MAT1100HF ALGEBRA: ASSIGNMENT II. Contents 1. Problem Problem Problem Problem Problem Problem
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1 MAT1100HF ALEBRA: ASSINMENT II J.A. MRACEK DEPARTMENT OF MATHEMATICS UNIVERSITY OF TORONTO Contents 1. Problem Problem Problem Problem Problem Problem Problem 7 4 Date: October 20, I
2 MAT1100HF ALEBRA: ASSINMENT II 1 1. Problem 1 Problem 1. Find the least integer n for which the symmetric group S n contains an element of order 18. Proof. We seek the smallest n such that S n contains a permutation of order 18. Permutations can be written in terms of a collection of disjoint cycles. Let σ be a permutation of order 18 consisting of q disjoint cycles, and let the lengths of these cycles be k 1,..., k q. The order of the element will be equal to the least common multiple of the lengths of its cycles. By the fundamental theorem of arithmetic, any number can be uniquely decomposed into a product of powers of prime numbers. It is easy to see that 18 = 2 3 2, so each of the k i s must have a prime decomposition consisting of only 3 s and 2 s. The maximum allowed powers of 3 and 2 are 2 and 1, respectively, because otherwise the LCM would be larger than 18. Here is a list all such possible decompositions: (1) (2) (3) (4) k 1 = 3 2, k 2 = 3 k 1 = 2, k 2 = 3, k 3 = 3 k 1 = k 1 = 3 2, k 2 = 2 Automatically we can rule out possibilities (1) and (2) because they have LCM({k i }) = 6. Options (3) and (4) are valid selections. The minimal order of the element is the sum of the cycle lengths, so clearly option 4 is the best selection because it gives a cycle of length 11; thus, S 11 is the smallest symmetric group containing an element with order 18. Problem 2. What is the maximal order of an element in S 26? Proof. As before, let σ S 26 be comprised of q disjoint cycles having length k 1,..., k q = 1,..., 26. To find the maximal order of σ S 26, we need to maximize LCM(k 1,..., k q ) subject to the constraint q i=1 k q = 26. A short script in MATLAB was written to carry out exactly this task: function [ O ] = MaximalOrder( n ) %INPUTS: Order of the symmetric group, Sn %OUTPUTS: Maximal order of an element in Sn p = primes(n); Ord = 1; LCM = 1; %i indexes the number of cycles for i = 1:n %l is an index to run through every possible way of summing the lengths %of i cycles to 26 for l = 1:i k = zeros(1,i); k(i) = n - i + 1; k(1:i-1) = 1; A = zeros(i,length(p)); while(k(i) >1+ max(k(1:i-1))) %Find the prime factorization of each k_i for j = 1:i A(j,:) = PrimeFactors(k(j),n); %Find the lowest common multiple alpha = max(a); for j = 1:length(p) LCM = p(j)^(alpha(j))*lcm;
3 2 J.A. MRACEK DEPARTMENT OF MATHEMATICS UNIVERSITY OF TORONTO O = Ord; %Keep track of the highest obtained LCM if(ord < LCM) Ord = LCM; LCM = 1; %Now change the k s slightly and try again for j = 1:i-1 if(k(j+1)>k(j)+1)&&(j>=i-l) k(j+1) = k(j+1) -1; k(j) = k(j) + 1; break I also wrote the short script, PrimeFactors.m to determine the multiplicities of the prime decomposition of the cycle lengths: function [ alpha ] = PrimeFactors( x,n ) %INPUTS: A natural number, x and a number n such that x < n %OUTPUTS: A row vector alpha containing the multiplicity of all the prime %factors less than n %x = \prod_{i}^{n} p_i^{alpha_i} g = primes(n); p = factor(x); a1 = zeros(1,length(g)); count = 0; for i = 1:length(g) for j = 1:length(p) if(p(j)==g(i)) count = count+1; a1(i) = count; count = 0; alpha = a1; The command prompt commands and output are provided below: >> OrderS26 = MaximalOrder(26) OrderS26 = 1260 So the maximal order of an element in S 26 is Problem 2 Proposition 1. If H < with [ : H] = 2, then H
