and GF(3) can be represented as the set (0,2,22 = 1). Another example is GF(5).
Exercise B.I shows that element 2 of GF(5) is primitive because the smallest n for
which 2n — 1 is?7, = 4 = 5 — 1. Hence, the exponential representation of GF(5) with
respect to 2 is (0,2, 22 - 4, 23 - 3,24 = 1).
o Exercise B.8: Show that 3 is also a primitive element of GF(5).
The exponential representation of Galois fields can be extended to fields GF(qm)
where m > 1.
An irreducible polynomial p{x) of degree m in GF(g) is said to be primitive if the
smallest positive integer n for which p{x) divides xn — 1 is n = qm — 1.
It can be shown that the roots ctj of an rath-degree primitive polynomial p(x) in
GF(q) have order qm — 1. This implies that the roots ctj of p{x) are primitive elements
in GF{qm). The exponential representation of GF(qm) can therefore be constructed
from any of these roots.
As an example, we show the construction of the exponential representation of
GF(23). The polynomial p{x) = x3 + x + 1 is primitive in GF(2). Let a be any
root of p(x) = x3 -f x + 1. From a3 + a + 1 = 0 we get a3 = a + 1 (this is done by