Math 410 - Abstract Algebra Discussions - Week 4

MATH 410 - ABSTRACT ALGEBRA DISCUSSIONS - WEEK 4

CAN OZAN OGUZ

Let’s start by showing that a left inverse is necessarily a right inverse in a group. Let (G, ∗) be our group. Take g ∈ G and let h ∈ G be its left inverse(i.e. h ∗ g = e).

h ∗ g = e ⇒ g ∗ h ∗ g = g we multiplied with g from left ⇒ g ∗ h ∗ g ∗ h = g ∗ h we multiplied with h from right It isn’t clear how we are going to show that g ∗ h = e with the above equation. We just ended up with three copies of g ∗h, 2 on the left hand side, and one on the right hand side. Maybe we need to cancel the one on the left with the one the right. To cancel an element in a group, we multiply the element by its inverse. Let’s denote the inverse of g ∗ h by (g ∗ h)−1. Since g ∗ h ∈ G, it must have an inverse.

g ∗ h ∗ g ∗ h = g ∗ h ⇒ (g ∗ h)−1 ∗ (g ∗ h) ∗ (g ∗ h) = (g ∗ h)−1(g ∗ h) ⇒ e ∗ (g ∗ h) = e since we have right inverses ⇒ g ∗ h = e since we have right identity

1. Symmetric group Sn Last week we saw many examples of groups. One direction we can take in studying groups is to seperate them into two distincs collections: Finite groups and infinite groups. Today we will study in detail a finite group, namely the symmetric group Sn whose elements are bijections from the set {1, 2, ··· , n} to itself. Our group operation will be composition of functions. We have the identity function, sending every element to itself. Function composition is associative and bijective functions are invertible, hence we have a group. The order of Sn is n!. The group Sn has a very special role in finite group theory: Theorem (Cayley’s theorem) Every finite group G with n elements can be seen as living in the symmetric group Sn. Therefore one might say: If we understand everything about the symmetric groups Sn, for all n, then we would understand everything about all finite groups. As you can see, this has put the symmetric group to the center of attention in finite group theory. To understand a group G with order n, we will have to understand a group with n! elements though, so there is a catch. In case you don’t know already, factorials grow too fast, even for todays supercomputers. Dealing with 20! elements is not an easy task, both in terms of memory and in terms of computation time. So we really need some theoretical understanding.

Date: September 14th, 2017. 1 2 CAN OZAN OGUZ

Notation: We have several notations to describe a permutation. Our two main options are two line notation, such as 1 2 3 4 a = 2 4 1 3 or cycle notation (1243) where in each permutation, 1 goes to 2, 2 goes to 4, 4 goes to 3 and 3 goes to 1. I prefer the cycle notation and will use it for the rest of the notes. However it is a good exercise to convert one notation to the other. 1 2 3 4 5 6 7 8 9 Exercise Write the permutation a = 2 6 1 3 9 5 7 4 8 We have a special name for a 2-cycle (ab), we call it a transposition.

2. Decomposition of permutations A permutation may not necessarily be a single cycle, but it can be a product of cycles such as (135)(35)(24)(321)(6). As you can see, the cycles don’t need to be disjoint a priori. However, we have a strong result for permutations.

Theorem Every permutation σ ∈ Sn can be written in a unique way as a product of disjoint cycles, up to reordering the cycles. This is in the same spirit of the fundamental theorem of arithmetic. In fact, it is same statement if you replace ”permutation” by ”natural number” and ”disjoint cycle” by ”prime number”. Therefore we can say that permutations are made of disjoint cycles, just like natural numbers were made of prime numbers. Note that for natural numbers, instead of decomposing them into prime factors, we could have written any natural number as a product of a power of 2 and an odd number. It is a less useful decomposition, but it is just another decomposition. Turns out we have other decompositions of permutations as well, and they turn out to be useful.

Proposition Every permutation σ ∈ Sn can be written as a product of transpositions. Read the proposition carefully. It doesn’t mention uniqueness at all. Yes, we can write any permutation as a product of transpositions, but not in a unique way. For example (13) = (12)(23)(12) = (23)(12)(23) Losing the uniqueness is a big deal. This means if two people write down a permutation as a product of transpositions, there is a high chance their answers will look diﬀerent even if they start with the same permutation. In order to compare two answers, disjoint cycle decomposition is the correct way to go. Then what is the use of decomposing a permutation into transpositions? We will use the decomposition into transposition to describe even and odd permutation, which in turn will help us describe an important subgroup of Sn. k Deﬁnition Let sgn : Sn → {1, −1} be the map that sends σ to (−1) where k is the number of transpositions appearing in the decomposition of σ into transpositions. MATH 410 - ABSTRACT ALGEBRA DISCUSSIONS - WEEK 4 3

We have just deﬁned the map sgn depending on the decompositon of a permutation into transpositions. However this decomposition is not unique. That is, given a permutation σ ∈ Sn, we might have

σ = α1α2 ··· αk = β1β2 ··· βl where α’s and β’s are transpositions, and there might be a diﬀerent number of them. Of course this would make our deﬁnition of the map sgn not a proper one. Fortunately, in a decomposition of transpositions, the parity of the number of transpositions remains the same. If sgn(σ) = 1, we say that σ is an even permutation. If sgn(σ) = −1, we say that σ is an odd permutation. Note that identity is an even permutation. This map sgn has another interesting property: sgn(στ) = sgn(σ)sgn(τ) That means product of even permutations is an even permutation. Therefore all even permutations in Sn form a group. We will denote this group by An. This is called the alternating subgroup of the symmetric group. n! In the homework, you will show that the number of even permutation is . 2