Mathematical Proof of Division Algorithm and Group Theory Concepts, Study notes of Local Government Studies

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Part II

Math 103A Lecture Notes

Lecture 1 (1/5/2009)

Notation 1.1 Introduce N := { 0 , 1 , 2 ,... } , Z, Q, R, and C. Also let Z+ := N \ { 0 }.

• Set notations.
• Recalled basic notions of a function being one to one, onto, and invertible. Think of functions in terms of a bunch of arrows from the domain set to the range set. To find the inverse function you should reverse the arrows.
• Some example of groups without the definition of a group:
  1. GL 2 (R) =

g :=

[

a b c d

]

: det g = ad − bc 6 = 0

1. Vector space with “group” operation being addition.
2. The permutation group of invertible functions on a set S like S = { 1 , 2 ,... , n}.

1.1 A Little Number Theory

Axiom 1.2 (Well Ordering Principle) Every non-empty subset, S, of N contains a smallest element.

We say that a subset S ⊂ Z is bounded below if S ⊂ [k, ∞) for some k ∈ Z and bounded above if S ⊂ (−∞, k] for some k ∈ Z.

Remark 1.3 (Well ordering variations). The well ordering principle may also be stated equivalently as:

1. any subset S ⊂ Z which is bounded from below contains a smallest element or
2. any subset S ⊂ Z which is bounded from above contains a largest element.

To see this, suppose that S ⊂ [k, ∞) and then apply the well ordering principle to S − k to find a smallest element, n ∈ S − k. That is n ∈ S − k and n ≤ s − k for all s ∈ S. Thus it follows that n + k ∈ S and n + k ≤ s for all s ∈ S so that n + k is the desired smallest element in S. For the second equivalence, suppose that S ⊂ (−∞, k] in which case −S ⊂ [−k, ∞) and therefore there exist a smallest element n ∈ −S, i.e. n ≤ −s for all s ∈ S. From this we learn that −n ∈ S and −n ≥ s for all s ∈ S so that −n is the desired largest element of S.

Theorem 1.4 (Division Algorithm). Let a ∈ Z and b ∈ Z+, then there exists unique integers q ∈ Z and r ∈ N with r < b such that

a = bq + r.

(For example,

5

2 | 12 10 2

so that 12 = 2 · 5 + 2.)

Proof. Let S := {k ∈ Z : a − bk ≥ 0 } which is bounded from above. Therefore we may define,

q := max {k : a − bk ≥ 0 }.

As q is the largest element of S we must have,

r := a − bq ≥ 0 and a − b (q + 1) < 0.

The second inequality is equivalent to r − b < 0 which is equivalent to r < b. This completes the existence proof. To prove uniqueness, suppose that a = bq′^ +r′^ in which case, bq′^ +r′^ = bq +r and hence, b > |r′^ − r| = |b (q − q′)| = b |q − q′|. (1.1) Since |q − q′| ≥ 1 if q 6 = q′, the only way Eq. (1.1) can hold is if q = q′^ and r = r′.

Axiom 1.5 (Strong form of mathematical induction) Suppose that S ⊂ Z is a non-empty set containing an element a with the property that; if [a, n) ∩ Z ⊂ S then n ∈ Z, then [a, ∞) ∩ Z ⊂ S.

Axiom 1.6 (Weak form of mathematical induction) Suppose that S ⊂ Z is a non-empty set containing an element a with the property that for ev- ery n ∈ S with n ≥ a, n + 1 ∈ S, then [a, ∞) ∩ Z ⊂ S.

Lecture 2 (1/7/2009)

Definition 2.1. Given a, b ∈ Z with a 6 = 0 we say that a divides b or a is a divisor of b (write a|b) provided b = ak for some k ∈ Z.

Definition 2.2. Given a, b ∈ Z with |a| + |b| > 0 , we let

gcd (a, b) := max {m : m|a and m|b}

be the greatest common divisor of a and b. (We do not define gcd (0, 0) and we have gcd (0, b) = |b| for all b ∈ Z\ { 0 } .) If gcd (a, b) = 1, we say that a and b are relatively prime.

Remark 2.3. Notice that gcd (a, b) = gcd (|a| , |b|) ≥ 0 and gcd (a, 0) = 0 for all a 6 = 0.

Lemma 2.4. Suppose that a, b ∈ Z with b 6 = 0. Then gcd (a + kb, b) = gcd (a, b) for all k ∈ Z.

Proof. Let Sk denote the set of common divisors of a + kb and b. If d ∈ Sk, then d|b and d| (a + kb) and therefore d|a so that d ∈ S 0. Conversely if d ∈ S 0 , then d|b and d|a and therefore d|b and d| (a + kb) , i.e. d ∈ Sk. This shows that Sk = S 0 , i.e. a + kb and b and a and b have the same common divisors and hence the same greatest common divisors. This lemma has a very useful corollary.

Lemma 2.5 (Euclidean Algorithm). Suppose that a, b are positive integers with a < b and let b = ka + r with 0 ≤ r < a by the division algorithm. Then gcd (a, b) = gcd (a, r) and in particular if r = 0, we have

gcd (a, b) = gcd (a, 0) = a.

