ALGEBRAIC SYSTEMS, FALL 2012
TODD COCHRANE
1. Notation
N = {1, 2, 3, 4, 5, . . . } = Natural numbers
Z = {0, ±1, ±, 2, ±3, . . . } = Integers
E = {0, ±2, ±4, ±6, . . . } = Even integers
O = {±1, ±3, ±5, . . . } = Odd integers
Q = {a/b : a, b ∈ Z, b 6= 0} = Rational numbers
R = Real numbers
C = Complex numbers
Zm = Ring of integers mod m
[a]m = {a + mx : x ∈ Z} = Residue class of a mod m
Um = Multiplicative group of units mod m
−1
a
(mod m) = “multiplicative inverse of a (mod m)”
φ(m) = Euler phi-function
(a, b) = gcd(a, b) = greatest common divisor of a and b
[a, b] = lcm[a, b] = least common multiple of a and b
a|b = “a divides b”
M2,2 (R) = Ring of 2 × 2 matrices over a given ring R
R[x] = Ring of polynomials over R
|S| = order or cardinality of a set S
Sn = n-th symmetric group
∩ intersection
∅
∪ union
⊆
empty set
∃ there exists
∀
⇔
∃!
⇒
for all
equivalent to
∈
subset
there exists a unique
implies
iff if and only if
≡
element of
Date: December 7, 2012.
1
congruent to
2
TODD COCHRANE
2. Math 511, Axioms for the set of Integers Z.
We shall assume the following properties as axioms for the set of integers.
1] Addition Properties. There is a binary operation + on Z, called addition,
satisfying
a) Addition is well defined, that is, given any two integers a, b, a + b is a uniquely
defined integer.
b) Substitution Law for addition: If a = b and c = d then a + c = b + d.
c) The set of integers is closed under addition. For any a, b ∈ Z, a + b ∈ Z.
d) Addition is commutative. For any a, b ∈ Z, a + b = b + a.
e) Addition is associative. For any a, b, c ∈ Z, (a + b) + c = a + (b + c).
f) There is a zero element 0 ∈ Z (also called the additive identity), satisfying
0 + a = a = a + 0 for any a ∈ Z.
g) For any a ∈ Z, there exists an additive inverse −a ∈ Z satisfying
a + (−a) = 0 = (−a) + a.
Properties a),b), and c) above are implicit in the definition of a binary operation.
Definition: Subtraction in Z is defined by a − b = a + (−b) for a, b ∈ Z.
2] Multiplication Properties. There is an operation · (or ×) on Z called multiplication, satisfying,
a) Multiplication is well defined, that is, given any two integers a, b, a · b is a
uniquely defined integer.
b) Substitution Law for multiplication: If a = b and c = d then ac = bd.
c) Z is closed under multiplication. For any a, b ∈ Z, a · b ∈ Z.
d) Multiplication is commutative. For any a, b ∈ Z, ab = ba.
e) Multiplication is associative. For any a, b, c ∈ Z, (ab)c = a(bc).
f) There is an identity element 1 ∈ Z satisfying 1 · a = a = a · 1 for any a ∈ Z.
3] Distributive law. This is the one property that combines both addition and
multiplication. For any a, b, c ∈ Z, a(b + c) = ab + ac. One can deduce (from the
given axioms) the additional distributive laws, (a + b)c = ac + bc, a(b − c) = ab − ac
and (a − b)c = ac − bc.
4] Trichotomy Principle. The set of integers can be partitioned into three disjoint
sets, Z = −N ∪ {0} ∪ N, where
N = {1, 2, 3, . . . } = Natural Numbers = Positive Integers,
−N = {−1, −2, −3, . . . } = Negative Integers.
One then defines the inequalities > and < by saying a > b if a − b ∈ N and
a < b if a − b ∈ −N. Thus we get the Law of Trichotomy which states that for any
two integers a, b exactly one of the following holds: a < b, a = b or a > b, (that is
a − b ∈ −N, a − b = 0 or a − b ∈ N.)
5] Positivity Axiom. The sum of two positive integers is positive. The product
of two positive integers is positive.
6] Discreteness Properties.
a) Well Ordering Property of N. Any nonempty subset of N has a smallest
element.
b) Principle of Induction. Let S be a subset of N such that
ALGEBRAIC SYSTEMS, FALL 2012
(i) 1 ∈ S and
Then S = N.
3
(ii) n ∈ S ⇒ n + 1 ∈ S.
Further Properties of Z. The properties below can all be deduced from the
axioms above. You may assume them in your homework unless specifically asked
to prove the property.
7] Cancellation law for addition: If a + x = a + y then x = y.
8] Cancellation law for multiplication: If ax = ay and a 6= 0 then x = y.
9] Subtraction-Equality principle. x = y if and only if x − y = 0.
10] Additive inverses are unique, that is, if a, b, c are integers such that a + b = 0
and a + c = 0 then b = c.
11] Zero multiplication property: a · 0 = 0 for any a ∈ Z.
12] Zero divisor property, or integral domain property: If ab = 0 then a = 0 or
b = 0.
13] Properties of negatives: (−a)b = −(ab) = a(−b), (−a)(−b) = ab, (−1)a = −a.
14] The product of two negative integers is positive.
15] “FOIL” Law (and all similar distributive laws): For any integers a, b, c, d,
(a + b)(c + d) = ac + ad + bc + bd.
16] Genassocomm Law: General Associative-Commutative Law:
a) Addition: When adding a collection of n integers a1 + a2 + · · · + an , the
numbers may be grouped in any way and added in any order. In particular, the
sum a1 +a2 +· · ·+an is well defined, that is, no parentheses are necessary to specify
the order of operations.
b) Multiplication: When multiplying a collection of n integers a1 a2 · · · · · an , the
numbers may be grouped in any way and multiplied in any order. In particular,
the product a1 a2 · · · · · an is well defined, that is, no parentheses are necessary to
specify the order of operations.
17] Binomial Expansion:
For anyintegers a, b and positive integer n we have
(a + b)n = an + n1 an−1 b + n2 an−2 b2 + · · · + bn .
In particular,
(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2 b + 3ab2 + b3 .
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TODD COCHRANE
3. Algebraic Properties of the Integers
Definition 3.1. A statement is a sentence that can be assigned a truth value. (In
general there is a subject, verb and object in the statement).
Ex. A: “x2 = 4”, B: “x = 2”, C: “x = ±2”
If A and B are statements, A ⇒ B means A implies B, that is, if A is true then
B is true. A ⇔ B means A is equivalent to B, that is, A is true iff B is true.
Ex. Which are true? A ⇒ C, A ⇒ B, B ⇒ A, A ⇔ C.
The symbols ⇒ and ⇔ are used between statements. The symbol = is used
between objects (numbers, functions, sets, etc. ).
Definition 3.2. 1) A binary operation ⊕ on Z is a function that assigns to each
ordered pair (a, b) of integers a unique integer denoted a ⊕ b.
2) It is called commutative if a ⊕ b = b ⊕ a for all a, b ∈ Z.
3) It is called associative if a ⊕ (b ⊕ c) = (a ⊕ b) ⊕ c for all a, b, c ∈ Z.
4) An element e ∈ Z is called an identity element with respect to ⊕ if a ⊕ e = a
and e ⊕ a = a for all integers a.
Example 3.1. Ordinary addition and multiplication are binary ops on Z; so is subtraction. Division fails. Addition and Multiplication are commutative and associative, and both have identities (what are they?).
Definition 3.3. A subset S of Z is said to be closed with respect to ⊕ if for any
two a, b ∈ S we have a ⊕ b ∈ S.
Example 3.2. (1) Let a ⊕ b = 2a + b. Is it binary op on Z? Is it commut? Is there
an identity? Is O closed
under ⊕? Is N closed under ⊕?
√
(2) Let a ⊕ b = ab. Is this a binary op on Z?
3.1. Deducing elementary properties of the integers from the axioms. In
the following we will provide examples of two styles of proofs. The first is “twocolumn” style, where the right column provides the justification for each step. The
second is text style, where the proof is written in paragraph form with complete
sentences following all the rules of grammar.
Example 3.3. Cancellation Law for Addition: Let a, x, y be integers such that a +
x = a + y. Then x = y.
Proof.
a + x = a + y,
⇒
− a + (a + x) = −a + (a + y),
⇒
(−a + a) + x = (−a + a) + y,
⇒ 0 + x = 0 + y,
⇒ x = y,
assumption
addition is well defined
associative law
additive inverse property
0 is additive identity
Note 3.1. Look at the axioms required to prove the cancellation law. Any algebraic
system satisfying those same axioms will also satisfy the cancellation law. “Rings”
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and “Additive Groups” are both examples of such systems that we will visit this
semester.
Example 3.4. Every integer has a unique additive inverse.
Proof. (We’ll do this one in text form.) By one of the axioms of Z, we know
that every integer has an additive inverse, so our task here is to show that it is
unique. Let a be a given integer. Suppose that b, c are additive inverses of a. Then
a + b = 0 and a + c = 0. By the transitive law for equality, a + b = a + c. Thus by
the cancellation law, b = c.
Example 3.5. Subtraction-Equality principle: For any integers x, y, x − y = 0 if
and only if x = y.
Proof.
x − y = 0,
assumption
⇔ (x − y) + y = 0 + y,
addition is well defined
⇔ (x + (−y)) + y = 0 + y,
definition of subtraction
⇔ x + (−y + y) = 0 + y,
associative law
⇔ x + 0 = 0 + y,
additive inverse property
⇔ x = y,
0 is additive identity
Note that because the statement was an if and only if statement we needed
left-right arrows at each step.
Example 3.6. For any integer n, n · 0 = 0.
Proof. The formal proof is homework but we’ll give you a hint. Since 0 is linked
with additive properties of Z and this theorem is a multiplicative statement, you
will need to make use of the one axiom linking addition and multiplication (what
is it?) Now start by writing 0 = 0 + 0 (what property have I just used?)
Example 3.7. Here is a proof written in text form.
Property of Negatives: For any integer a, (−1)a = −a.
Proof. (Here, we’ll start with text form and then go to two-column form.) Our
goal is to show that (−1)a satisfies the property of an additive inverse, that is,
(−1)a + a = 0. Now,
(−1)a + a = (−1)a + 1(a),
= (−1 + 1)a,
= 0a,
= 0,
1 is the multiplicative identity
distributive law
property of additive inverses
by preceding example.
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TODD COCHRANE
Why do we just use = here but ⇒ in the earlier proofs.