4 MAT1100HF ALEBRA: ASSINMENT II 3 Proof. H if and only if for every g, ghg 1 H. Since [ : H] = 2, the group can be written as a disjoint union of two partitions. Now pick some g. There are only two possibilities for g: Either it is in H or it is not. If g H, then clearly ghg 1 = H because H is a subgroup. Otherwise, we could have had that g / H. Let h, h H. Suppose for the sake of contradiction that ghg 1 gh, then: ghg 1 = gh hg 1 = h g 1 = h 1 h H A contradiction, since it was assumed that g / H. The only possibility is that our original assumption was wrong, and ghg 1 H. But now we have seen that for any g, ghg 1 H; thus, H. 3. Problem 3 We may write σ = (a 1 1, a 1 2, a 1 3, a 1 4, a 1 5)(a 2 1, a 2 2, a 2 3)(a 3 1, a 3 2, a 3 3)(a 4 1, a 4 2). Let τ C S20 (σ), then τστ 1 = σ. In other words, τ maps each of the cycles to a cycle with equivalent action on the underlying set. To find the order of the centralizer, we just have to count how many ways it is possible to map each cycle to itself. Note that two cycles are equivalent to each other if their action on the underlying set is identical. This can clearly only happen when the cycles are cyclic permutations of each other. For example, in the 5-cycle, the following are equivalent: (a 1 1, a 1 2, a 1 3, a 1 4, a 1 5) = (a 1 2, a 1 3, a 1 4, a 1 5, a 1 1) = (a 1 3, a 1 4, a 1 5, a 1 1, a 1 2) = (a 1 4, a 1 5, a 1 1, a 1 2, a 1 3) = (a 1 5, a 1 1, a 1 2, a 1 3, a 1 4) So there are 5 possible ways for this cycle to get mapped to itself. By an identical argument, the two cycle can get mapped to itself in only 2 ways. A slight modification of the above argument is necessary for the three cycles. There are 2 copies of a 3-cycle in σ. Let s call these cycles A and B. One way to map to an equivalent cycle is to have τ map the A cycle to itself, and the B cycle to itself. Each cycle can be mapped to itself in 3 ways, so in total there are 3 3 = 9 ways to find an equivalent cycle. Additionally, we could have that τ maps the A cycle to the B cycle, and vice versa, providing another 9 ways to create an equivalent cycle. In total, there are = 18 ways to map the 3-cycles to themselves. Finally, there are seven 1-cycles that can be freely permuted amongst one another. There are 7! ways to map the 1-cycles to themselves. We compute the order of the centralizer by counting the total number of ways to map σ to itself. This will just be the product of the number of ways to map each cycle to itself: C S20 (σ) = 7! = Problem 4 Proposition 2. Let be a group of odd order, and let g, then x = gx 1 g 1 x = e Proof. Suppose for the sake of contradiction that x is conjugate x 1, then g such that x = gx 1 g 1. Let me denote the order of g by g. Note that g must be odd, because the order of the cyclic subgroup generated by g must divide the order of the group. x = gx 1 g 1 x 1 = (gx 1 g 1 ) 1 = gxg 1 Plugging back into the conjugacy relation, then substituting the original definition of x, we obtain: x = gx 1 g 1 = g(gxg 1 )g 1 = g 2 xg 2 = g 2 (gx 1 g 1 )g 2 = g 3 x 1 g 3 We may continue this process inductively until arriving at the result: x = g g x 1 g g x = x 1 x 2 = e So the order of the cyclic subgroup generated by x is two. If x e then we have a contradiction because the order of a subgroup must divide the order of the group. The only other possibility is that x = e, else, x cannot be conjugate to x 1. Proposition 3. If Proof. Since Z() Z() 5. Problem 5 is cyclic, then is Abelian. Z() may be written partition, so both of x, y may be written is cyclic, it has a generator. Let g, then a general element of as [g k ] = g k Z(). Pick any two x, y. The cosets Z()