Example 2.6. Suppose that a = 15 = 3 · 5 and b = 28 = 2^2 · 7. In this case it is easy to see that gcd (15, 28) = 1. Nevertheless, lets use Lemma 2.5 repeatedly as follows;

28 = 1 · 15 + 13 so gcd (15, 28) = gcd (13, 15) , (2.1) 15 = 1 · 13 + 2 so gcd (13, 15) = gcd (2, 13) , (2.2) 13 = 6 · 2 + 1 so G gcd (2, 13) = gcd (1, 2) , (2.3) 2 = 2 · 1 + 0 so gcd (1, 2) = gcd (0, 1) = 1. (2.4)

Moreover making use of Eqs. ( 2.1–2.3) in reverse order we learn that,

1 = 13 − 6 · 2 = 13 − 6 · (15 − 1 · 13) = 7 · 13 − 6 · 15 = 7 · (28 − 1 · 15) − 6 · 15 = 7 · 28 − 13 · 15.

Thus we have also shown that

1 = s · 28 + t · 15 where s = 7 and t = − 13.

The choices for s and t used above are certainly not unique. For example we have, 0 = 15 · 28 − 28 · 15 which added to 1 = 7 · 28 − 13 · 15 implies,

1 = (7 + 15) · 28 − (13 + 28) · 15 = 22 · 28 − 41 · 15

as well.

Example 2.7. Suppose that a = 40 = 2^3 · 5 and b = 52 = 2^2 · 13. In this case we have gcd (40, 52) = 4. Working as above we find,

52 = 1 · 40 + 12 40 = 3 · 12 + 4 12 = 3 · 4 + 0

so that we again see gcd (40, 52) = 4. Moreover,

4 = 40 − 3 · 12 = 40 − 3 · (52 − 1 · 40) = 4 · 40 − 3 · 52.

So again we have shown gcd (a, b) = sa + tb for some s, t ∈ Z, in this case s = 4 and t = 3.

12 2 Lecture 2 (1/7/2009)

Example 2.8. Suppose that a = 333 = 3^2 · 37 and b = 459 = 3^3 · 17 so that gcd (333, 459) = 3^2 = 9. Repeated use of Lemma 2.5 gives,

459 = 1 · 333 + 126 so gcd (333, 459) = gcd (126, 333) , (2.5) 333 = 2 · 126 + 81 so gcd (126, 333) = gcd (81, 126) , (2.6) 126 = 81 + 45 so gcd (81, 126) = gcd (45, 81) , (2.7) 81 = 45 + 36 so gcd (45, 81) = gcd (36, 45) , (2.8) 45 = 36 + 9 so gcd (36, 45) = gcd (9, 36) , and (2.9) 36 = 4 · 9 + 0 so gcd (9, 36) = gcd (0, 9) = 9. (2.10)

Thus we have shown that

gcd (333, 459) = 9.

We can even say more. From Eq. (2.10) we have, 9 = 45 − 36 and then from Eq. (2.10), 9 = 45 − 36 = 45 − (81 − 45) = 2 · 45 − 81.

Continuing up the chain this way we learn,

9 = 2 · (126 − 81) − 81 = 2 · 126 − 3 · 81 = 2 · 126 − 3 · (333 − 2 · 126) = 8 · 126 − 3 · 333 = 8 · (459 − 1 · 333) − 3 · 333 = 8 · 459 − 11 · 333

so that 9 = 8 · 459 − 11 · 333.

The methods of the previous two examples can be used to prove Theorem 2.9 below. However, we will two different variants of the proof.

Theorem 2.9. If a, b ∈ Z\ { 0 }, then there exists (not unique) numbers, s, t ∈ Z such that gcd (a, b) = sa + tb. (2.11)

Moreover if m 6 = 0 is any common divisor of both a and b then m| gcd (a, b).

Proof. If m is any common divisor of a and b then m is also a divisor of sa + tb for any s, t ∈ Z. (In particular this proves the second assertion given the truth of Eq. (2.11).) In particular, gcd (a, b) is a divisor of sa + tb for all s, t ∈ Z. Let S := {sa + tb : s, t ∈ Z} and then define

d := min (S ∩ Z+) = sa + tb for some s, t ∈ Z. (2.12)

By what we have just said if follows that gcd (a, b) |d and in particular d ≥ gcd (a, b). If we can snow d is a common divisor of a and b we must then have d = gcd (a, b). However, using the division algorithm,

a = kd + r with 0 ≤ r < d. (2.13)

As r = a − kd = a − k (sa + tb) = (1 − ks) a − ktb ∈ S ∩ N, if r were greater than 0 then r ≥ d (from the definition of d in Eq. (2.12) which would contradict Eq. (2.13). Hence it follows that r = 0 and d|a. Similarly, one shows that d|b.

Lemma 2.10 (Euclid’s Lemma). If gcd (c, a) = 1, i.e. c and a are relatively prime, and c|ab then c|b.

Proof. We know that there exists s, t ∈ Z such that sa+tc = 1. Multiplying this equation by b implies, sab + tcb = b. Since c|ab and c|cb, it follows from this equation that c|b.

Corollary 2.11. Suppose that a, b ∈ Z such that there exists s, t ∈ Z with 1 = sa + tb. Then a and b are relatively prime, i.e. gcd (a, b) = 1.