Comment on the General Associative/Commutative Law. What does 2 · 5 ·
3 · 7 mean? Note there are many different groupings one can give. The general
associative law says all these values are equal.
Comment on Discreteness Axioms for Z: 1) Well ordering property. Compare
R. 2) Induction Principle.
Proof by Induction: Let P (n) be a statement involving a natural number n.
Suppose that
(i) P (1) is true. (Base Case)
(ii) If P (n) is true for a given n then P (n + 1) is true. (Note induction assumption.)
Then P (n) is true for all n ∈ N.
Example 3.8. 1. Sum of first n odd numbers. 2. Sum of first n numbers.
Example 3.9. Prove that for any positive integer n,
1 3 + 2 3 + · · · + n3 =
(3.1)
n2 (n + 1)2
.
4
2
2
Proof. Proof by induction. For n = 1 we have 13 = 1 4·2 , a true statement. Suppose
that statement (3.1) is true for a given n. Then for n + 1 we have
13 + 23 + · · · + n3 + (n + 1)3 = (13 + 23 + · · · + n3 ) + (n + 1)3
n2 (n + 1)2
+ (n + 1)3 , by induction assumption (3.1),
4
(n + 1)2 2
=
[n + 4(n + 1)],
4
(n + 1)2 2
(n + 1)2
(n + 1)2 ((n + 1) + 1)2
=
[n + 4n + 4] =
[n + 2]2 =
.
4
4
4
=
QED.
Example 3.10. n3 − n is a multiple of 3 for any integer n.
Proof. Proof by induction. For n = 1 we note that 13 − 1 = 0 = 0 · 3, a multiple of
3. Suppose that the statement is true for a given n, that is, n3 − n = 3k for some
k ∈ Z. Then for n + 1 we have
(n + 1)3 − (n + 1) = n3 + 3n2 + 3n + 1 − n − 1 = (n3 − n) + 3n2 + 3n
= 3k + 3n2 + 3n,
by induction assumption,
= 3(k + n2 + n) = 3 · integer,
since the integers are closed under addition and multiplication. QED.
Example 3.11. Let {Fn } = 1, 1, 2, 3, 5, 8, 13, . . . , the Fibonacci sequence. Prove that
(3.2)
F1 + F3 + · · · + F2k−1 = F2k ,
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for any k ∈ N.
Proof. Proof by induction on k. For k = 1 we have F1 = 1 = F2 , so the statement
is true. Suppose that the statement (3.2) is true for a given k. Then for k + 1 we
have
F1 + F3 + · · · + F2k−1 + F2k+1 = (F1 + F3 + · · · + F2k−1 ) + F2k+1
= F2k + F2k+1 ,
by the induction hypothesis,
= F2k+2 = F2(k+1) ,
by the defining property of the Fibonacci sequence. QED.
Definition 3.4. Let a, b ∈ Z, a 6= 0. We say a divides b, written a|b, if ax = b for
some integer x.
Ex. 3|12 since.., 5 - 12 since... Distinguish 3|12 from 3/12.
Equivalent terms: a divides b. a is a divisor of b. a is a factor of b. b is divisible
by a. b is a multiple of a.
Example 3.12. 1) What are the divisors of 6? What are the divisors of 0?
Goal: Fundamental Theorem of Arithmetic.
Theorem 3.1. Basic divisibility properties. Let a, b, d be integers.
(i) If d|a and d|b then d|(a + b).
(ii) If d|a and d|b then d|(a − b).
(iii) If d|a and d|b then for any integers x, y, d|(ax + by).
Proof. (iii) Suppose that d|a, d|b and that x, y ∈ Z. Then a = dk and b = dl for
some integers k, l. Thus,
ax + by = (dk)x + (dl)y = d(kx) + d(ly) = d(kx + ly) = d(integer),
since Z is closed under addition and multiplication. Thus d|ax + by.
Example 3.13. Another way to think about them, is to use the word multiple. If a
and b are multiples of d then so is a + b, etc. Let S be the set of all multiples of 5.
Note S is closed under addition and subtraction.
Theorem 3.2. Transitive law for divisibility. For any integers a, b, c, if a|b and
b|c, then a|c.
Proof. Homework
Definition 3.5. Let a, b be integers not both 0. The greatest common divisor of
a, b, denoted gcd(a, b) is the largest integer that divides both a and b.
ii) Two numbers are called relatively prime if gcd(a, b) = 1.
Note 3.2. 1. gcd(0,0) is undefined. Why?
2. If a, b are not both zero, gcd(a,b) exists and is unique. (Why? Let S be the
set of common divisors. It is a finite nonempty set, so it has a maximum element.)
3. gcd(0, n) = |n|.
4. gcd(a, b) = gcd(b, a)=gcd(−a, b)= gcd(−a, −b).
Example 3.14. 1) gcd(-16,-28)=4.
2) gcd(6,-16,-28) = 2.
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TODD COCHRANE
Lemma 3.1. Subtraction Principle for GCDs. For any a, b ∈ Z, not both zero,
and any integer q, gcd(a, b) = gcd(a − qb, b).
Proof. S, T be set of common divisors. Show S ⊆ T and T ⊆ S.
Example 3.15. Find gcd(1023, 1026). By subtraction principle this equals gcd(1023, 3).
The latter equals 3 since 3|1023.
Division of Integers with remainder. Ex. 38÷5 = 7R2, that is, 38 = 5·7+2.
Quotient, remainder, divisor, dividend.
Theorem 3.3. Division Algorithm. Let a, b be integers with b > 0. Then there
exist integers q, r such that a = qb + r with 0 ≤ r < b. Moreover q, r are unique.
q=quotient and r= remainder in dividing a by b.
Proof. Existence: We let q be the greatest integer such that qb ≤ a, so that qb ≤
a < (q + 1)b. Then set r = a − qb.
Euclidean Algorithm.
Example 3.16. Find d = gcd(126, 49).
(1)
126 = 2 · 49 + 28,
(2)
49 = 28 + 21,
(3)
28 = 21 + 7,
(4)
21 = 3 · 7,
d = gcd(28, 49)
d = gcd(28, 21)
d = gcd(7, 21)
d = gcd(7, 0) = 7, ST OP
Definition 3.6. A linear comb. of two integers a, b is an integer of the form ax+by
where x, y ∈ Z.
Claim: If d = gcd(a, b) then d can be expressed as a linear comb. of a and b.
Example 3.17. gcd(20,8)=4. By trial and error, 4 = 1 · 20 + (−2)8.
gcd(21,15)=3. By trial and error, 3 = 3 · 21 − 4 · 15.
Back Substitution: A method of solving the equation d = ax + by (with
d = gcd(a, b)) by working backwards through the steps of the Euclidean algorithm.
Example 3.18. Use example above for gcd(126,49) to express 7 as a LC of 126 and
49. Use the method of back substitution. Start with equation (3): 7 = 28 − 21. By
(2) we have 21 = 49−28. Substituting this into previous yields 7 = 28−(49−28) =
2 · 28 − 49. By (1) we have 28 = 126 − 2 · 49. Substituting this into previous yields
7 = 2 · (126 − 2 · 49) − 49 = 2 · 126 − 5 · 49, QED.
Array Method.
Example 3.19. Redo example using array method. Perform Euclidean Alg. on the
numbers in top row, but do column operations on the array. Let C1 be the column
with top entry 126, C2 the column with top entry 49, etc. Then C3 = C1 − 2C2 .
C4 = C2 − C3 , C5 = C3 − C4 .
126x + 49y 126 49 28 21
7
x
1
0
1 −1 2 Thus, 7 = 7 · 126 − 5 · 49.
y
0
1 −2 3 −5
ALGEBRAIC SYSTEMS, FALL 2012
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Example 3.20. Find gcd(83, 17) and express it as a LC of 83 and 17.
83x + 17y 83 17 15
2
1
x
1 0
1 −1
8 Thus gcd = 1 and 1 = 8 · 83 − 39 · 17.
y
0 1 −4 5 −39
Theorem 3.4. GCDLC. Let a, b be integers not both zero, d = gcd(a, b). Then d
can be expressed as a LC of a and b.
Note 3.3. (i) The set of all linear combinations of a, b is just the set of multiples
of d.
(ii) The gcd of a and b is the smallest positive LC of a and b.
(iii) Every common divisor of a and b is a divisor of gcd(a, b).
Solving Linear Equations in integers: Solve ax + by = c. GCDLC theorem
tells us that this equation can be solved iff c is a multiple of d, that is d|c.
Theorem 3.5. Solvability of a Linear Equation. The linear equation ax + by = c
has a solution in integers x, y iff d|c where d = gcd(a, b).
Example 3.21. Solve the following equations or show that there is no solution.
120x − 75y = 150,
,
120x − 75y = 11.
By the array method we obtain 120(2) − 75(3) = 15, the gcd of 120 and 75. Multiplying by 10 gives the solution (20, 30) to the first equation above. Since 15 - 11
the second equation has no solution.
Example 3.22. A parcel costs $2 and we only have 13 cent and 17 cent stamps.
How can we do it? 13x + 17y = 200. We know 200 is a lc since gcd=1. Use array
to get (-50,50) then note that you can add (17,-13) to get another solution.
Definition 3.7. Two integers a, b are called relatively prime if gcd(a, b) = 1.
Lemma 3.2. Euclid’s Lemma. If d|ab and gcd(d, a) = 1 then d|b.
Note: This lemma fails if gcd(d, a) 6= 1. For example 4|(2 · 2), but 4 - 2. Thus
d|ab does not imply that d|a or d|b.
Note 3.4. Applications of Euclid’s Lemma.
(i) Every rational number can be uniquely expressed as a fraction in reduced
form. Proof. Homework.
√
(ii) If n is not a perfect square, then n is irrational. Proof. Homework.
Definition 3.8. i) A positive integer p > 1 is called a prime if its only positive
factors are 1 and itself. 2,3,5,7,...
ii) A positive integer n > 1 is called a composite if it is not a prime, that is,
n = ab for some positive integers a, b with a > 1 and b > 1. 4,6,8,9,...
Note 3.5. 1 is not a prime or a composite. It is the multiplicative identity element.
(Later, we will call it a “unit”.) Why? If 1 is a prime then we would violate unique
factorization, eg 6 = 2 · 3 = 1 · 2 · 3.
Lemma 3.3. a) Let p be a prime such that p|ab. Then p|a or p|b.
b) Let p be a prime such that p|a1 a2 . . . ak where ai are integers. Then p|ai for
some i.
