Math 3121 Abstract Algebra I

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# Math 3121 Abstract Algebra I - PowerPoint PPT Presentation

Math 3121 Abstract Algebra I. Lecture 5 Finish Sections 6 + Review: Cyclic Groups, Review. Questions on HW. Don’t hand in: Pages 66-67: 3, 5, 11, 17, 23, 45. Finish Section 6: Cyclic Groups.

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### Math 3121Abstract Algebra I

Lecture 5

Finish Sections 6 + Review: Cyclic Groups, Review

Questions on HW
• Don’t hand in:

Pages 66-67: 3, 5, 11, 17, 23, 45

Finish Section 6: Cyclic Groups
• Main Theorem: A finite cyclic group is isomorphic to ℤn, and an infinite cyclic group is isomorphic to ℤ.
• GCD(n, m) = the positive generator of <n, m>.
• Size of a subgroup of finite cyclic group must divide the order of that group.
More on ℤn
• The group ℤncan be defined as follows: The set is the set of equivalence classes of the relation: x = y mod n. Addition is defined by adding members of equivalence class. This is well-defined: x1 = y1 mod n and x2 = y2 mod n ⇒ x1 + x2 = y1 + y2 mod n.
Theorem

Theorem: A cyclic group is isomorphic to ℤ or to ℤn. In the first case it is of infinite order, and in the second case it is of finite order.

Proof: Let G be a cyclic group. Pick a generator a of G, and let f be the power function:

f(n) = an

Then f is a homomorphism from ℤ to G. It is onto. Let K be the subset of all k in ℤ such that f(k) = e. This is called the kernel of the homomorphism f. It is easy to show that K is a subgroup of ℤ, and hence of the form nℤ, for some n in ℤ.

If n = 0, then f is one-to-one: for each distinct x, y in ℤ, x-y is not in K, hence f(x-y) ≠ e, hence ax-y ≠ e, hence ax ≠ ay , hence f(x) ≠ f(y). Thus f is an isomorphism.

Otherwise, we can assume n is positive, and f(x) = f(y) ⇔ x = y mod n. Thus f induces a one-to-one map from the equivalence classes onto G. The equivalence classes of mod n are the elements of ℤn. Thus G is isomorphic to ℤn.

GCD

Definition (normal definition): The greatest common divisor gcd of two integers is the largest integer that divides both (in the sense that any divisor of both integers is a divisor of their gcd).

Theorem: If n and m are integers, then there are integers a and b such that gcd(n, m) = a n + b m.

Proof: Let H = {a n + b m | a, b in ℤ}. Then H = d ℤ, for some d in ℤ. We can assume d is nonnegative. Otherwise, replace d by –d.

d divides n and m, since they are in H. If x divides n and m, then x divides any element of H, hence d.

Relatively Prime

Definition: Two integers are relatively prime if their gcd is 1.

Theorem: If n and m are relatively prime, and n divides m s, then n divides s.

Theorem

Theorem: Let G be a cyclic group of finite order n and generated by a. Let b = as, for some integer s. Then b generates a cyclic group of order n/gcd(n, s). Furthermore, <as> = <at> iff gcd(s, n) = gcd(t, n).

Proof: Use ℤn as a model. Suppose s in ℤ represents an element of ℤn. Let d = gcd(s, n). Then d divides n. Thus the multiples k d for 0 ≤ k < n/d are less than n and thus distinct mod n. Thus the cyclic group <d> mod n has n/d elements.

Let w = s/d. Find u, v in ℤ such that u s + v n = d. Then u s = d mod n. Also w u = 1 mod n. Thus the maps

g(x) = w x mod n and h(y) = u y mod n

are inverse to each other. Now g takes elements of <d> mod n to elements of <s> mod n: g(k d) = s/d k d = k s. Thus both sets have the same number of elements, namely n/d. In fact, the members of <s> mod n are also members of <d> mod n.

Furthermore, if d = gcd(s, n) = gcd(t, n), then members of <t> mod n are also members of <d> mod n.

Review
• Topics
• Sample questions