Proof. If m > 0 is a divisor of a and b, then m| (sa + tb) , i.e. m|1 which implies m = 1. Thus the only positive common divisor of a and b is 1 and hence gcd (a, b) = 1.

2.1 Ideals (Not covered in class.)

Definition 2.12. As non-empty subset S ⊂ Z is called an ideal if S is closed under addition (i.e. S + S ⊂ S) and under multiplication by any element of Z, i.e. Z · S ⊂ S.

Example 2.13. For any n ∈ Z, let

(n) := Z · n = nZ := {kn : k ∈ Z}.

I is easily checked that (n) is an ideal. The next theorem states that this is a listing of all the ideals of Z.

Theorem 2.14 (Ideals of Z). If S ⊂ Z is an ideal then S = (n) for some n ∈ Z. Moreover either S = { 0 } in which case n = 0 for S 6 = { 0 } in which case n = min (S ∩ Z+).

Proof. If S = { 0 } we may take n = 0. So we may assume that S contains a non-zero element a. By assumption that Z · S ⊂ S it follows that −a ∈ S as well and therefore S ∩ Z+ is not empty as either a or −a is positive. By the well ordering principle, we may define n as, n := min S ∩ Z+.

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Lecture 3 (1/9/2009)

3.1 Prime Numbers

Definition 3.1. A number, p ∈ Z, is prime iff p ≥ 2 and p has no divisors other than 1 and p. Alternatively put, p ≥ 2 and gcd (a, p) is either 1 or p for all a ∈ Z.

Example 3.2. The first few prime numbers are 2, 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 ,....

Lemma 3.3 (Euclid’s Lemma again). Suppose that p is a prime number and p|ab for some a, b ∈ Z then p|a or p|b.

Proof. We know that gcd (a, p) = 1 or gcd (a, p) = p. In the latter case p|a and we are done. In the former case we may apply Euclid’s Lemma 2.10 to conclude that p|b and so again we are done.

Theorem 3.4 (The fundamental theorem of arithmetic). Every n ∈ Z with n ≥ 2 is a prime or a product of primes. The product is unique except for the order of the primes appearing the product. Thus if n ≥ 2 and n = p 1... pn = q 1... qm where the p’s and q’s are prime, then m = n and after renumbering the q’s we have pi = qi.

Proof. Existence: This clearly holds for n = 2. Now suppose for every 2 ≤ k ≤ n may be written as a product of primes. Then either n + 1 is prime in which case we are done or n + 1 = a · b with 1 < a, b < n + 1. By the induction hypothesis, we know that both a and b are a product of primes and therefore so is n + 1. This completes the inductive step. Uniqueness: You are asked to prove the uniqueness assertion in 0.#25. Here is the solution. Observe that p 1 |q 1... qm. If p 1 does not divide q 1 then gcd (p 1 , q 1 ) = 1 and therefore by Euclid’s Lemma 2.10, p 1 | (q 2... qm). It now follows by induction that p 1 must divide one of the qi, by relabeling we may assume that q 1 = p 1. The result now follows by induction on n ∨ m.

Definition 3.5. The least common multiple of two non-zero integers, a, b, is the smallest positive number which is both a multiple of a and b and this number will be denoted by lcm (a, b). Notice that m = min ((a) ∩ (b) ∩ Z+).

Example 3.6. Suppose that a = 12 = 2^2 ·3 and b = 15 = 3· 5. Then gcd (12, 15) = 3 while

lcm (12, 15) =

· 5 = 2^2 · (3 · 5) =

Observe that

gcd (12, 15) · lcm (12, 15) = 3 ·

This is a special case of Chapter 0.#12 on p. 23 which can be proved by similar considerations. In general if

a = pn 1 1 · · · · · pn k kand b = pm 1 1... pm k kwith nj , ml ∈ N

then

gcd (a, b) = pn 1 1 ∧m^1 · · · · · pn k k^ ∧mkand lcm (a, b) = pn 1 1 ∨m^1 · · · · · pn kk ∨mk.

Therefore,

gcd (a, b) · lcm (a, b) = pn 1 1 ∧m^1 +n^1 ∨m^1 · · · · · pn kk^ ∧mk^ +nk^ ∨mk = pn 1 1 +m^1 · · · · · pn kk +mk= a · b.

3.2 Modular Arithmetic

Definition 3.7. Let n be a positive integer and let a = qan+ra with 0 ≤ ra < n. Then we define a mod n := ra. (Sometimes we might write a = ra mod n – but I will try to stick with the first usage.)

Lemma 3.8. Let n ∈ Z+ and a, b, k ∈ Z. Then:

1. (a + kn) mod n = a mod n.
2. (a + b) mod n = (a mod n + b mod n) mod n.
3. (a · b) mod n = ((a mod n) · (b mod n)) mod n.

Proof. Let ra = a mod n, rb = b mod n and qa, qb ∈ Z such that a = qan+ra and b = qbn + rb.

16 3 Lecture 3 (1/9/2009)

1. Then a + kn = (qa + k) n + ra and therefore,

(a + kn) mod n = ra = a mod n.