Proof. Use Euclid’s lemma for part (a) and induction for (b).
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TODD COCHRANE
Theorem 3.6. FTA: Fundamental Theorem of Arithmetic. Any positive integer
n > 1 can be expressed as a product of primes, and this expression is unique up to
the order of the primes.
Note 3.6. (i) 12 = 2·2·3 = 2·3·2 = 3·2·2, are all considered the same factorization.
(ii) We say that a prime p has a trivial factorization as a product of primes.
Strong Form of Induction Let P (n) be a statement involving the natural
number n. Suppose
(i) P (1) is true, and
(ii) If P (k) is true for all k < n then P (n) is true.
Then P (n) is true for all natural numbers n.
Proof of FTA. Existence. Proof is by strong form of induction. Let P (n) be the
statement that n has a factorization as a product of primes. P (2) is true. Suppose
P (k) is true for all values k smaller than a given n. Consider P (n). If n is prime
we are done. Otherwise n = ab for some integers a, b with 1 < a < n, 1 < b < n.
By the induction assumption, a and b can be expressed as products of primes, say
a = p1 · · · pk , b = q1 · · · q` . Then ab = p1 · · · pk q1 · · · q` , a product of primes. QED
Uniqueness. Suppose that n is a positive integer with two representations as a
product of primes, say,
(3.3)
n = p1 · · · pk = q1 · · · qr
for some primes pi , qj , 1 ≤ i ≤ k, 1 ≤ j ≤ r. We may assume WLOG that k ≤ r.
Then p1 |q1 . . . qr , so by lemma, p1 |qi1 for some i1 ∈ {1, 2, . . . , r}. Since p1 and qi1
are primes, we must have p1 = qi1 . Cancelling p1 in (3.3) yields
(3.4)
p2 p3 · · · pk = q1 · · · q̂i1 · · · qr ,
where q̂i1 indicates that this factor has been removed. We can then repeat the
argument with p2 in place of p1 . After repeating this process k times we have that
(3.5)
p1 = qi1 , p2 = qi2 , . . . , pk = qik
for some distinct integers i1 , i2 , . . . , ik ∈ {1, 2, . . . , r}. Moreover, after cancelling
each of the pi from (3.3) we are left with 1 on the LHS. If r > k then (3.3) would
say that 1 is a product of primes, a contradiction. Therefore r = k, and so by (3.5),
the primes pi are just a permutation of the primes qi .
Theorem 3.7. There exist infinitely many primes.
Proof. (Euclid) Proof by contradiction. Suppose that there are finitely many
primes, say {p1 , p2 , . . . , pk }. Let N = p1 p2 · · · pk +1. By FTA, N has a prime factor
pi , for some i ≤ k. Thus, pi |N and pi |(p1 p2 · · · pk ). Therefore pi |(N − p1 · · · pk ),
that is, pi |1, a contradiction.
Theorem 3.8. Basic primality test. √
Let a > 1 be a positive integer such that a is
not divisible by any prime p with p ≤ a. Then a is a prime.
Proof. Homework.
Note 3.7. Sieve of Eratosthenes: This is the method of finding all
√ of the primes in
a given interval [a, b] by crossing out all multiples of primes p ≤ b.
ALGEBRAIC SYSTEMS, FALL 2012
11
4. Modular Arithmetic and the Ring of Integers (mod m)
Example 4.1. What’s the pattern? 3+5=8, 6+4=10, 7+6=1, 9+8=5, 9+2=11
Let m ∈ N. m =modulus.
Definition 4.1. We say that two integers a, b are congruent modulo m, written
a ≡ b (mod m), if a and b differ by a multiple of m, that is m|(a − b).
Note: a ≡ b (mod m) is equivalent to a = b + mk for some integer k.
Example 4.2. Let m = 12. Then 16 ≡ 4 (mod 12) since 16 − 4 = 12. 13 ≡ 1
(mod 12). In the example above we see 9 + 8 = 17 ≡ 5 (mod 12). How about 256
what is it (mod 12). 256 = 21 · 12 + 4, so 256 ≡ 4 (mod 12).
Definition 4.2. The least residue of a (mod m) is the smallest nonnegative integer
that a is congruent to (mod m).
Note: The least residue of a (mod m) is the remainder in dividing a by m. Since
0 ≤ r < m l.r. is always in {0, 1, 2, 3, . . . , m − 1}.
Example 4.3. m = 5 Wrap the integers around a five hour clock.
Theorem 4.1. Congruence is an equivalence relation. That is (i) Reflexive, (ii)
Symmetric and (iii) Transitive.
Theorem 4.2. Important properties of congruences. The substitution laws. Suppose a ≡ b (mod m), and c ≡ d (mod m). Then
(i) a ± c ≡ b ± d (mod m).
(ii) a · c ≡ b · d (mod m).
(iii) an ≡ bn (mod m) for any positive integer n.
Example 4.4. 281 · 717 (mod 7). 544 + 27 · 392 (mod 5).
Proof. Two types for (i) and (ii). Induction for (iii).
Example 4.5. Explore powers of 2 (mod 3), (mod 6), (mod 7), (mod 8), (mod 9).
Note repeating pattern of length ≤ m. Use for finding 2100 (mod 6).
Note 4.1. Trick for calculating an (mod m) if gcd(a, m) = 1. First find a power
k such that ak ≡ ±1 (mod m). Find 4750 (mod 5), 2100 (mod 7), 21110 (mod 7),
2100 (mod 17).
A few applications of congruences:
Example 4.6. Day of the week. What day of the week is it 10 years from today?
What time will it be 486 hours from now?
Divisibility tests: For numbers written in the base-10 (decimal) number system.
What does 2715 mean? Is it divisible by 9?
Theorem 4.3. Divisibility tests for 3,9 and 11. Let n be a positive integer with
decimal rep. n = ak · 10k + . . . a0 , where the ai ∈ {0, 1, 2, . . . , 9}.
(i) 3|n iff 3|(ak + · · · + a0 ).
(ii) 9|n iff 9|(ak + · · · + a0 ).
(iii) 11|n iff 11|ak − ak−1 + ak−2 − · · · + (−1)k a0 .
Example 4.7. UPC symbols. A 12 digit code d1 , d2 , . . . , d12 . d12 is the check digit.
3(d1 + d3 + · · · + d11 ) + (d2 + . . . d12 ) ≡ 0 (mod 10).
12
TODD COCHRANE
Definition 4.3. An integer x is called a multiplicative inverse of a (mod m) if
ax ≡ 1 (mod m). We write x ≡ a−1 (mod m) in this case. Avoid fractions.
Example 4.8. Find mult. inverse of 3 (mod 5), 4 (mod 6), by trial and error. Which
numbers have mult. inverse (mod 10).
Theorem 4.4. a has a mult inverse (mod m) iff gcd(a, m) = 1.
Example 4.9. Find mult. inverse of 12 (mod 17). Then solve 12x ≡ 5 (mod 17).
Example 4.10. Solve 3x ≡ 5 (mod 6).
Theorem 4.5. The congruence ax ≡ b (mod m) is solvable iff d|b where d =
gcd(a, m).
Definition 4.4. The (residue class) congruence class of a (mod m), denoted [a]m
is the set of all integers congruent to a (mod m). Thus [a]m = {a + km : k ∈ Z}.
Example 4.11. [2]5 = {2, 7, 12, . . . } ∪ {−3, −8, . . . }. Note [7]5 , [12]5 also represent
the same class. Draw five hour clock.
Note 4.2. [a]m = [b]m iff a ≡ b (mod m). Thus eg. [2]5 = [12]5 . The values 2,7,12,
etc. are called representatives for the class [2]5 .
Definition 4.5. (i) Let m be a positive integer. The ring of integers (mod m)
(residue class ring (mod m)) denoted Zm , is the set of all congruence classes
(mod m). Zm = {[0]m , . . . , [m − 1]m }.
(ii) We define addition and multiplication on Zm as follows: For [a]m , [b]m ∈ Zm ,
[a]m + [b]m := [a + b]m ,
[a]m [b]m := [ab]m .
Example 4.12. [3]5 + [4]5 = [2]5 . [3]5 [4]5 = [2]5 .
Note 4.3. Addition and multiplication are well defined on Zm , that is, if [a]m = [b]m
and [c]m = [d]m then [a + c]m = [b + d]m and [ac]m = [bd]m . (That is, the sum and
product do not depend on the choice of representatives for the congruence classes.)
Proof. We’ll do multiplication. The proof for addition is similar. First, the definition of multiplication in Zm is [x]n [y]m = [xy]m , for any [x]m , [y]m ∈ Zm . To show
that the product is well defined we must show that the product does not depend on
the choice of representatives for the congruence classes. Now lets begin the proof.
Suppose that [a]m = [a0 ]m and [b]m = [b0 ]m . Our goal is to show that [ab]m =
0 0
[a b ]m . By the definition of a congruence classes, we have a ≡ a0 (mod m) and
b ≡ b0 (mod m). By the substitution property of congruences this implies that
ab ≡ a0 b0 (mod m), that is, [ab]m = [a0 b0 ]m . QED.
Note 4.4. The laws for Z hold for Zm as well: Commutative, Associative, Distributive, zero element, additive inverses
(ii) Note one important property that Z has that Zm doesn’t have. Integral
domain property.
Convention. If it is understood that we are working in Zm then the bracket
notation can be dropped, and abbreviated Zm = {0, 1, 2, . . . , m − 1}. We can say,
in Z6 , 3 · 7 = 3. What is 3 + 4 in Z5 ? Find mult table for Z4 . Note 2 · 2 = 0 in Z4 .
Definition 4.6. The group of units Um = {x ∈ Zm : gcd(x, m) = 1}.
ALGEBRAIC SYSTEMS, FALL 2012
13
Note 4.5. (i) Um is the set of elements of Zm that have multiplicative inverses.
(ii) Um is closed under multiplication.
Example 4.13. U9 and multiplication table. Note closed, each row and column,
mult inverse.
Definition 4.7. For any set S we define the cardinality of S, |S|, to be the number
of elements in S. Write |S| = ∞ is S is infinite.
Example 4.14. |Z9 | =, |U9 | = 6, |Z| = ∞.
Definition 4.8. Euler phi-function.
Note 4.6. By theorem above, φ(m) = |Um |.
Find a formula for φ(m): Test p, pe , pe q f . Use Inclusion/Exclusion. Note
φ(pe q f ) = |U | − |S| − |T | + |S ∩ T | = m − m/p − m/q + m/pq = m(1 − 1/p)(1 − 1/q).
