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Abstract Algebra Discussions - Week 3

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Abstract Algebra Discussions - Week 3 MATH 410 - ABSTRACT ALGEBRA DISCUSSIONS - WEEK 3 CAN OZAN OGUZ 1. Sets A set is the most basic algebraic structure one can have. It is a collection of elements, and that is that. We can't say anything else about a set. It has no other structure. In fact, there is a little more to it. Set might look like a very basic concept, and this is why todays mathematics is all based on sets, however that means set theory is not based on anything. Therefore we don't have good definitions to answer the questions "What is a set?" and "What is an element?". Surely these questions caused a lot of trouble within the mathematics community especially since the beginning of 20th century. But once you don't ask these two questions, you should be on a safe ground. For example, we can answer the question "What is a vector space?" easily by referring to sets. Similarly we can answer the question "What is a function?" by saying it is a special relation, and a relation is a subset of a particular set. We are not going to deal with set theory too much, but let's say a few things about them since after all a group and a ring is a set with some properties. One important example is the empty set ?. It has no elements, and it is the only set with this property. It is easy work with the empty set. Suppose we want to prove that all elements of the empty set are even integers. We will prove this by contradiction. Suppose it has an element which is not an even integer. But the empty set cannot have any elements, therefore we get a contradiction. Hence all elements of the empty set are even integers. The same proof wouuld of course work for odd integers. Actually the same proof works for any property you ask from the elements of the empty set. 1.1. Equivalence Relation. We have an operation on sets called Cartesian prod- uct. Given two sets A and B, we can form a new set A × B by taking the Cartesian product of A and B. The elements of A×B are ordered pairs (a; b) such that a 2 A and b 2 B. Definition A relation R from A to B is a subset of A × B. There are tons of relations. Example We can define a relation from R to R using the order given by ≤: • 0 is related to all positive real numbers. • 1 is not related to −1. • Any real number a is related to itself. Example Another example is relating two integers a and b if b is divisible a. • For example 2 and 4 are related in this relation, but 2 and 5 are not. Date: September 7th, 2017. 1 2 CAN OZAN OGUZ • Notice that 2 is related to 4, but 4 is not related to 2. • Note that any number a is related to itself. • 2 is related to 4 and 6, but 4 and 6 are not related. Example Another is relation is being congruent modulo n. This is what modular arithmetic is all about. • a ≡ a(mod n) • If a ≡ b(mod n), then b ≡ a(mod n) • If a ≡ b(mod n) and b ≡ c(mod n), then a ≡ c(mod n) These three properties are called being reflexive, symmetric and transitive. Any relation with such properties is called an equivalence relation. Equivalence relations will play an important role in group theory. 2. Groups 2.1. Origins. We already saw the definition of a group. It was a pair (G; ∗) where G is a set, and ∗ : G×G ! G is a binary operation with certain properties(identity, inverses, associativitiy). If our operation is also commutative, that is if a ∗ b = b ∗ a, then we say it is a commutative group, or an abelian group after the Norwegian mathematician Abel. To get a better understanding of this concept, I think we should see how it was created in the first place. Let me start by saying that "Group theory" is a terrible name for this beautiful subject. I would much rather prefer "Symmetry theory", where our main tool is a group. Because a group is an abstraction of symmetry of some object. 2.2. Geometric symmetry. Think about it, say geometrically. Given a geometric shape, you can immediately realize some of its symmetries, such as if you flip the shape this way, it looks the same, or if you rotate it that way, it looks the same...etc. There is always the obvious symmetry, which does nothing to your geometric object, leaves it as it is. This is the "identity" in our group. Also no matter which operation you apply to realize a symmetry, you can just undo it, hence "invertible" elements of our group. Associativitiy is a little bit more tricky, and later mathematicians realized some non-associative "symmetries". But we will work with associative ones in this class. 2.3. Algebraic symmetry. We also have algebraic symmetries, which is how groups first appeared in the mind of young French mathematician Evariste Galois(1811- 1832), who was killed at a duel about a woman. An important question in that era was solving polynomial equations of degree 5 or higher. Quadratic formula was dis- covered a long time ago, and mathematicians of 16th century(Cardano) had found formulas for solving cubic and quartic equations. However no progress was made about degree 5 or higher equations, until the Norwegian mathematician Niels Hen- rik Abel(1802-1829) showed there cannot be such general formulas for equations of degree ≥ 5. Galois's ideas was used to show when does a polynomial equation of degree ≥ 5 has solutions in terms of its coefficients. When I say an algebraic symmetry, you should be thinking about a polynomial, say (x − 2)(x − 5) = 0; MATH 410 - ABSTRACT ALGEBRA DISCUSSIONS - WEEK 3 3 where two solutions are x1 = 2 and x2 = 5, or x1 = 5 and x2 = 2. We have a choice in enumarating the solutions(n! choices for degree n polynomials), and all these enumerations give us the same polynomial, hence the word symmetry. 2.4. Examples of groups. Let's list some examples of groups we already saw. • (Z; ?): We can see that the integers form a group, but a set is not enough to get a group. We need to specify the operation as well. For the addition operation, integers form a group, with 0 as the identity element. • (Z; ×): Is integers a group under multiplication? We have an identity, 1, and multiplication is associative. The problem is with inverses. 2 does not have a multipliative inverse in integers, that is there is no g 2 Z such that 2 × g = 1. Hence this is not a group. • (R; +): Real numbers under addition form a group, with 0 as the identity. • (R; ×): Real numbers under multiplication has the identity 1, multiplication 1 is associative, and given r 2 R, r 2 R is its multiplicative inverse. This is a much better situation than the integers with multiplication. The only problem is the element 0. It is not invertible. But without 0, we would have a group, hence (R; ×) is not a group, but (R∗; ×) = (R − f0g; ×) is a group. • (M2×2(R); +): The set of 2 by 2 matrices with real coefficients describe all linear transformation of a 2-dimensinal real vector space. This set is a group under the addition operation, where zero matrix is the identity of the group. • (M2×2(R; ×)): Is the same set a group under matrix multiplication? We have the identity matrix, and matrix multiplication is associative. However we get into trouble with inverses again, not every 2 by 2 matrix is invertible. If you recall, a matrix is invertible if and only if it has non-zero determinant. To solve this problem, we will use a simple idea: Just work with invertible matrices! • (GL2(R); ×): General Linear group, which is the set of invertible matrices, forms a group under multiplication since we have dealt with the problem of non-invertible matrices by throwing them away. This is a group since det(AB) = det(A)det(B) is cannot be zero unless A or B is not invertible. • (Z=nZ; +): We go back to modular arithmetic to see another example of a group. This time, our set has finitely many elements(even though each element can be seen as an infinite set, the set of integers congruent modulo n). This is an example of a finite group. • (Z=nZ; ×): We certainly have an identity, 1, and multiplication is asso- ciative. Let's see if every element has an inverse. For n = 4, in Z=4Z, 2 is not invertible since 2 times any number will be either divisible by 4, or will have a remainder 2 when divided by 4. So there is no solution to 2×x ≡ 1(mod n). We already saw a condition as to when a×x ≡ 1(mod n). There is a solution if and only if gcd(a; n) = 1. So (Z=nZ; ×) is not a group, but we can use our old trick to get a group out of it: Restrict yourself to invertible elements. ((Z=nZ)∗; ×) will be a group. As Tony mentioned in discussion, this is a group whose elements are integers modulo n that are relatively prime to n. If n is a prime itself, then we just need to exclude 0. If n is not a prime, we have to identify the integers that are relatively prime to n.
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