1. a + b = (qa + qb) n + ra + rb and hence by item 1 with k = qa + qb we find,

(a + b) mod n = (ra + rb) mod n. = (a mod n + b mod n) mod n.

1. For the last assertion,

a · b = [qan + ra] · [qbn + rb] = (qaqbn + raqb + rbqa) n + ra · rb

and so again by item 1. with k = (qaqbn + raqb + rbqa) we have,

(a · b) mod n = (ra · rb) mod n = ((a mod n) · (b mod n)) mod n.

Example 3.9. Take n = 4, a = 18 and b = 7. Then 18 mod 4 = 2 and 7 mod 4 =

1. On one hand,

(18 + 7) mod 4 = 25 mod 4 = 1 while on the other, (2 + 3) mod 4 = 1.

Similarly, 18 · 7 = 126 = 4 · 31 + 2 so that

(18 · 7) mod 4 = 2 while (2 · 3) mod 4 = 6 mod 4 = 2.

Remark 3.10 (Error Detection). Companies often add extra digits to identi- fication numbers for the purpose of detecting forgery or errors. For example the United Parcel Service uses a mod 7 check digit. Hence if the identification number were n = 354691332 one would append

n mod 7 = 354691332 mod 7 = 2 to the number to get 354691332 2 (say).

See the book for more on this method and other more elaborate check digit schemes. Note, 354691332 = 50 670 190 · 7 + 2.

Remark 3.11. Suppose that a, n ∈ Z+ and b ∈ Z, then it is easy to show (you prove) (ab) mod (an) = a · (b mod n).

Example 3.12 (Computing mod 10). We have,

123456 mod 10 = 6 123456 mod 100 = 56 123456 mod 1000 = 456 123456 mod 10000 = 3456 123456 mod 100000 = 23456 123456 mod 1000000 = 123456

so that an... a 2 a 1 mod 10k^ = ak... a 2 a 1 for all k ≤ n.

Solution to Exercise (0.52). As an example, here is a solution to Problem

0.52 of the book which states that

k times ︷ ︸︸ ︷ 111... 1 is not the square of an integer except when k = 1. As 11 is prime we may assume that k ≥ 3. By Example 3.12, 111... 1 mod 10 = 1 and 111... 1 mod 100 = 11. Hence 1111... 1 = n^2 for some integer n, we must have

n^2 mod 10 = 1 and

n^2 − 1

mod 100 = 10.

The first condition implies that n mod 10 = 1 or 9 as 1^2 = 1 and 9^2 mod 10 = 81 mod 10 = 1. In the first case we have, n = k · 10 + 1 and therefore we must require,

n^2 − 1

mod 100 =

[

(k · 10 + 1)^2 − 1

]

mod 100 =

k^2 · 100 + 2k · 10

mod 100 = (2k · 10) mod 100 = 10 · (2k mod 10)

which implies 1 = (2k mod 10) which is impossible since 2k mod 10 is even. For the second case we must have,

n^2 − 1

mod 100 mod 100 =

[

(k · 10 + 9)^2 − 1

]

mod 100

=

k^2 · 100 + 18k · 10 + 81 − 1

mod 100 = ((10 + 8) k · 10 + 8 · 10) mod 100 = (8 (k + 1) · 10) mod 100 = 10 · 8 k mod 10

which implies which 1 = (8k mod 10) which again is impossible since 8k mod 10 is even.

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Lecture 4 (1/12/2009)

Theorem 4.1. Let R or ∼ be an equivalence relation on S and for each a ∈ S, let [a] := {x ∈ S : a ∼ x}

be the equivalence class of a.. Then S is partitioned by its distinct equivalence classes.

Proof. Because ∼ is reflexive, a ∈ [a] for all a and therefore every element a ∈ S is a member of its own equivalence class. Thus to finish the proof we must show that distinct equivalence classes are disjoint. To this end we will show that if [a] ∩ [b] 6 = ∅ then in fact [a] = [b]. So suppose that c ∈ [a] ∩ [b] and x ∈ [a]. Then we know that a ∼ c, b ∼ c and a ∼ x. By reflexivity and transitivity of ∼ we then have, x ∼ a ∼ c ∼ b, and hence b ∼ x,

which shows that x ∈ [b]. Thus we have shown [a] ⊂ [b]. Similarly it follows that [b] ⊂ [a].

Exercise 4.1. Suppose that S = Z with a ∼ b iff a mod n = b mod n. Identify the equivalence classes of ∼. Answer,

{[0] , [1] ,... , [n − 1]}

where [i] = i + nZ = {i + ns : s ∈ Z}.

Exercise 4.2. Suppose that S = R^2 with a = (a 1 , a 2 ) ∼ b = (b 1 , b 2 ) iff |a| = |b| where |a| := a^21 + a^22. Show that ∼ is an equivalence relation and identify the equivalence classes of ∼. Answer, the equivalence classes consists of concentric circles centered about the origin (0, 0) ∈ S.

4.1 Binary Operations and Groups – a first look

Definition 4.2. A binary operation on a set S is a function, ∗ : S × S → S. We will typically write a ∗ b rather than ∗ (a, b).

Example 4.3. Here are a number of examples of binary operations.