Theorem 4.6. Let m = pe11 . . . pekk .
(i) φ(m) = φ(pe11 )φ(pe22 ) . . . φ(pekk ) = (pe11 − p1e1 −1 ) . . . (pekk − pkek −1 ).
(ii) φ(m) = m(1 − p11 ) . . . (1 − p1k ).
Example 4.15. φ(1500).
Euler’s Theorem and Fermat’s Little Theorem.
Recall its useful for
modular arith to find exponent k such that ak ≡ 1 (mod m).
Theorem 4.7. Eulers Theorem. Let m ∈ N, and a ∈ Z with gcd(a, m) = 1. Then
aφ(m) ≡ 1 (mod m).
Example 4.16. Find 171602 (mod 1500).
Theorem 4.8. Fermats Little Theorem. Let p be a prime, and a ∈ Z, p - a. Then
ap−1 ≡ 1 (mod p).
Example 4.17. Find 2150 (mod 37).
Note 4.7. FLT is a special case of Euler’s Theorem. If p|a the theorem fails, but it
can be restated ap ≡ a (mod p) for any a ∈ Z.
Lemma 4.1. Permutation Lemma. Let m ∈ N and Um = {x1 , x2 , . . . , xr } where
r = φ(m). Let a ∈ Z with gcd(a, m) = 1. Then Um = {ax1 , ax2 , . . . , axr }, that is
ax1 , . . . , axr is just a permutation of the values x1 , . . . , xr .
Example 4.18. U9 = {1, 2, 4, 5, 7, 8}. Test a = 2, a = 4. Note failure if a = 3.
Proof. Note (i) for 1 ≤ i ≤ r, axi ∈ Um . (ii) The values axi are distinct, by
cancellation law. Thus{ax1 , . . . , axr } is a set of r distinct elements in Um , and so
it must equal all of Um .
Proof. . Proof of Eulers Theorem Standard.
Public Key Cryptography. Idea is to send a secure message over a public
medium such as radio, tv, cell phone, internet, etc. in such a way that only the
intended recipient can decipher the message.
First words are converted to numbers: A=01, B=02, etc. Hello = 805,121,215
Each person selects their own modulus m, encoding exponent e, and decoding
exponent d. The first two are public and the latter top secret. e, d are chosen so
that for any integer M with gcd(M, m) = 1, M de ≡ M (mod m).
14
TODD COCHRANE
Example 4.19. Say John wishes to send the message M to Mary. He looks up
Marys m and e in the phone book. Assume that M < m and gcd(M, m) = 1.
John calculates Me ≡ M e (mod m) (encoded message). Me is then sent publicly
to Mary. Mary then calculates Med (mod m). Note Med ≡ M de ≡ M (mod m).
Thus Mary recovers the original message! Say M = 805,m = 1147 = 31 · 37, e = 23,
d = 47. Note φ(m) = 30 · 36 = 1080. If (M, m) = 1 by Euler’s theorem M φ(m) ≡ 1
(mod m). Thus M de ≡ M 1081 ≡ M (mod m). Me ≡ 805e ≡ 743 (mod 1147).
Md ≡ 743d ≡ 805 (mod m).
In practice m is chosen to be a huge number (200 digits) that cannot be factored,
and so φ(m) cannot be determined from the phone book information. Thus d
remains secure. Security depends on the fact that we have no factoring algorithms
for 200 digit numbers that can run in less time than the age of the universe.
ALGEBRAIC SYSTEMS, FALL 2012
15
5. Rings, Integral Domains and Fields
Definition 5.1. A ring is a set R with two binary operations +, · satisfying
(1) Closed under + and ·
(2) Associative law for both addition and multiplication.
(3) Commutative law for addition.
(4) Distributive laws hold.
(5) R has a zero element 0.
(6) Every element of R has an additive inverse.
If R is a ring with commutative multiplication then R is called a commutative
ring. If R is a ring with unity element 1 then R is called a ring with unity. (We
require 1 6= 0, so that R 6= {0}.)
Example 5.1. Z, R, Q, Zm are all rings. What type?
Definition 5.2. Let R be a given ring. A subset S of R is called a subring if S is
a ring under the same two binary operations.
Example 5.2. Let E be the set of even numbers, O, the set of odd numbers. Is
either of these a subring of Z?
Example 5.3. Show that the set 3Z, of all multiples of 3 is a subring of Z. (1) Closed
under addition: Let 3n, 3m ∈ 3Z, where m, n ∈ Z. Then 3n + 3m = 3(n + m) ∈ 3Z.
Also, 3n·3m = 3(3nm) ∈ 3Z. (2)-(4) The associative, commutative and distributive
laws are inherited from Z. (5) 0 = 3·0 ∈ 3Z. (6) If 3n ∈ 3Z then −3n = 3(−n) ∈ 3Z.
Thus all 6 properties hold, so 3Z is a subring of Z.
Note 5.1. To show a subset S of a given ring R is a subring of R it suffices to verify
(1) S is closed under + and ·, (5) 0 ∈ S, and (6) If x ∈ S then −x ∈ S. All other
properties are inherited from R.
Example 5.4. E is a subring of Z. Z is a subring of Q. Q is a subring of R.
Example 5.5. The subrings of Z are of the form nZ := {nx : x ∈ Z}, with n a fixed
integer. For instance E = 2Z, or 3Z = {0, ±3, ±6, . . . }.
Example 5.6. If d|m we say dZm = {0, d, 2d, . . . , ( m
d − 1)d}. Every subring of Zm
is of the form dZm with d|m. Consider Z12 . Find all subrings.
5.1. Polynomials.
Definition 5.3. Let R be a given ring.
a) A polynomial over R in the variable x is an expression of the form
f (x) = an xn + an−1 xn−1 + · · · + a0 ,
where the ai are elements of R.
b) The values ai are called coefficients of the polynomial.
c) If an 6= 0 then an is called the leading coefficient of the polynomial and the
polynomial is said to be of degree n.
d) A polynomial of the form f (x) = a with a ∈ R, is called a constant polynomial. If a 6= 0 then it has degree 0. The zero polynomial, f (x) = 0, is not
assigned a degree.
16
TODD COCHRANE
Definition 5.4. Let R be given ring. The polynomial ring in (the variable) x over
R, denoted R[x], is the set of all polynomials in x with coefficients in R,
R[x] = {an xn + · · · + a0 : ai ∈ R, 0 ≤ i ≤ n, n ≥ 0}.
Pn
Pn
Addition and multiplication are standard: Let f (x) = i=0 ai xi , g(x) = j=0 bj xj .
Pn
Addition: f (x) + g(x) := i=0 (ai + bi )xi .
Pn Pn
P2n P
Multiplication: f (x) · g(x) := i=0 j=0 ai bj xi+j = k=0 ( i+j=k ai bj )xk .
Note that since R is a ring, the coefficients P
of f (x) + g(x) and f (x)g(x) are again
n
in R. We also have 0 ∈ R[x] and −f (x) = i=0 (−ai )xi ∈ R[x], so properties (5)
and (6) are satisfied. It is routine, but tedious to verify that properties (2), (3) and
(4) hold.
Note 5.2. i) If R is ring with unity then so is R[x]. Indeed, if 1 ∈ R then 1 is a
constant polynomial in R[x].
ii) If R is commutative then so is R[x]. This follows from the fact that ai bj = bj ai
for all terms in the product definition above.
Example 5.7. In Z2 [x] find (1 + x)2 . In Z3 [x] find (x + 1)3 .
Definition 5.5. A nonzero element a ∈ R is called a zero divisor if ab = 0 or
ba = 0 for some nonzero b ∈ R.
Example 5.8. 3 is a zero divisor in Z6 since 3 · 2 = 0 in Z6 .
Example 5.9. Find all zero divisors in Z9 . Note that the remaining values are units.
Note 5.3. (i) If p(x) = an xn + an−1 xn−1 + · · · + a0 , with an 6= 0, then the degree of
p(x) is n, the leading term of p(x) is an xn and the leading coefficient of p(x) is an .
ii) If p(x) = an xn + · · · + a0 , q(x) = bm xm + · · · + b0 , with a0 6= 0, b0 6= 0, then
p(x)q(x) = an bm xm+n + · · · + a0 b0 . Note that if an , bm are not zero divisors then
an bm 6= 0 and so the degree of p(x)q(x) is m + n.
Recall, the group of units for Zm , Um .
Definition 5.6. Let R be a ring with unity. An element a ∈ R is called a unit if
a has a multiplicative inverse in R, that is, ab = 1 = ba for some b ∈ R.
Example 5.10. Find all units in Z, Q, Z6
Theorem 5.1. Let a ∈ Zm , a 6= 0. Then a is a unit if (a, m) = 1 and a is a zero
divisor if (a, m) > 1.
Definition 5.7. An integral domain is a commutative ring with unity having no
zero divisors, that is, if ab = 0 then either a = 0 or b = 0.
Example 5.11. Z is an integral domain.
Theorem 5.2. Zm is an integral domain iff m is a prime.
Note 5.4. The importance of integral domain is that we can solve equations in the
same manner that you are used to: Solve x2 − 3x + 2 = 0 in an integral domain R.
Now, solve (x − 1)2 = 0 in Z8 , and note the difference because Z8 is not an integral
domain.
Lemma 5.1. Let R be an integral domain and f (x), g(x) ∈ R[x] be nonzero polynomials of degrees n, m respectively. Then deg(f (x)g(x)) = n + m.
ALGEBRAIC SYSTEMS, FALL 2012
Proof. Homework.
17
Theorem 5.3. If R is an integral domain, then R[x] is an integral domain.
Proof. We already observed above that R[x] is a commutative ring with unity in
this case, so we only need to show that R[x] has no zero divisors. This is a homework
problem.
Example 5.12. More standard examples of integral domains: R[x] where R is a
given integral domain, such as Z[x], R[x], etc. Note that the product of two nonzero
polynomials with coefficients in R is always nonzero.
Definition 5.8. A ring R is called a field if (i) R has a unity, (ii) R is commutative,
(iii) Every nonzero element of R is a unit.
Example 5.13. Standard examples of fields: Q, R, C, Zp where p is a prime. Also,
F (x) the set of all rational functions p(x)/q(x) with coefficients in a given field F .
Theorem 5.4. If R is a field then R is an integral domain.
Definition 5.9. A 2 by 2 matrix with entries in a given ring R is an array of the
form
a b
,
c d
where a, b, c, d ∈ R. The entry position is given by specifying the row number first,
column number second. Thus, a is the entry in the 1, 1 position, b the 1, 2 position,
c the 2, 1 position and d the 2, 2 position.
