MATH 47A FALL 2008 INTRO TO MATH RESEARCH 73. more group theoryToday we talked about the main properties of the elements of agroup (although I didn’t define a group yet). I just said that these areall permutations. I changed the format of the class to problem solving,similar to problem sessions, instead of lecturing.3.1. inverse.Definition 3.1. The inverse of a permutation σ is given byσ−1(x) = y where σ(y) = xa) Show that this defines a permutation σ−1.b) Find a formula for the inverse of τ = (a1, a2, ··· , ak).c) Show that (στ )−1= τ−1σ−1.Students found Question (a) confusing so we first did (b) and (c).3.1.1. inverse of a k-cycle. The inverse of a cycle is given by writingthe cycle backwards:τ−1= (ak, ak−1, ··· , a2, a1)This is supposed to obvious, but a proof would go like this:Proof. Let x = ai. Then τ (ai−1) = ai= x. So, τ−1(ai) = ai−1. Specialcare is needed in the case i = 1. Then the equation τ (ak) = a1means(by definition) that τ−1(a1) = ak. If x is not any of the aithen τ (x) = x.So, τ−1(x) = x. This shows that τ−1(x) is given by the cycle above forall x. !3.1.2. inverse of a product. The derivation of the formula(στ )−1= τ−1σ−1used a lot more stuff than I thought. The concepts that were used bystudents familiar with groups were the following:(1) associativity: (ab)c = a(bc ). This implies that parentheses canbe placed arbitrarily.(2) identity: id(x) = x is called the identity function and it is alsocalled e = id. This has the property that eσ = σ = σe since:eσ(x) = e(σ(x)) = σ(x), σe(x) = σ(e(x)) = σ(x)(3) The group theoretic definition of the inverse which is:σ−1σ = e = σσ−1.8 NOTESThe proof of the inverse formula that students came up with, usingthese properties, was:(στ )(τ−1σ−1) = σ (τ τ−1)! "# $e=idσ−1by associativity= σeσ−1by group theoretic def of inverse= σσ−1by property of e= e by def of inverseWe needed to know that, when the product of two things is the identity,the two things are inverse to each other:Lemma 3.2. If ab = e then b = a−1.Proof. Multiply both sides by a−1:a−1ab = a−1eThe left hand side (LHS) is a−1ab = eb = b, the RHS is a−1. !I actually skipped this lemma. What I verified in class was the grouptheoretic definition of inverse using my definition.Lemma 3.3. If σ−1is defined as in Def. 3.1 (the “inverse function”definiton) then we get the formulas:σσ−1(x) = x ∀x, i.e., σσ−1= idσ−1σ(y) = y ∀y, i.e., σ−1σ = idProof. My definition was that σ−1(x) = y if σ(y) = x. If we insert σ(y)in for x in the first equation we get:σ−1(σ(y)) = yThus σ−1σ = id. If we insert the first equation into the second we get:x = σ(y) = σ(σ−1(x))i.e., σσ−1= id. !3.1.3. definition of a function. I had to explain the first question be-cause students had no idea even what it was asking.Define: When I say that this formula defines a function I mean thatfor every x there is a unique y (so that σ(y) = x).I used the fact that σ is a permutation ⇒ bijection ⇒ 1-1 and onto.(1) (existence) y exists since σ is onto: This means for any x ∈ Xthere is a y ∈ X so that σ(y) = x. So, σ−1(x) exists.MATH 47A FALL 2008 INTRO TO MATH RESEARCH 9(2) (uniqueness) We need to know that for each x there is only oney, otherwise we don’t have a function. I gave the example of:√x = y if y2= xThis formula does not define the square root function since thereare two y’s for each positive x.Uniqueness of y follows from the fact that σ is 1-1: If I hadtwo y’s say y1and y2(in other words, σ−1(x) = y1and σ−1(x) =y2) then I would have σ(y1) = x = σ(y2) which implies y1= y2since σ cannot send to y’s to the same thing.This show that σ−1: X → X is a function. To show it is a permu-tation, we have to show it is 1-1 and onto. We decided in class thatboth statements are obvious when you write down what they are inequation form:σ−1(x1) = y = σ−1(x2) ⇒ σ(y) = x1, σ(y) = x2So, σ−1is 1-1. To show it is onto, we need to take any y in our set Xand find some x so thatσ−1(x) = y, i.e., σ(y) = x.Well, there it is!3.2. commutativity. Students brought up the subject ofcommutativity.Definition 3.4. a and b commute if ab = ba. A group G is commutativeif ab = ba for all a, b ∈ G.I pointed out that “commute” is a verb and “commutative” is anadjective.Problem: If a, b commute, show that (ab)−1= a−1b−1.Proof. To show this you need to showaba−1b−1= eThe proof that students gave is to switch a, b on the left since theycommute:aba−1b−1= baa−1b−1= bb−1= e.!10 NOTES3.3. conjugation.Definition 3.5. We say that a is conjugate to b if there exists a c sothata = cbc−1I used the following notation. First, a ∼ b for “a is conjugate to b” andφc(b) := cbc−1The function φcis a “homomorphism” (a concept that I will explainlater).(1) Show that conjugation is an equivalence relation, i.e., it is(a) reflective: (∀a)a ∼ a(b) symmetric: (∀a, b)a ∼ b ⇒ b ∼ a(c) transitive: (∀a, b, c)a ∼ b, b ∼ c ⇒ a ∼ c(2) If τ = (a1, a2, ··· , ak) then find a formula for στ σ−1(3) Show that every permutation of n letters is conjugate to itsinverse.This is taking too long. I will write the answers