1. S = Z and ∗ = “ + ”
2. S = {odd integers} and ∗ = “ + ” is not an example of a binary operator since 3 ∗ 5 = 3 + 5 = 8 ∈/ S.
3. S = Z and ∗ =“·”
4. S = R\ { 0 } and ∗ =“·”
5. S = R\ { 0 } with ∗ = “\” = “ ÷ ”.
6. Let S be the set of 2 × 2 real (complex) matrices with A ∗ B := AB.

Definition 4.4. Let ∗ be a binary operation on a set S. Then;

1. ∗ is associative if (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c ∈ S.
2. e ∈ S is an identity element if e ∗ a = a = a ∗ e for all a ∈ S.
3. Suppose that e ∈ S is an identity element and a ∈ S. We say that b ∈ S is an inverse to a if b ∗ a = e = a ∗ b.
4. ∗ is commutative if a ∗ b = b ∗ a for all a, b ∈ S.

Definition 4.5 (Group). A group is a triple, (G, ∗, e) where ∗ is an associa- tive binary operation on a set, G, e ∈ G is an identity element, and each g ∈ G has an inverse in G. (Typically we will simply denote g ∗ h by gh.)

Definition 4.6 (Commutative Group). A group, (G, e) , is commutative if gh = hg for all h, g ∈ G.

Example 4.7 ((Z, +)). One easily checks that (Z, ∗ = +) is a commutative group with e = 0 and the inverse to a ∈ Z is −a. Observe that e ∗ a = e + a = a for all a iff e = 0.

Example 4.8. S = Z and ∗ =“·” is an associative, commutative, binary oper- ation with e = 1 being the identity. Indeed e · a = a for all a ∈ Z implies e = e · 1 = 1. This is not a group since there are no inverses for any a ∈ Z with |a| ≥ 2.

Example 4.9 ((R\ { 0 } , ·)). G = R\ { 0 } =: R∗, and ∗ =“·” is a commutative group, e = 1, an inverse to a is 1/a.

Example 4.10. S = R\ { 0 } with ∗ = “\” = “ ÷ ”. In this case ∗ is not associative since

a ∗ (b ∗ c) = a/ (b/c) = ac b

while

(a ∗ b) ∗ c = (a/b) /c = a bc

It is also not commutative since a/b 6 = b/a in general. There is no identity element e ∈ S. Indeed, e ∗ a = a = a ∗ e, we would imply e = a^2 for all a 6 = 0 which is impossible, i.e. e = 1 and e = 4 at the same time.

Example 4.11. Let S be the set of 2 × 2 real (complex) matrices with A ∗ B := AB. This is a non-commutative binary operation which is associative and has an identity, namely

e :=

[

]

It is however not a group only those A ∈ S with det A 6 = 0 admit an inverse.

Example 4.12 (GL 2 (R)). Let G := GL 2 (R) be the set of 2 × 2 real (complex)

matrices such that det A 6 = 0 with A ∗ B := AB is a group with e :=

[

]

and

the inverse to A being A−^1. This group is non-abeliean for example let

A :=

[

]

and B =

[

]

then

AB =

[

] [

]

[

]

while

BA =

[

] [

]

[

]

6 = AB.

Example 4.13 (SL 2 (R)). Let SL 2 (R) = {A ∈ GL 2 (R) : det A = 1}. This is a group since det (AB) = det A · det B = 1 if A, B ∈ SL 2 (R).

(brackets involving g 1... gn)·gn+1 = Mn (g 1 ,... , gn) gn+1 = Mn+1 (g 1 ,... , gn+1) ,

wherein we used induction in the first equality and the definition of Mn+1 in the second. Now suppose the assertion holds for some k ≥ 0 and consider the case where there are k + 1 parentheses appearing on the right of this expression,

i.e.... g

k+ n

)... ). Using the associativity law for the last bracket on the right we can transform this expression into one with only k parentheses appearing

on the right. It then follows by the induction hypothesis, that... g

k+ n

Mn+1 (g 1 ,... , gn+1).

Notation 5.8 For n ∈ Z and g ∈ G, let gn^ :=

n times ︷ ︸︸ ︷ g... g and g−n^ :=

n times ︷ ︸︸ ︷

(^ g−^1... g−^1 = g−^1

)n if n ≥ 1 and g^0 := e.

Observe that with this notation that gmgn^ = gm+n^ for all m, n ∈ Z. For example,

g^3 g−^5 = gggg−^1 g−^1 g−^1 g−^1 g−^1 = ggg−^1 g−^1 g−^1 g−^1 = gg−^1 g−^1 g−^1 = g−^1 g−^1 = g−^2.

5.2 More Examples of Groups

Example 5.9. Let G be the set of 2 × 2 real (complex) matrices with A ∗ B := A + B. This is a group. In fact any vector space under addition is an abelian group with e = 0 and v−^1 = −v.

Example 5.10 (Zn). For any n ≥ 2 , G := Zn = { 0 , 1 , 2 ,... , n − 1 } with a ∗ b = (a + b) mod n is a commutative group with e = 0 and the inverse to a ∈ Zn being n − a. Notice that (n − a + a) mod n = n mod n = 0.