Definition 5.10. Matrix Rings. Let R be a given ring. The ring of 2 by 2 matrices
over R is given by
a b
M2,2 (R) =
: a, b, c, d ∈ R .
c d
Addition andmultiplication
are
standard.
a b
e f
a+e b+f
Addition:
+
=
.
c d
g h c +
g d+h
a b e f
ae + bg af + bh
Multiplication:
=
c d g h
ce + dg cf + dh
Note 5.5. Matrix multiplication is obtained by taking dot products of the rows of
the left matrix with columns of the right matrix. Let A, B be the two matrices
above. Let R1 , R2 be the two rows of A and C1 , C2 the two columns of B. Then
the ij-th entry of AB is equal to Ri · Cj .
Note 5.6. M2,2 (R) is in fact a ring.
(1) Since R is closed under +, it follows that so is the matrix ring. Since R is
closed under addition and mult, the product of any two matrices over R again has
entries in R.
(2) The associative law for addition follows immediately from the assoc. law for
addition in R. The associative law for multiplication is not trivial. For people with
more background withP
matrices:
Let A = [aij ], B = [bij ], C = [cij ]. The ij-th entry
P
of
(AB)C
is
given
by
(a
b
ik
kl )clj while the ij-th entry of A(BC) is given by
k
l
P P
a
(b
c
).
Thus
they
are
equal
by the associative law for R.
ik
kl
lj
k
l
(3) The commutative law for addition is immediate.
18
TODD COCHRANE
(4) The distributive law: The ij-th entry of A(B + C) is given by
2
X
k=1
aik (bkj + ckj ) =
2
X
(aik bkj + aik ckj ) =
k=1
2
X
k=1
aik bkj +
2
X
aik ckj
k=1
which is just the ij-th entry of AB + AC.
0 0
.
0 0
(6) The additive inverse of A = [aij ] is the matrix −A = [−aij ], which is in
M2,2 (R).
1 0 0 0
Note 5.7. (i) Matrix multiplication is not commutative. eg. compare
,
0 0 1 0
and its reverse.
(ii) M2,2 (R) has zero divisors. Indeed, for any a, b, c, d ∈ R,
a 0 0 0
0 0
=
.
b 0 c d
0 0
(5) The zero element in M2,2 (R) is the matrix 0 =
(iii) If R is a ring with unity 1, then M2,2 (R) is a ring with unity I2 given by
1 0
I2 :=
.
0 1
Example 5.14. M2,2 (Z2 ), is a ring with 16 elements.
a b
∈
Theorem 5.5. Let R be a commutative ring with unity, and A =
c d
M2,2 (R). Put ∆ = ad − bc, the determinant of A. Then A is a unit in M2,2 (R) if
and only if ∆ is a unit in R.
Proof. In your homework you will show that if ∆ is a unit in R then
d −b
−1
−1
A =∆
.
−c a
The converse is done in a matrix theory class.
Definition 5.11. The complex numbers C is the set of numbers,
C := {a + bi : a, b ∈ R},
√
where i is the imaginary unit i = −1. (Draw complex plane with real and
imaginary axes and indicate the point a + bi).
ii) Let z = a + bi. Then a is called the real part of z and b is called the
imaginary part.
iii) Two complex numbers are equal iff they have the same real and imaginary
parts.
iv) The complex conjugate of z = a + bi, denoted z, is given by z = a − bi. It
is the reflection of z in the real axis.
v) Addition in C is defined by (a + bi) + (c + di) = (a + c) + (b + d)i.
vi) Multiplication in C is defined by (a + bi)(c + di) = (ac − bd) + (bc + ad)i.
√ vii) The modulus or absolute value of z = a + bi, denoted |z|, is given by |z| =
a2 + b2 .
ALGEBRAIC SYSTEMS, FALL 2012
19
Note 5.8. i) One can verify that C is a commutative ring with unity 1.
ii) Every nonzero complex number has a multiplicative inverse in C, indeed, if
z = a + bi then
a − bi
z
z −1 = 2
= 2.
2
a +b
|z|
Thus C is a field.
iii) For any z ∈ C, zz = |z|2 .
Definition 5.12. Polar coordinates r, θ, of a complex number z.
i) The polar angle or argument of z, denoted θ, is the angle formed with
respect to the positive real axis, (draw picture). It is not unique. One can add any
multiple of 2π.
ii) r = |z|, the modulus of z. It is unique and nonnegative, (unlike polar
coordinates in R2 .)
Definition 5.13. i) The polar form of a complex number is given by
z = r(cos(θ) + i sin(θ)),
where r = |z| and θ is the polar angle of z. This identity follows from definition of
the trig functions (cos θ is the x-coordinate on unit circle, sin θ is the y-coordinate.
Illustrate).
ii) The exponential polar form of a complex number is given by
z = reiθ .
To obtain the exponential polar form we need the following theorem.
Theorem 5.6. For any real number t we have eit = cos(t) + i sin(t).
Proof. Recall the Taylor expansions
ez =
∞
X
zk
k=0
k!
,
sin(t) =
∞
X
k=1
(−1)k−1
t2k−1
,
(2k − 1)!
cos(t) =
it
Insert z = it, to get e = cos(t) + i sin(t).
∞
X
k=0
(−1)k
t2k
.
(2k)!
Note 5.9. eiθ represents a complex number on the unit circle at polar angle θ. eg.
eiπ/2 = i, eiπ/4 = √12 + √i2 .
Example 5.15. A beautiful relationship. eiπ + 1 = 0. This equation has all the
fundamental values, 0, 1, e, π and i in one equation.
Theorem 5.7. A geometric interpretation of multiplication and division
of complex numbers.
a) If z, w ∈ C then zw is a complex number whose modulus is the product of the
moduli of z, w, that is, |zw| = |z||w|, and whose polar angle is the sum of the polar
angles of z and w.
b) If w 6= 0, the quotient z/w is a complex number whose modulus is |z|/|w| and
whose polar angle is the difference of the polar angles of z and w.
Theorem 5.8. de Moivre’s Formula for n-th powers. Let z be a complex
number with exp. polar form z = reiθ . Then for any natural number n,
z n = rn einθ = rn (cos(nθ) + i sin(nθ)).
20
TODD COCHRANE
Example 5.16. (1 + i)10 . Start by writing 1 + i in exp. polar form 1 + i =
Thus
√ π 10
5
π
= 25 ei 2 π = 25 ei 2 = 32i.
(1 + i)10 =
2ei 4
√
π
2ei 4 .
Definition 5.14. Let n ∈ N, z ∈ C. The n-th roots of z denoted z 1/n are the set
of complex numbers w satisfying wn = z.
z 1/n = {w ∈ C : wn = z}.
Recall convention that if x is a nonnegative real number then
nonnegative n-th root of x.
√
n
x denotes the
Example 5.17.
41/2 = {−2,
2}. √
11/4 = {1, −1, i, −i}.
√
√
4
4
4
1/4
1/4
2
= 2·1
= {± 2, ± 2i}.
Theorem 5.9. de Moivre’s Formula for n-th roots: Let z be a complex number
with exp. polar form z = reiθ . Then
√
θ
2π
z 1/n = n rei( n + n k) , with k = 0, 1, 2 . . . , n − 1.
(Technically, it is the set of these values, but the convention is to omit the set
brackets.)
inα
Proof. Let w = ρeiα . Then wn = z is equivalent to ρn e√
= reiθ , which means,
n
n
ρ = r and nα = θ + 2πk, for some k ∈ Z. Thus ρ = r and α = nθ + 2π
n k, for
some k ∈ Z. Although k is allowed to be any integer, the polar angle for w repeats
once k reaches n. Thus the distinct angles are obtained by letting k run from 0 to
n − 1.
Note 5.10. Every nonzero complex number
√ has n distinct n-th roots. They are
equally spaced around the circle of radius n r, centered at the origin.
Example 5.18. a) Find i1/4 . Start with the general exponential polar form of i,
π
i = ei( 2 +2πk) , k ∈ Z. In the general form one allows all possible polar angles for i.
Thus
π
i1/4 = ei( 2 +2πk)
1/4
π
1
π
π
= ei( 2 +2πk) 4 = ei( 8 + 2 k) ,
π
5π
9π
13π
with k = 0, 1, 2,
in these values of k, gives i1/4
= {ei 8 , ei 8 , ei 8 , ei 8 }.
√ 3. Plugging
√
b) Find (− 3 + i)1/5 . √
By plotting the point z = − 3 + i we see that its polar
angle is 65 π. Also, |z| = 3 + 1 = 2. Thus the general exp. polar form of z is
5
2ei( 6 π+2πk) and we obtain,
√
√
5
1
1
2
5
5
z 1/5 = 2ei( 6 π+2πk) 5 = 2ei( 6 π+ 5 πk) ,
with k = 0, 1, 2, 3, 4.
c) Find all solutions of the equation x5 + 2 = 0, with x ∈ C. This is equivalent
to solving the equation x5 = −2, that is x = (−2)1/5 . The general exp. polar form
of −2 is −2 = 2ei(π+2πk) , k ∈ Z. Thus
√
√
1
π
2π
5
5
(−2)1/5 = 2e(iπ+2πk) 5 = 2ei( 5 + 5 k) ,
with k = 0, 1, 2, 3, 4.
ALGEBRAIC SYSTEMS, FALL 2012
21
6. Factoring Polynomials
Definition 6.1. Let F be a field, and F [x] be the set of polynomials with coefficients in F .
a) If f (x) ∈ F [x] we call f (x) a polynomial “over” F .
b) The zero polynomial is the polynomial f (x) = 0 (with all coeff equal to zero).
c) Say f (x) = an xn + · · · + a0 with an 6= 0. Then an is the leading coeff. of f (x),
an xn is the leading term, and n is the degree of f (x).
d) f (x) is called monic if an = 1.
Definition 6.2. Let F be a field.
a) A poly f (x) over F is called reducible over F if f (x) = g(x)h(x) for some
nonconstant polys g(x), h(x). In particular 1 ≤ deg(g), deg(h) < deg(f ).
b) A poly f (x) over F is a called irreducible over F if deg(f ) ≥ 1 and f (x) is
not reducible.
Note 6.1. Thus there are four types of polys in F [x]: 1) Zero, 2) Nonzero constant
polys (these are the units), 3)Reducibles, 4) Irreducibles. Note analogy with Z.
