MATH 241 Homework 9, Fall 2011

Math 241 Problem Set 9 solution manual Exercise. A9.1 a- Dodecahedron: vertices =20, edges=30 , faces=12 , and 3 edges emanate from each vertex. Icosahedron: vertices =12, edges=30, faces=20 , and 5 edges emanate from each vertex. b- For the dodecahedron: Notice that the stabilizer of a vertex is a cyclic group of order 3, and hence we can deduce that the order of the group Gd is equal to 3×number of vertices, i.e |Gd|=60. If we allow reflections then the stabilizer of the vertex of a dodecahedron is isomorphic to S3, and hence the number of elements would be 120. For the icosahedron: Similarly as above, one can notice that the stabilizer of a vertex is the cyclic group of order 5, hence |Gi|=60, and allowing reflections then the stabilizer of a vertex would be isomorphic to D5, hence the number if elements would be 120. c- For the icosahedron:( the stab of a vertex is computed in part (b) ) For the face we have the stabilizer is a cyclic group of order 3. For the edge we have the stabilizer is a cyclic group of order 2. For the dodecahedron :( the stab of a vertex is computed in part (b) ) For the face we have the stabilizer is a cyclic group of order 5. For the edge we have the stabilizer is a cyclic group of order 2. Now it is easy to check that those results are compatible with the number of elements in Gi , and Gd computed in part (b). d- To see that the two groups Gi, and Gd are isomorphic, we have to consider each vertex of the icosahedron as a face of dodecahedron, and each face of the icosahedron as a vertex of the dodecahedron. As for seeing the isomorphism between Gd and A5, you can either use the action of Gi on the 5 squares that can be embedded inside the dodecahedron, or you can find a a subgroup H of order 12 in Gd, and use the action of Gd on Gd/H. Exercise. A9.2 For S5: Since we know that f(ijk..)f−1 = (f(i)f(j)f(k)...), then it is easy to see that all the m-cycles in S5 are conjugate. Similarly the f(ij...)(kl...)f−1 = (f(i)f(j)...)(f(k)f(l)...) , and hence all prod- ucts of m-cycles by n-cycles in S5 are conjugate. Hence we have the following: 1 x Cx |Cx| |Zx| idS5 {id} 1 120 (ij), transposition { all transpositions } 10 12 (ijk) 3-cycle { all 3-cycles } 20 6 (ij)(kl) product of two { all products of two 15 8 disjoint transpositions disjoint transpositions } (ijkl) 4-cycle {all 4-cycles} 30 4 (ijk)(lm) product of disjoint { all disjoint products of 20 6 transposition and 3-cycle transpositions with 3-cycles } (ijklm), 5-cycle { all 5-cycles } 24 5 Lemma 1. If τ commutes with σ, then (fτ)σ(fτ)−1 = fσf−1. Proof. (fτ)σ(fτ)−1 = (fτ)στ −1f−1 = f(τστ)f−1 = fττσf−1 = fσf−1. Corollary 1. If ∃ τ (transposition) that commutes with σ then the conjugacy class of σ in A5 is the same as the conjugacy class in S5. Proof. It is easy to see that any element conjugate to σ in A5 is conjugate to σ is S5. Now suppose σ′ is conjugate to σ in S5, then ∃ f ∈ S5 such that fσf−1 = σ, if f ∈ A5 then σ and σ′ are conjugate in A5, if not ( i.e f /∈ A5) then fτσ1τf−1 = σ2 and fτ is an element of A5, hence σis conjugate to σ′ is A5. So for A5 we have only 5 orbits. In A5 we have 3-cycles, product of disjoint transpositions, and 5-cycles. For any 3-cycle we can find a transposition τ that commutes with it, and hence by above corol- lary the conjugacy class for the 3-cycles is the same as in S5. For the product of two disjoint transpositions, Notice that (ij) commutes with (ij)(kl) for dis- tinct i, j, k, l, and hence their conjugacy class is the same as in S5. For the 5-cycles they are not all conjugate. Let Cσ be the conjugacy class in S5, and C′ σ be the conjugacy class in A5. Fix a transposition τ , it is easy to see that S5 = A5 ∪ A5τ, and we can then deduce that Cσ = C′ σ ∪ C′ τστ −1 since for f ∈ S5 with fσf−1 either f ∈ A5 , and hence fσf−1 ∈ C′ σ or f /∈ A5, then ∃ g ∈ A5 such that f = gτ, with fσf−1 = (gτ)σ(gτ)−1 , so fσf−1 ∈ C′ τστ −1. Next we find |C′ σ| = |A5|/|Z′ σ|, where Z′ σ = Zσ ∩ A5. But Z′ σ =< σ > since from above table we have |Zσ| = 5, and we know that < σ >⊂ Z′ σ. Hence we deduce that |C′ σ| = 12, and similarly |C′ τστ −1|=12. Hence we deduce that we have two conjugacy classes for the 5-cycles. Exercise. A9.3 Consider the function : f : GLn(Zp) −→ Z∗ p defined by f(g) = |g|. It is easy to see that f is a surjective group homomorphism. The kernel of f is equal to the subgroup SLn(Zp). So we deduce that GLn(Zp)/SLn(Zp) ∼= Z∗ p, hence |SLn(Zp)| = |GLn(ZP )|/|Z∗ p| = (pn −1)(pn − p)...