Example 5.11. Suppose that S = { 0 , 1 , 2 ,... , n − 1 } with a ∗ b = ab mod n. In this case ∗ is an associative binary operation which is commutative and e = 1 is an identity for S. In general it is not a group since not every element need have an inverse. Indeed if a, b ∈ S, then a ∗ b = 1 iff 1 = ab mod n which we have seen can happen iff gcd (a, n) = 1 by Lemma 9.8. For example if n = 4, S = { 0 , 1 , 2 , 3 } , then

2 ∗ 1 = 2, 2 ∗ 2 = 0, 2 ∗ 0 = 0, and 2 ∗ 3 = 2,

none of which are 1. Thus, 2 is not invertible for this operation. (Of course 0 is not invertible as well.)

Lecture 6 (1/16/2009)

Theorem 6.1 (The groups, U (n)). For n ≥ 2 , let

U (n) := {a ∈ { 1 , 2 ,... , n − 1 } : gcd (a, n) = 1}

and for a, b ∈ U (n) let a ∗ b := (ab) mod n. Then (U (n) , ∗) is a group.

Proof. First off, let a ∗ b := ab mod n for all a, b ∈ Z. Then if a, b, c ∈ Z we have

(abc) mod n = ((ab) c) mod n = ((ab) mod n · c mod n) mod n = ((a ∗ b) · c mod n) mod n = ((a ∗ b) · c) mod n = (a ∗ b) ∗ c.

Similarly one shows that

(abc) mod n = a ∗ (b ∗ c)

and hence ∗ is associative. It should be clear also that ∗ is commutative. Claim: an element a ∈ { 1 , 2 ,... , n − 1 } is in U (n) iff there exists r ∈ { 1 , 2 ,... , n − 1 } such that r ∗ a = 1. ( =⇒ ) a ∈ U (n) ⇐⇒ gcd (a, n) = 1 ⇐⇒ there exists s, t ∈ Z such that sa + tn = 1. Taking this equation mod n then shows,

(s mod n · a) mod n = (s mod n · a mod n) mod n = (sa) mod n = 1 mod n = 1

and therefore r := s mod n ∈ { 1 , 2 ,... , n − 1 } and r ∗ a = 1. (⇐=) If there exists r ∈ { 1 , 2 ,... , n − 1 } such that 1 = r ∗ a = ra mod n, then n| (ra − 1) , i.e. there exists t such that ra − 1 = kt or 1 = ra − kt from which it follows that gcd (a, n) = 1, i.e. a ∈ U (n). The claim shows that to each element, a ∈ U (n) , there is an inverse, a−^1 ∈ U (n). Finally if a, b ∈ U (n) let k := b−^1 ∗ a−^1 ∈ U (n) , then

k ∗ (a ∗ b) = b−^1 ∗ a−^1 ∗ a ∗ b = 1

and so by the claim, a ∗ b ∈ U (n) , i.e. the binary operation is really a binary operation on U (n).

Example 6.2 (U (10)). U (10) = { 1 , 3 , 7 , 9 } with multiplication or Cayley table given by a\b 1 3 7 9 1 3 7 9

where the element of the (a, b) row indexed by U (10) itself is given by a ∗ b = ab mod 10.

Example 6.3. If p is prime, then U (p) = { 1 , 2 ,... , p}. For example U (5) = { 1 , 2 , 3 , 4 } with Cayley table given by,

a\b 1 2 3 4 1 2 3 4

Exercise 6.1. Compute 23−^1 inside of U (50).

Solution to Exercise. We use the division algorithm (see below) to show 1 = 6 · 50 − 13 · 23. Taking this equation mod 50 shows that 23−^1 = (−13) = 37. As a check we may show directly that (23 · 37) mod 50 = 1. Here is the division algorithm calculation:

50 = 2 · 23 + 4 23 = 5 · 4 + 3 4 = 3 + 1.

So working backwards we find,

1 = 4 − 3 = 4 − (23 − 5 · 4) = 6 · 4 − 23 = 6 · (50 − 2 · 23) − 23 = 6 · 50 − 13 · 23.

Lecture 7 (1/21/2009)

Definition 7.1 (Sub-group). Let (G, ·) be a group. A non-empty subset, H ⊂ G, is said to be a subgroup of G if H is also a group under the multiplication law in G. We use the notation, H ≤ G to summarize that H is a subgroup of G and H < G to summarize that H is a proper subgroup of G.

Theorem 7.2 (Two-step Subgroup Test). Let G be a group and H be a non-empty subset. Then H ≤ G if

1. H is closed under ·, i.e. hk ∈ H for all h, k ∈ H,
2. H is closed under taking inverses, i.e. h−^1 ∈ H if h ∈ H.

Proof. First off notice that e = h−^1 h ∈ H. It also clear that H contains inverses and the multiplication law is associative, thus H ≤ G.

Theorem 7.3 (One-step Subgroup Test). Let G be a group and H be a non-empty subset. Then H ≤ G iff ab−^1 ∈ H whenever a, b ∈ H.

Proof. If a ∈ H, then e = a a−^1 ∈ H and hence so is a−^1 = ae−^1 ∈ H. Thus it follows that for a, b ∈ H, that ab = a

b−^1

∈ H and hence H ≤ G. and the result follows from Theorem 7.2.

Example 7.4. Here are some examples of sub-groups and not sub-groups.