Example 6.1. Determine whether the following are irreducible over the given field,
and if not, factor.
a) 2x + 4 over Q, R
b) x2 − 2 over Q, R, C
c) x2 + 2 over Q, R, C
Definition 6.3. Let f (x), g(x) ∈ F [x]. We say that f (x) divides g(x) in F [x],
written f (x)|g(x) if f (x)h(x) = g(x) for some h(x) ∈ F [x]. f (x) is called a factor
or divisor of g(x), etc. (same language as in Z.).
Example 6.2. Factor x2 + 1 in R, C, Z5 .
Theorem 6.1. Let F be a field and f (x), g(x) ∈ F [x] with g(x) 6= 0. Then there
exist polynomials q(x), r(x) such that f (x) = q(x)g(x) + r(x) with either r(x) = 0
or deg(r(x)) < deg(g(x)). q(x) is called the quotient and r(x) the remainder.
Proof. Sketch. case i: Suppose deg(f ) < deg(g). case ii: Suppose deg(f ) ≥ deg(g).
Say f = an xn + . . . , g = bm xm + . . . , with bm 6= 0. Then in the first step of long
n−m
division we have an (b−1
. Subtract to get smaller degree etc.
m )x
Example 6.3. 2x3 + 3x2 + 1 ÷ x2 − 1, (x2 + 2) ÷ (x − i) in C[x]. (x4 − x + 1) ÷ (x2 + 2)
in Z3 [x].
Note 6.2. f (x)|g(x) iff the remainder in dividing f (x) by g(x) is zero.
Example 6.4. x3 − 1 = (x − 1)(x2 + x + 1) over any field F . Thus (x − 1) and
(x2 + x + 1) are factors of x3 − 1.
Definition 6.4. Let f (x) ∈ F [x]. An element a ∈ F is called a zero or root of f if
f (a) = 0.
Theorem 6.2. Factor Theorem. Let F be a field, f (x) ∈ F [x], a ∈ F . a is a zero
of f iff (x − a) is a factor of f (x).
Proof. Know this one. If (x − a) is a factor then... Converse. Suppose a is a zero.
Strategy, to show (x − a) is a factor show remainder is zero.
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TODD COCHRANE
Example 6.5. Given graph of 4-th degree polynomial with x-intercepts at -2,0,2,
and tangent to x-axis at 0, determine the equation.
Example 6.6. a) Given that x = 3 is a zero of f (x) = x3 − x2 − 4x − 6, factor f (x)
completely over R, and over C.
b) Factor x3 + x + 1 completely over Z3 .
Example 6.7. a) Factor x2 + x + 1 over R, C.
b) Factor x5 + 2 over C.
c) Factor x5 + x2 + x + 1 over Z2 .
Note analogy between Z and F [x]: 1)The four types, primes, composites,
units, zero. 2)Definition of factor. 3) GCD. 4) Division algorithm. 5) Euclidean
Algorithm. 6) GCDLC theorem. 7) Euclid’s Lemma. 8) If p|ab then p|a or p|b. 9)
Unique factorization.
Theorem 6.3. Unique Factorization Theorem for F [x]: Let F be a field and f (x)
be a polynomial over F of degree ≥ 1. Then f (x) can be expressed as a product of
irreducible polynomials over F and this factorization is unique up to the order of
the factors and unit multiples.
Proof. Sketch. Existence: By strong form induction on the degree of f . If f is
irreducible done, otherwise f = gh with g, h of smaller degree.
Uniqueness: Key Lemma. If p(x)|f (x)g(x) and p(x) is irreducible, then p(x)|f (x)
or p(x)|g(x). Then do exactly same proof as for Z. But how to get key Lemma:
Need GCDLC: If f (x), g(x) ∈ F [x] and d(x) = gcd(f, g) then there exist polys
a(x), b(x) such that f (x)a(x) + g(x)b(x) = d(x).
Example 6.8. What do we mean by unique up to unit multiples. Factor x2 − 3x + 2
over R. x2 − 3x + 2 = (x − 1)(x − 2) = (x − 2)(x − 1) = (1 − x)(2 − x) = .. =
(7x − 7)( 71 x − 27 )
Definition 6.5. Let F be a field and f (x) ∈ F [x]. A zero a of f (x) is said to have
multiplicity m if (x − a)m |f (x), but (x − a)m+1 - f (x).
Example 6.9. Suppose f (x) = (x + 1)3 (x − 2)4 (x2 + 1). Over R f (x) has a zero at
-1 of mult 3 and zero at 2 of mult 4. Over C it has additional zeros at ±i each of
mult. 1.
Theorem 6.4. Number of zeros of a polynomial. Let F be any field, f (x) ∈ F [x]
of degree n. Then the total number of zeros of f (x) in F counted with multiplicity
is at most n.
Proof. Let r1 , . . . , rk be the zeros of f (x) in F of mult. m1 , m2 , . . . , mk . Then
f (x) = (x − r1 )m1 (x − r2 )m2 . . . (x − rk )mk g(x) for some polynomial g(x) having no
zero in F . Thus deg(f ) = m1 + m2 + · · · + mk + deg(g) ≥ m1 + m2 + · · · + mk . Theorem 6.5. Some useful factoring formulas for any field F .
a) For any n ∈ N, xn − an = (x − a)(xn−1 + axn−2 + · · · + an−1 ).
n
n−1
n−2
b) For any odd n ∈ N, xn +
+ · · ·√
− an−1 ). √
√a = (x + a)(x 2 − ax
2
c) If F is a field in which −1 exists, then x + a = (x + a −1)(x − a −1).
(ex. F = C, or Zp , with p a prime, p ≡ 1 (mod 4). ex. In Z5 , 22 = −1. In Z13 ,
52 = −1, etc.)
ALGEBRAIC SYSTEMS, FALL 2012
23
√
√
2 exists, then x4 + a4 = (x2 − 2ax + a2 )(x2 +
√ d) If F2 is a field in which
√
2ax + a ), provided that 2 ∈ F . (ex. F = R, Zp with p ≡ ±1 (mod 8). In Z7 ,
32 = 2. In Z17 , 62 = 2. etc.
Proof. a,b,c are basic. For d, suppose that a > 0. Then by de Moivre the zeros are
aw, −aw, aw, −aw, where w = e2πi/8 = √12 + √12 i. Pair the conjugate factors to get
the formula.
Note 6.3. Here√is a trick for sums of 4-th powers: x4 +a4 = x4 +2a2 x2 +a4 −2a2 x2 =
(x2 + a2 )2 − ( 2ax)2 , which is a difference of two squares, and so can be factored
easily.
Example 6.10. Factor x4 + 1 over R and Z7 . Note, there are no zeros, and yet the
polynomial is not irreducible. This can’t happen for cubic or quadratic polys.
Theorem 6.6 (Conjugate Pair Theorem.). Let f (x) be a polynomial with real
coefficients and z be a complex zero of f (x). Then z is also a zero of f (x).
Note 6.4. 1. If z is a real number then z = z and so the conclusion of the theorem
is trivial.
2. The theorem generalizes
√ to other fields. For instance, F = Q. Suppose
f (x) ∈ Q[x]√and that a + b m is a zero of f (x), where m is not a perfect square.
Then a − b m is a zero of f (x). You’ve seen this for quadratic equations.
Theorem 6.7. Irreducibility of a Quadratic or Cubic polynomials: Let f (x) be a
quadratic or cubic polynomial over a field F having no zero in F . Then f (x) is
irreducible over F .
Note: This does not generalize to higher degree polynomials.
Factoring over Q.
Theorem 6.8. Rational Root Test: (Descartes’ Criterion) Let f (x) = an xn + · · · +
a0 be a polynomial over Z and rs be a rational root of f (x) with r, s relatively prime
integers. Then r|a0 and s|an .
Example 6.11. What are the possible rational zeros of 4x3 + 7x − 9.
Example 6.12. Let m ∈ Z such that m is not a perfect cube. Prove that
irrational.
√
3
m is
Example 6.13. Test whether x4 + 2x3 + 17x + 1 is irreducible over Q. Note that
the graph has two x-intercepts (using calculator). Describe the factorization over
R and C. Use Gauss’ test to show it cannot factor as a product of two quadratics
over the rationals.
Theorem 6.9. Gauss’ Test for irreducibility. Let f (x) be a polynomial over Z
such that f (x) is irreducible over Z that is f (x) 6= g(x)h(x) for any polynomials of
positive degree with coeff. in Z. Then f (x) is irreducible over Q.
Factoring over C
Theorem 6.10. Fundamental Theorem of Algebra: Let f (x) be a nonconstant
polynomial over C. Then f (x) has a zero in C.
24
TODD COCHRANE
Proof. Done in Complex Analysis. You first prove that if f (z) is differentiable on
C and bounded |f (z)| ≤ C, then f (z) is a constant function. Apply this result to
1/f (z). If f (z) has no zero in C then it is differentiable everywhere. Furthermore
1/|f (z)| → 0 as |z| → ∞ so it is bounded. Thus it would have to be constant, a
contradiction.
Theorem 6.11. Linear Factorization Theorem for C[x] (Also called FTA) Any
nonconstant polynomial over C can be expressed as a product of linear polynomials
over C. More precisely, if f (x) is a polynomial over C of degree n ≥ 1 with leading
coefficient an , then there exist complex numbers r1 , r2 , . . . , rn such that
f (x) = an (x − r1 )(x − r2 ) . . . (x − rn ).
Corollary 6.1. The only irreducible polynomials over C are linear polynomial.
Factoring over R.
Theorem 6.12. Odd degree over R theorem. Let f (x) be a polynomial of odd
degree over R. Then f (x) has a zero in R.
This is easy to see by looking at the graph, since f (x) → ±∞ as x → ∞, and does
just the opposite as x → −∞. Thus the graph must cross the x-axis.
Theorem 6.13. Factorization Theorem for R[x]: Let f (x) be a polynomial over
R. Then
i) f (x) is irreducible if and only if f (x) is linear, or quadratic with no zero in
R.
ii) In general, if f (x) is of degree n with leading coefficient an and roots r1 , ..., rj ∈
R (allowing repetition), then f (x) has factorization over R,
f (x) = an (x − r1 )(x − r2 ) . . . (x − rj )q1 (x)q2 (x) . . . qk (x),
for some monic irreducible quadratic polynomials q1 (x), ..., qk (x) over R.
Summary of irreducible factors:
1. Over C: only linear polynomials are irreducible.
2. Over R: linear or quadratics with no real zeros, that is, negative discriminants.
3. Over Q and Zp . There are irreducible polynomials of every degree. In general
it is very difficult to tell whether the polynomial is irreducible.