(pn − pn−1)/(p − 1). 2 Exercise. A9.4 Lemma 2. If L1, L2, and L3 are three different lines in the plane then ∃ v1 ∈ L1 , and v2 ∈ L2 such that v3 = v1 + v2 ∈ L3 and L1 = span{v1}, L2 = span{v2}, and L3 = span{v3}. Proof. Suppose that L1 = span{u1}, L2 = span{u2}, and L3 = span{u3}, then u1, and u2 are linearly independent, and hence u3 = iu1 + ju2 ( i, j both non-zero). Then we let v1 = iu1, and v2 = ju3, and we get v3 = v1 + v2. A one dimensional subspaces of R2 is the span of some vector (a, b). Let L1, L2, and L3, be 3 different lines in R2, hence we can find 3 vector v1, v2, and v3 such that span{vi} = Li, with v3 = v1 + v2. Similarly for the three lines m1, m2, and m3, we can find 3 vectors w1, w2, w3 such that w3 = w1 + w2, and mi = span{wi}. Next we can find a matrix g ∈ GL(2, R) such that gv1 = w1, and gv2 = w2, where g is the transition matrix from the basis {v1, v2} to {w1, w2}. So we have gv3 = g(v1 + v2) = w1 + w2 = w3. So we get our result. Exercise. A9.5 Let us consider first the L = span{u} , where u = � 1 0 � ( Note that L = span{ � a 0 � } for any a ̸= 0). The orbit of L is OL = {g.L | g ∈ B}. Let g = � a b 0 c � with a, c non-zero, then g.L = span(gu) = span( � a 0 � ) = span(u) = L , so OL = {L}. This orbit has one element whose stabilizer is B. For any other line L such that L = span(v) , where v = � α β � with α, β ∈ R, and β ̸= 0, we have g.L = span(gv) = span( � aα + bβ cβ � . Notice that for any vector � i j � ∈ R2 with j ̸= 0, we can choose g = � 1 i−α β 0 j β � , then g.L = span( � i j � ). We deduce that OL is equal to all one dimensional subspaces of R2 except the one spanned by � 1 0 � . The stabilizer of L = span(v) where v = � 0 1 � is the set of diagonal matrices. Now generalizing for the action of the group of upper triangular matrices B of GL(n, R). We start by u =   α 0 : 0   and L = span(u). The orbit of L is OL = {L} . Next the we consider the vector v =   α1 α2 0 : 0   where α1, α2 ∈ R with α2 ̸= 0 , and we let 3 L = span(v), the orbit of L is equal to all the one dimensional subspace generated by vectors w =   β1 β2 0 : 0   where β1, β2 ∈ R with β2 ̸= 0. We keep doing this and hence we get n different orbits, where the ith orbit is the set of the lines spanned by a vector of the form   α1 α2 : αi 0 : 0   , with αi ̸= 0. Section. 36 Exercise. 1 p = 3, and |G| = 12 = 22 × 3, hence the order of the Sylow 3-subgroup is 3. Exercise. 2 p = 3, and |G| = 54 = 2 × 33, hence the order of the Sylow 3-subgroup is 27. Exercise. 3 p = 2, and |G| = 24 = 24 × 3, let s be the number of the Sylow 2-subgroups, we know from the third Sylow theorem that s/24 ,and s ≡ 1 mod(2), the divisors of 24 are 1,2,3,4,6,8,12,24, but s must be congruent to 1 mod(2), so our only choices are 1, and 3. Exercise. 4 |G| = 255 = 3 × 5 × 17, following the same argument as in ex 3, the number of Sylow 3-subgroups can be either 1 or 85, and the number if the Sylow 5-subgroups can either be 1 or 51. Exercise. 11 GH{g ∈ G | gHg−1 = H}. a- identity is in GH since eHe−1 = eHe = H. b- GH is closed under multiplication. Let g, l ∈ GH, then gHg−1 = H, and lHl−1 = H, and hence (gl)H(gl)−1 = g(lHl−1)g−1) = gHg−1 = H. So gl ∈ GH. c- Let a ∈ GH then aHa−1 = H, but also H = (a−1a)H(a−1a) = a−1(aHa−1)a = a−1Ha, hence a−1 ∈ Gh. Hence GH is a subgroup of G. 4 Exercise. 12 G has a unique Sylow p-subgroup called P. Let g ∈ G be any element gPg−1 is another Sylow p-subgroup , hence gPg−1 = P, since P is unique, and hence all the conjugates of P are equal to P, so P is normal. Then G has a non-trivial normal subgroup, so G is not simple. Exercise. 13 |G| = 45 = 32 × 5. G has a Sylow 3-subgroup of order 9, using the same argument used in number 3 we can deduce that it is a unique subgroup, and hence by number 12 it is normal. So G has a normal subgroup of order 9. 5