1. 2Z < Z while 3Z ⊂ Z but is not a sub-group.
2. Zn = { 0 , 1 , 2 ,... , n − 1 } ⊂ Z is not a subgroup of Z since they have different group operations.
3. {e} ≤ G is the trivial subgroup and G ≤ G.

Example 7.5. Let us find the smallest sub-group, H containing 7 ∈ U (15). Answer, 72 mod 15 = 4, 73 mod 15 = 13, 74 mod 15 = 1

so that H must contain, { 1 , 7 , 4 , 13 }. One may easily check this is a subgroup and we have | 7 | = 4.

Proposition 7.6. The elements, O (2) := {Sα, Rα : α ∈ R} form a subgroup GL 2 (R) , moreover we have the following multiplication rules:

RαRβ = Rα+β , SαSβ = R2(α−β), (7.1) Rβ Sα = Sα+β/ 2 , and SαRβ = Sα−β/ 2. (7.2)

for all α, β ∈ R. Also observe that

Rα = Rβ ⇐⇒ α = β mod 360 (7.3)

while, Sα = Sβ ⇐⇒ α = β mod 180. (7.4)

Proof. Equations (7.1) and (7.2) may be verified by direct computations using the matrix representations for Rα and Sβ. Perhaps a more illuminating way is to notice that all linear transformations on R^2 are determined by there actions on u (θ) for all θ (actually for two θ is typically enough). Using this remark we find,

RαRβ u (θ) = Rαu (θ + β) = u (θ + β + α) = Rα+β u (θ) SαSβ u (θ) = Sαu (2β − θ) = u (2α − (2β − θ)) = u (2 (α − β) + θ) = R2(α−β) u (θ) , Rβ Sαu (θ) = Rβ u (2α − θ) = u (2α − θ + β) = u (2 (α + β/2) − θ) = Sα+β/ 2 u (θ) , and SαRβ u (θ) = Sαu (θ + β) = u (2α − (θ + β)) = u (2 (α − β/2) − θ) = Sα−β/ 2 u (θ)

which verifies equations (7.1) and (7.2). From these it is clear that H is a closed under matrix multiplication and since R−α = R− α 1 and S− α 1 = Sα it follows H is closed under taking inverses. To finish the proof we will now verify Eq. (7.4) and leave the proof of Eq. (7.3) to the reader. The point is that Sα = Sβ iff

u (2α − θ) = Sαu (θ) = Sβ u (θ) = u (2β − θ) for all θ

which happens iff

[2α − θ] mod 360 = [2β − θ] mod 360

which is equivalent to α = β mod 180.

Lecture 8 (1/23/2009)

Notation 8.1 The order of a group, G, is the number of elements in G which we denote by |G|.

Example 8.2. We have |Z| = ∞, |Zn| = n for all n ≥ 2 , and |D 3 | = 6 and |D 4 | = 8.

Definition 8.3 (Euler Phi – function). For n ∈ Z+, let

ϕ (n) := |U (n)| = # { 1 ≤ k ≤ n : gcd (k, n) = 1}.

This function, ϕ, is called the Euler Phi – function.

Example 8.4. If p is prime, then U (p) = { 1 , 2 ,... , p − 1 } and ϕ (p) = p − 1. More generally U (pn) consists of { 1 , 2 ,... , pn} {multiples of p in { 1 , 2 ,... , pn}}. Therefore,

ϕ (pn) = |U (pn)| = pn^ − # {multiples of p in { 1 , 2 ,... , pn}}

Since

{multiples of p in { 1 , 2 ,... , pn}} =

kp : k = 1, 2 ,... , pn−^1

it follows that # {multiples of p in { 1 , 2 ,... , pn}} = pn−^1 and therefore,

ϕ (pn) = pn^ − pn−^1 = pn−^1 (p − 1)

valid for all primes and n ≥ 1.

Example 8.5 (ϕ (pmqn)). Let N = pmqn^ with m, n ≥ 1 and p and q being distinct primes. We wish to compute ϕ (N ) = |U (N )|. To do this, let let Ω := { 1 , 2 ,... , N − 1 , N } , A be the multiples of p in Ω and B be the multiples of q in Ω. Then A ∩ B is the subset of common multiples of p and q or equivalently multiples of pq in Ω so that;

(A) = N/p = pm−^1 qn,

(B) = N/q = pmqn−^1 and

(A ∩ B) = N/ (pq) = pm−^1 qn−^1.

Therefore,

ϕ (N ) = # (Ω \ (A ∪ B)) = # (Ω) − # (A ∪ B) = # (Ω) − [# (A) + # (B) − # (A ∩ B)]

= N −

[

N

p

N

q

N

p · q

]

= pm^ · qn^ − pm−^1 · qn^ − pm^ · qn−^1 + pm−^1 · qn−^1

pm^ − pm−^1

qn^ − qn−^1

which after a little algebra shows,

ϕ (pmqn) =

pm^ − pm−^1

qn^ − qn−^1

= N

p

q

The next theorem generalizes this example.