Cardano’s Solution of the Cubic Equation in the year 1545
We wish to solve
x3 + ax2 + bx + c = 0
over C. If we substitute x = y − a/3 we obtain a cubic of the form y 3 + Ax + B = 0
2
2
3
a3
where A = a3 − 2a3 + b, B = a9 − ab
3 + c − 27 . Thus we may assume there is no
x2 term.
Note 6.5. Recall that every complex number z has three cube roots {α, αω, αω},
where α is a particular cube root of z and ω = e2πi/3 . Indeed, if z = reiθ then
√
√
1
θ
2kπ
1
z 3 = 3 rei( 3 + 3 ) , k = 0, 1, 2, and so letting α = 3 reiθ/3 , we see that z 3 =
2
2
{α, αω, αω }. Note that ω = ω.
ALGEBRAIC SYSTEMS, FALL 2012
25
Example 6.14. Solve x3 + x − 1 = 0. Trick. Let x = u + v, to get u3 + v 3 + (3uv +
1)(u + v) = 1. Set 3uv + 1 = 0, u3 + v 3 = 0. The first becomes 27u3 v 3 = −1. Set
U = u3 , V = v 3 , so that we have a system U + V = 1, 27U V = −1, which results
in the quadratic equation 27U 2 − 27U
− 1 = 0. By√symmetry, U, V are the distinct
√
roots of this quadratic: U = 12 + 1893 , V = 21 − 1893 . u, v are cube roots of U, V
such that 3uv = −1, so that uv is real. Let ω = e2πi/3 be a primitive cube root of
unity, and α denote the real cube root of U , β the real cube root of V . Then, in
order to make uv real, we need u = αω k , v = βω −k , k = 0, 1, 2. Note that with
this pairing of u and v we have (using U V = −1/27)
√
3
3uv = 3αω k βω −k = 3αβ = 3 U V = −1.
Finally, x = u + v = α + β, αω + βω, αω + βω.
Cardano’s Solution of the Quartic Equation in 1545 Cardano succeeded
in solving the quartic equation
ax4 + bx3 + cx2 + dx + e = 0,
by reducing it to a cubic equation and then using his formula for the solution of a
cubic.
For the next few hundred years, no further progress was made, that is, no formula
could be obtained for the solution of a fifth degree or higher equation. It was finally
proved by Abel and Ruffini in 1824, that there does not exist a formula for solving
a fifth degree or higher polynomial. In order to succeed in proving this they needed
to create a whole new branch of mathematics, called Group Theory.
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TODD COCHRANE
7. Group Theory
Definition 7.1. A group is a set G with binary operation ∗ such that
i) G is closed under ∗, that is for any x, y ∈ G, x ∗ y ∈ G.
ii) ∗ is associative: For any x, y, z ∈ G, (x ∗ y) ∗ z = x ∗ (y ∗ z).
iii) G has an identity element e satisfying x ∗ e = e ∗ x = x for all x ∈ G.
iv) Inverses exist: For any element x ∈ G there is an element y ∈ G such that
x ∗ y = y ∗ x = e.
If in addition
v) ∗ is commutative, then G is called an abelian group.
Notation: 1. (G, ∗) denotes a group G with binary operation ∗.
2. If + is used, generally 0 is used to denote the identity and −a the inverse of
a.
3. If · is used, 1 is commonly used to denote the identity and a−1 the inverse.
4. Unless indicated otherwise, we shall use multiplicative notation for groups
when stating theorems. Thus a product of two elements a, b ∈ G will simply be
denoted ab, no matter what the binary operation is.
Example 7.1. Examples of additive groups: For any ring R, (R, +) is an abelian
group. For example, (Zm , +), (Z, +), or (M2,2 (R), +).
Example 7.2. Examples of multiplicative groups:
1) (Um , ·), for any m ∈ N. Um is the multiplicative group of units (mod m).
2) (F∗ , ·) where F is any field.
Definition 7.2. A subset H of a group (G, ∗) is called a subgroup of G if H is a
group wrt ∗.
Note: 1. To show a subset is a subgroup it suffices to check properties (i), (iii)
and (iv). Associativity is inherited.
2. If G is a finite set, then suffices to check just (i). One can prove that if (i)
holds then so do (iii) and (iv).
Example 7.3. Find all subgroups of (Z6 , +). 2Z6 = {0, 2, 4}, 3Z6 = {0, 3}, {0} and
Z6 .
Definition 7.3. If (G, ∗) is a group and a ∈ G then
a) For any n ∈ N, an = a ∗ a ∗ · · · ∗ a, n-times and a−n = (an )−1 = a−1 ∗ · · · ∗ a−1 .
b) a0 = e where e is the identity element in G.
c) < a >= {an : n ∈ Z}, called the subgroup of G generated by a.
Lemma 7.1. Laws of Exponents. Let (G, ∗) be a group.
a) For any integers m, n and element a ∈ G, we have an ∗ am = an+m .
b) For any integers m, n and element a ∈ G, we have (an )m = anm .
c) If G is an abelian group, then for any a, b ∈ G and integer n we have (a∗b)n =
an ∗ bn . (Note, this is false for nonabelian groups.)
Lemma 7.2. For any group G and element a ∈ G, the set < a > is a subgroup of
G.
Note 7.1. If + is the binary operation, then < a >= {na : n ∈ Z}.
ALGEBRAIC SYSTEMS, FALL 2012
27
Example 7.4. a) In (Z6 , +), find < 1 >, < 2 >, etc.
b) In (Z, +) find < 3 >.
c) In (U5 , ·), find < 1 >, < 2 >, < 3 >, ..
Definition 7.4. Let G be a group with identity e.
a) The order of a group G is the number of elements in G, denoted |G|; it is
also called the cardinality of G.
b) The order of an element a of a group G, denoted ord(a) is the smallest
positive integer n such that an = e, (if such an n exists.). If no such n exists, a is
said to have infinite order.
Note: In additive notation the definition reads: If (G, +) is a group and a ∈ G
then the order of a is the smallest positive integer n such that na = 0.
Example 7.5. In (U5 , ·), find ord(2). In (Z, +) find ord(2). In (C, ·), find ord(i).
The next theorem gives the connection between the two different usages of the
word “order”.
Theorem 7.1. If G is a group and a ∈ G, then ord(a) = | < a > |. That is, the
order of the element a is the same as the order of the subgroup generated by a.
Definition 7.5. G is called a cyclic group if G =< a > for some a ∈ G. a is called
a generator of G.
Example 7.6. a) Cyclic groups of order 4: 1) (U5 , ·): U5 =< 2 >= {1, 2, 4, 3}. Note
also, U5 =< 3 >= {1, 3, 4, 2}. Thus we see that a cyclic group can have more than
one generator.
2) (Z4 , +): Z4 =< 1 >= {0, 1, 2, 3}.
3) < i >= {1, i, −1, −i} in C.
b) Cyclic groups of order 6: (U7 , ·), (U9 , ·), (Z6 , +), < ω > where ω = e2πi/6 .
Theorem 7.2. Subgroups of Cyclic Groups: Let Cn =< a > be a cyclic group of
order n (under multiplication). Then Cn is an abelian group and
(i) For any positive divisor d of n, there is a unique subgroup of order d given
by Cd =< an/d >. (For an additive group we would have Cd =< nd a >.)
(ii) Every subgroup of Cn is of the type given in part (i).
Example 7.7. Find all subgroups of C12 =< a > and place in a tree diagram. Find
all subgroups of (Z12 , +) and place in a tree diagram. The subgroups are
< a >= C12 , < a2 >= C6 , < a3 >= C4 , < a4 >= C3 , < a6 >= C2 , < e >= C1 ,
one for each divisor of 12. In the tree diagram, a group is placed below another, if
it is a subset of the one above.
Example 7.8. Example of a noncyclic group. Klein 4-group: K4 = Z2 × Z2 , under
addition.
Theorem 7.3. Lagrange’s Theorem: If G is a finite group and H is a subgroup of
G then |H| is a divisor of |G|.
In order to prove Lagrange’s Theorem we need the concept of a coset.
Definition 7.6. Let (G, ·) be a group and H be a subgroup of G. A right coset
of H is a set of the form
Ha := {ha : h ∈ H},
28
TODD COCHRANE
with a a fixed element of G. (Similar definition for left coset.) In additive notation,
if (G, +) is an additive group, then a right coset is denoted
H + a := {h + a : h ∈ H}.
We will just work with right cosets and so will drop the word “right” and just
call them cosets.
Lemma 7.3. a) For any two cosets Ha, Hb of a subgroup H, we either have Ha =
Hb or Ha ∩ Hb = ∅.
2) If H is a subgroup of G, then G can be expressed as a disjoint union of cosets
of H.
Proof. a) Suppose Ha ∩ Hb 6= ∅, say x ∈ Ha ∩ Hb, x = h1 a = h2 b for some
h1 , h2 ∈ H. In particular, ab−1 = h2 h−1
∈ H. We claim that Ha = Hb. Let
1
ha ∈ Ha, with h ∈ H. Note, h(ab−1 ) = h0 for some h0 ∈ H, since H is closed under
multiplication. Thus ha = ha(b−1 b) = (hab−1 )b = h0 b ∈ Hb. Therefore Ha ⊆ Hb.
In a similar manner, Hb ⊆ Ha.
b) We just need to observe that every element g ∈ G is in some coset of H,
namely Hg. Thus the union of all of the cosets of H is G. In part (a) we showed
that distinct cosets are disjoint.
Proof of Lagrange’s Theorem. By the lemma, G = Ha1 ∪ Ha2 · · · ∪ Hak , for some
distinct cosets Ha1 , . . . , Hak of H. We know the number of cosets is finite, since G
is a finite group. Now for any coset Ha of H it is clear that |Ha| = |H| since there
is a 1-to-1 correspondence ha ↔ h, between Ha and H. Thus |G| = k|H|, that is,
|H| is a divisor of |G|.
Corollary 7.1. Suppose that G is a group of order p where p is a prime. Then G
has no nontrivial subgroup, and G is a cyclic group.
Proof. Suppose that H is a subgroup of G. By Lagrange’s Theorem |H| divides p,
and so |H| = 1 or p. If |H| = 1 then H = {e}. If |H| = p then H = G. Thus H is
a trivial subgroup. Now let a ∈ G, a 6= e and let H =< a >. Then H is a subgroup
of G and H 6= {e}, and so H = G. Thus G =< a >, a cyclic group.