5 4 J.A. MRACEK DEPARTMENT OF MATHEMATICS UNIVERSITY OF TORONTO as x = g k z 1, y = g l z 2 for some z 1, z 2 Z(). Since z 1, z 2 are in the center, they commute with every element of ; thus: xy = g k z 1 g l z 2 = g k g l z 1 z 2 = g l g k z 2 z 1 = g l z 2 g k z 1 = yx So is an abelian group. Lemma 1. Z() Inn Proof. I must find an isomorphism between map: ψ : 6. Problem 6 Z() and Inn. The most obvious choice is to define the Z() Inn gz() φ g Where φ g (x) = gxg 1 for any x. This is a well defined map. Pick two representatives g, h of the same coset, gz() = hz(), then g = hz for some z Z(). Recall that elements of the center commute with every g. ψ(gz())(x) = φ g (x) = gxg 1 = (hz)x(hz) 1 = hzxz 1 h 1 = hxh 1 = φ h (x) = ψ(hz())(x) Thus, the map ψ is well defined. It is a homomorphism because: ψ(gz() hz())(x) = ψ(ghz())(x) = φ gh (x) = ghxh 1 g 1 = φ g φ h (x) = ψ(gz()) ψ(hz())(x) Surjectivity of the map is a triviality following directly from the definition of ψ. Injectivity is clear since ker ψ = { z gxg 1 = x } = Z() is trivial. Lemma 2. A subgroup of a cyclic group is cyclic. Proof. Since is cyclic, it has a generator. Every element x may be written x = g k for some integer k. But now it is clear that since H, every element h H must also satisfy h = g l for some integer l; thus, H is cyclic. Proposition 4. If Aut is cyclic, then is abelian. Proof. Since Aut is cyclic and Inn < Aut, the second lemma has shown that Inn must also be cyclic. However, the first lemma showed that Z() Inn. This means that Z() is a cyclic group. By the result shown in problem 5, we know that must be abelian. 7. Problem 7 Proposition 5. Let be a group and H a subgroup of finite index, then there exists some normal subgroup N of with N H. Proof. Let x, g, and define a left action on the cosets H L : by : H H L x (gh) = (xg)h Let e,..., g n be representatives from each of the n cosets, and let L g (g i H) = gg i H = g j H - i.e. we can use the left action to define a homomorphism from to the symmetric group S n : ψ : This map is a homomorphism because: S n g [L g (eh), L g (g 2 H),..., L g (g n H)] ψ(xy) = [L xy (eh), L xy (g 2 H),..., L xy (g n H)] = [(xyh), (xyg 2 H),..., (xyg n H)] = [L x L y (eh), L x L y (g 2 H),..., L x L y (g n H)] = ψ(x) ψ(y) The kernel of this homomorphism is clearly contained in H, because for any h, L h (eh) = hh = eh clearly requires h H; thus, a necessary condition for h ker ψ is that h H. But now we are done, since ker ψ H is a normal subgroup of.
6 MAT1100HF ALEBRA: ASSINMENT II 5 Proposition 6. Let be a group and H 1 and H 2 be subgroups of. If [ : H 1 ] < and [ : H 2 ] <, then [ : H 1 H 2 ] <. Proof. By proposition 5, we know we can find N 1 and N 2 such that N 1 H 1 and N 2 H 2. It is clear that N 1 N 2 H 1 H 2, implying that N 1 N 2 H 1 H 2. Let [ : N 1 ] = m < and [ : N 2 ] = n <, and define homomorphisms ϕ 1 : S m and ϕ 2 : S n with ker ψ 1 = N 1 and ker ψ 2 = N 2, as was done in the proof of proposition 5. Now, define another homomorphism: ψ : S m S n ψ(g) = (ψ 1 (g), ψ 2 (g)) Clearly, this map is a homomorphism because both of its components are homomorphisms. The kernel of this map can be easily seen to be N 1 N 2, because ψ(g) = (e, e) requires that both ψ 1 (g) = e and ψ 2 (g) = e. By the first isomorphism theorem, Imψ S m S n N 1 N 2 But since S m S n is finite, so must be N 1 N 2 because N 1 N 2 S m S n. But now we re done because: > N 1 N 2 H 1 H 2 [ : H 1 H 2 ] <
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