Theorem 8.6 (Euler Phi function). Suppose that N = pk 11... pk nn with ki ≥ 1 and pi being distinct primes. Then

ϕ (N ) = ϕ

pk 11... pk nn

∏^ n

i=

pk i i− pk ii−^1

= N ·

∏^ n

i=

pi

Proof. (Proof was not given in class!) Let Ω := { 1 , 2 ,... , N } and Ai := {m ∈ Ω : pi|m}. It then follows that U (N ) = Ω \ (∪ni=1Ai) and therefore,

ϕ (N ) = # (Ω) − # (∪ni=1Ai) = N − # (∪ni=1Ai).

To compute the later expression we will make use of the inclusion exclusion formula which states,

(∪ni=1Ai) =

∑^ n

l=

(−1)l+^

1 ≤i 1 So in all cases |g| = |〈g〉|.

Proof. 1. If gi^ = gj^ for some i < j, then

e = gig−i^ = gj^ g−i^ = gj−i

so that gm^ = e with m = j − i ∈ Z+ from which we would conclude that |g| < ∞. Thus if |g| = ∞ it must be that all elements in the list, {gn^ : n ∈ Z} , are distinct. In particular 〈g〉 = {gn^ : n ∈ Z} has an infinite number of elements and therefore |〈g〉| = ∞.

1. Now suppose that n = |g| < ∞. Since gn^ = e, it also follows that g−n^ = (gn)−^1 = e−^1 = e. Therefore if m ∈ Z and m = sn + r where r := m mod n, then gm^ = (gn)s^ gr^ = gr^ , i.e. gm^ = gm^ mod^ n^ for all m ∈ Z. Hence it follows that 〈g〉 =

e, g, g^2 ,... , gn−^1

. Moreover if gi^ = gj^ for some 0 ≤ i ≤ j < n, then gj−i^ = e with j − i < n and hence j = i. Thus the list in Eq. (8.2) consists of distinct elements and therefore |〈g〉| = n. Lastly, if k, l ∈ Zn, then

gkgl^ = gk+l^ = g(k+l) mod^ n.

Lecture 9 (1/26/2009)

Corollary 9.1. Let a ∈ G. Then ai^ = aj^ iff |a| divides (j − i). Here we use the convention that ∞ divides m iff m = 0. In particular, ak^ = e iff |a| |k.

Corollary 9.2. For all g ∈ G we have |g| ≤ |G|.

Proof. This follows from the fact that |g| = |〈g〉| and 〈g〉 ⊂ G.

Theorem 9.3 (Finite Subgroup Test). Let H be a non-empty finite subset of a group G which is closed under the group law, then H ≤ G.

Proof. To each h ∈ H we have

hk

k=1 ⊂^ H^ and since # (H)^ <^ ∞,^ it follows that hk^ = hl^ for some k 6 = l. Thus by Theorem 8.12, |h| < ∞ for all h ∈ H and 〈h〉 =

e, h, h^2 ,... , h|h|−^1

⊂ H. In particular h−^1 ∈ 〈h〉 ⊂ H for all h ∈ H. Hence it follows by the two step subgroup test that H ≤ G.

Definition 9.4 (Centralizer of a in G). The centralizer of a ∈ G, denoted C(a), is the set of g ∈ G which commute with a, i.e.

C (a) := {g ∈ G : ga = ag}.

More generally if S ⊂ G is any non-empty set we define

C (S) := {g ∈ G : gs = sg for all s ∈ S} = ∩s∈S C (s).

Lemma 9.5. For all a ∈ G, 〈a〉 ≤ C(a) ≤ G.

Proof. If g ∈ C (a) , then ga = ag. Multiplying this equation on the right and left by g−^1 then shows,

ag−^1 = g−^1 gag−^1 = g−^1 agg−^1 = g−^1 a

which shows g−^1 ∈ C (a). Moreover if g, h ∈ C (a) , then gha = gah = agh which shows that gh ∈ C (a) and therefore C (a) ≤ G.

Example 9.6. If G is abelian, then C (a) = G for all a ∈ G.

Example 9.7. Let G = GL 2 (R) we will compute C (A 1 ) and C (A 2 ) where

A 1 =

[

]

and A 2 :=

[

]

1. We have B =

[

a b c d

]

∈ C (A 1 ) iff,

[ b a d c

]

[

a b c d

] [

]

[

] [

a b c d

]

[

c d a b

]

which means that b = c and a = d, i.e. B must be of the form,

B =

[

a b b a

]

and therefore, C (A 1 ) =

{[

a b b a

]

: a^2 − b^2 6 = 0

1. We have B =

[

a b c d

]

∈ C (A 2 ) iff,

[ a −b c −d

]

[

a b c d

] [

]

[

] [

a b c d

]

[

a b −c −d

]

which happens iff b = c = 0. Thus we have,

C (A 2 ) =

{[

a 0 0 d

]

: ad 6 = 0

Lemma 9.8. If {Hi} is a collection of subgroups of G then H := ∩iHi ≤ G as well.

Proof. If h, k ∈ H then h, k ∈ Hi for all i and therefore hk−^1 ∈ Hi for all i and hence hk−^1 ∈ H.

Corollary 9.9. C (S) ≤ G for any non-empty subset S ⊂ G.

Definition 9.10 (Center of a group). Center of a group, denoted Z(G), is the centralizer of G, i.e.

Z (G) = C (G) := {a ∈ G : ax = xa for all x ∈ G}