Theorem 7.4. Order of elements: If G is a finite group of order n and a ∈ G then
ord(a)|n.
Proof. We simply apply Lagrange’s Theorem to the subgroup H =< a >. By a
theorem above, ord(a) = |H|, and by Lagrange’s Theorem, |H| is a divisor of n.
Thus ord(a) is a divisor of n.
As an immediate consequence of this theorem, we obtain Euler’s Theorem and
Fermat’s Little Theorem.
Theorem 7.5. Euler’s Theorem. Let m be a positive integer and Um be the group
of units (mod m). Then, for any a ∈ Um , we have aφ(m) = 1, where φ(m) is the
Euler phi-function.
Proof. Recall that |Um | = φ(m). Let a ∈ Um . Say ord(a) = n. Then, by the
preceding theorem, n|φ(m). Say nk = φ(m) for some k ∈ N. Thus aφ (m) = ank =
(an )k = 1k = 1.
ALGEBRAIC SYSTEMS, FALL 2012
29
Note that a ∈ Um implies that gcd(a, m) = 1. Thus, in the language of congruences, Euler’s Theorem states that for any integer a with gcd(a, m) = 1, we have
aφ(m) ≡ 1 (mod m). Fermat’s Little Theorem is just the special case that m = p,
a prime.
Theorem 7.6. Fermat’s Little Theorem. Let p be a prime, and Up be the group of
units (mod p). Then, for any a ∈ Up , we have ap−1 = 1.
7.1. Permutation Groups.
Definition 7.7. Let S = {1, 2, . . . , n}.
1. A permutation of S is a 1-to-1 function σ from S into itself. (Recall σ is
1-to-1 if σ(i) 6= σ(j) for i 6= j.)
2. The identity function on S, denoted ι, is the function satisfying i(k) = k for
all k ∈ S.
3. The n-th symmetric group Sn is the set of all permutations of S, with binary
operation being function composition and identity ι.
Note: i) The composition symbol generally is dropped when working in Sn . Thus
στ = σ ◦ τ .
ii) Function composition is not commutative, that is, στ 6= τ σ, in general, as the
following example shows.
1 2 3 4 5
1 2 3 4 5
Example 7.9. Let σ =
,τ =
. Then σ, τ ∈ S5
2 3 5 4 1
2 1 4 3 5
and
1 2 3 4 5
1 2 3 4 5
στ =
, τσ =
.
3 2 4 5 1
1 4 5 3 2
In particular στ 6= τ σ. Next, lets find σ −1 :
1 2 3 4 5
σ −1 =
.
5 1 2 4 3
Theorem 7.7. For any natural number n, Sn is a group with binary operation
being function composition, and identity element ι.
Proof. Check the 4 axioms: 1. Composition of 1-to-1 functions is 1-to-1, so Sn is
closed under composition.
2. Function composition is always associative: (f ◦ g) ◦ h(x) = (f ◦ g)(h(x)) =
f (g(h(x))) while f ◦ (g ◦ h)(x) = f (g ◦ h(x)) = f (g(h(x)), the same thing.
3. ι is the identity element, satisfying ισ = σ = σι for any σ ∈ Sn .
4. Any 1-to-1 function f has an inverse function denoted f −1 .
Theorem 7.8. Sn is a group of order n!. For n ≥ 3, Sn is nonabelian.
Proof. Let σ ∈ Sn . There are n choices for σ(1), leaving (n − 1) choices for σ(2),
(n − 2) choices for σ(3) and so on. Thus altogether there are n! choices for σ. To
show Sn is nonabelian for n ≥ 3, let σ = (1, 2, 3), τ = (1, 2) (in cycle-notation).
Then στ 6= τ σ.
Cycle Notation: A k-cycle is a cyclical permutation of the form σ = (n1 , n2 , . . . , nk ),
meaning σ(n1 ) = n2 , . . . , σ(nk ) = n1 and all other integer values n are fixed by
σ, that is , σ(n) = n. 2-cycles are called transpositions. A set of cycles are called
disjoint if no number is repeated more than once in all of the cycles.
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TODD COCHRANE
Note: 1. Convention: If a number is not presented in a cycle, it is understood
to be fixed.
2. Cycles have multiple representations: For example, (1, 2, 3) = (2, 3, 1) =
(3, 1, 2).
Example 7.10. 1. What does σ = (1, 4, 3) mean, viewed as an element of S4 ; as
element of S6 . Find σ −1 , σ 2 , σ 3 . Note ord(σ) = 3.
2. Express the following as a product of disjoint cycles.
1 2 3 4 5 6 7 8 9
σ=
4 7 3 1 6 5 9 2 8
σ = (1, 4)(2, 7, 9, 8)(5, 6). Note, disjoint cycles can be placed in any order. They
commute since they consist of distinct integers. Thus we also have
σ = (1, 4)(5, 6)(2, 7, 9, 8), etc.
Example 7.11. If a set of cycles is not disjoint, then their product can be simplified.
Find product (1, 3, 5)(2, 4, 5, 6)(3, 5) in S6 . Answer: (1, 3, 6, 2, 4).
Theorem 7.9. Every element in Sn can be expressed as a product of disjoint kcycles.
Example 7.12. S3 = {ι, σ, σ 2 , τ, στ, σ 2 τ }, where σ = (1, 2, 3) and τ = (1, 2). (The
transposition (1,2) could be replaced with (1,3) or (2,3) here.) Note S3 is a nonabelian group of order 6=3!.
Theorem 7.10. (i) If σ is a k-cycle then ord(σ)=k.
(ii) More generally, for any permutation σ, ord(σ) is the least common multiple
of the length of its cycles when σ is written as a product of disjoint cycles.
Theorem 7.11. i) Every element of Sn can be expressed as a product of transpositions.
ii) The number of transpositions in such an expression is not unique, but the
parity (even/odd) of the number of transpositions is unique.
Example 7.13. (1, 2, 3) = (1, 3)(1, 2) = (2, 3)(1, 2)(2, 3)(1, 2),
(3, 5, 2, 7, 4) = (3, 4)(3, 7)(3, 2)(3, 5). In general, first express a permutation as a
product of disjoint cycles and then break up the cycles into a product of transpositions.
Definition 7.8. i) A permutation is called even if it can be expressed as a product
of an even number of transpositions, and odd if it can be expressed as a product of
an odd number of transpositions.
ii) The set of all even permutations, denoted An , is a subgroup of Sn called
the alternating group of degree n. To see that it is a group note that the product
of two even permutations is even, since an even number plus an even number is
even. Also, if σ = τ1 τ2 · · · τk , a product of an even number of transpositions, then
σ −1 = τk τk−1 · · · τ2 τ1 , also a product of an even number of transpositions. The
identity element ι is a product of zero transpositions, and so it is in An .
Note: i) |An | = n!/2, for n ≥ 1. An is nonabelian for n ≥ 4.
ii) It is the study of the alternating group A5 that led to the Abel-Ruffini Theorem that there is no formula in radicals for solving a fifth degree polynomial
equation. One needs to study the permutations of the zeros of a fifth degree polynomial.
ALGEBRAIC SYSTEMS, FALL 2012
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Definition 7.9. Dihedral group: The dihedral group Dn is the group of symmetries
of a regular n-gon.
Example 7.14. 1. D3 is the group of symmetries of an equilateral triangle. We have
D3 =< σ, τ >, where σ = (1, 2, 3) clockwise rotation 120◦ and τ = (2, 3) is flip or
reflection about vertical axis, etc. Note D3 = S3 .
2. D4 is the group of symmetries of a square.
D4 =< σ, τ >= {ι, σ, σ 2 , σ 3 , τ, στ, σ 2 τ, σ 3 τ },
where σ = (1, 2, 3, 4), τ = (1, 2)(3, 4). Note |D4 | = 8, D4 6= S4 .
Example 7.15. Rectangle, not a square. Here the group of symmetries is {<
ι, σ, τ, στ } where σ = (1, 3)(2, 4), 180 deg rotation, τ = (1, 2)(3, 4) a reflection
in the vertical axis, στ = (1, 4)(2, 3) a reflection in the horizontal axis. This is the
Klein-4 group, not a dihedral group. It is an abelian group of order 4.
The Dihedral group Dn . Let n ≥ 3 and P be a regular n-gon with vertices labeled 1,2,3,...,n running in a clockwise direction. Then Dn =< σ, τ >=
{ι, σ, σ 2 , . . . , σ n−1 , τ, στ, . . . , σ n−1 τ } where σ = (1, 2, 3, . . . , n), a clockwise rotation
of P through 360/n degrees, and τ is a reflection of P through any one of its axes of
symmetry. There are n rotation symmetries, σ j , j = 0, 1, 2, ..., n−1 and n reflection
symmetries σ j τ , j = 0, 1, 2, ..., n − 1.
Theorem 7.12. Dn is a nonabelian group of order 2n.
Isomorphism. Note we have seen a number of different examples of cyclic
groups of order 4: (Z4 , +) =< 1 >, (< i >, ·) in C, (U5 , ·), (U10 ·), < (1, 2, 3, 4) >
in S4 , etc. These are called isomorphic groups, meaning they have all the same
algebraic properties. On the other hand the Klein 4 group is really different, it has
no element of order 4.
Definition 7.10. Two groups G, H are said to be isomorphic if there is a 1-to-1
function f from G onto H such that f (x ∗ y) = f (x) ∗ f (y) for all x, y ∈ G.
Necessary conditions for two groups G, H to be isomorphic:
1. |G| = |H|
2. H and G have the same number of elements of order n, for any positive integer
n.
3. H and G have the same number of subgroups of order n for any positive
integer n.
Example 7.16. Show < i > is isomorphic to (Z4 , +). Pf. Let f (ik ) = [k]4 .
Theorem 7.13. Any two cyclic groups of the same order are isomorphic.
The goal is to classify all the different types of groups of a given order:
1. If p is a prime there is only one type of group of order p, a cyclic group.
2. There are two types of groups of order 4: cyclic and Klein-4 groups. We’ve
seen several examples of Klein-4 groups, such as U12 , U8 , Symmetries of a Rectangle
< σ, τ >, Z2 × Z2 .
3. There are two types of groups of order 6: cyclic and S3 .
4. There are five types of groups of order 8 :
Abelian: C2 × C2 × C2 , C2 × C4 , C8
Nonabelian: D4 ; Q = Quaternion group={±1, ±i, ±j, ±k} where i2 = j 2 = k 2 =
−1, ij = k, jk = i, ki = j.