Solutions 02 1. Let 퐻 and 퐾 be two subgroups of a group 퐺. For any element 푔 ∈ 퐺, the set 퐻 푔 퐾 ∶= {푓 ∈ 퐺 ∣ 푓 = ℎ 푔 푘 for some ℎ ∈ 퐻, 푘 ∈ 퐾} is called a double coset. (i) Prove that the double cosets partition 퐺. (ii) Do all double cosets have the same cardinality? (iii) When 퐺 has finite order, must the cardinality of a double coset divide |퐺|? Solution. (i) We claim that the following is an equivalence relation on 퐺: for all 푓, 푔 ∈ 퐺, we set 푓 ≊ 푔 if 푓 = ℎ 푔 푘 for some ℎ ∈ 퐻 and 푘 ∈ 퐾. (transitive) Suppose that 푓 ≊ 푔 and 푔 ≊ 푔′. By definition, there exists elements ℎ, ℎ′ ∈ 퐻 and 푘, 푘′ ∈ 퐾 such that 푓 = ℎ 푔 푘 and 푔 = ℎ′ 푔′ 푘′. Hence, we obtain 푓 = (ℎ ℎ′)푔′(푘′ 푘). Since 퐻 and 퐾 are subgroups, it follows that ℎ ℎ′ ∈ 퐻 and 푘′ 푘 ∈ 퐾. Thus, we deduce that 푓 ≊ 푔′. (symmetric) If 푓 ≊ 푔 then there exists ℎ ∈ 퐻 and 푘 ∈ 퐾 such that 푓 = ℎ 푔 푘 for some ℎ ∈ 퐻 and 푘 ∈ 퐾. Since 퐻 and 퐾 are subgroups, we have ℎ−1 ∈ 퐻 and 푘−1 ∈ 퐾. It follows that 푔 = ℎ−1 푓 푘−1 and 푔 ≊ 푓. (reflexive) Consider an element 푔 ∈ 퐺. Since the identity element 푒 ∈ 퐺 belongs to both 퐻 and 퐾, we have 푔 = 푒 푔 푒 and 푔 ≊ 푔. Since the double cosets are the equivalence classes for the binary relation ≊, it follows that they partition the underlying set of 퐺. (ii) All double cosets need not have the same cardinality. For example, consider the group 픖3 and the subgroups 퐻 = 퐾 ∶= {id3, (2 1)}. The double cosets are 퐻 id3 퐻 = {id3, (2 1)} = 퐻 (2 1) 퐻 퐻(3 1)퐻 = {(3 1), (3 2), (3 1 2), (3 2 1)} = 퐻 (3 2) 퐻 = 퐻 (3 1 2) 퐻 = 퐻 (3 2 1) 퐻 . (iii) The example in part (ii) satisfies |픖3| = 6 and |퐻 (3 1) 퐻| = 4. Therefore, the cardinality of a double coset does not have to divide the order of the group. □ 2. Let 퐺 be a group and let Aut(퐺) be its automorphism group. Given an element −1 푔 ∈ 퐺, consider the map 훾푔 ∶ 퐺 → 퐺 defined, for all 푓 ∈ 퐺, by 훾푔(푓)∶= 푔푓푔 . (i) For all 푔 ∈ 퐺, show that 훾푔 is an automorphism. (ii) Prove that the map Γ∶ 퐺 → Aut(퐺) defined, for all 푔 ∈ 퐺, by Γ(푔) ∶= 훾푔 is a group homomorphism. (iii) Show that Ker(Γ) = 푍(퐺). (iv) Prove that Inn(퐺)∶= Im(Γ) is a normal subgroup of Aut(퐺). Solution. (i) For all 푓, 푓′ ∈ 퐺, we have ′ ′ −1 −1 ′ −1 ′ 훾푔(푓푓 ) = 푔(푓푓 )푔 = (푔푓푔 )(푔푓 푔 ) = 훾푔(푓)훾푔(푓 ), so the map 훾푔 is a group homomorphism. For all 푓, 푔, ℎ ∈ 퐺, we also have −1 −1 −1 −1 (‡) (훾푔 ∘ 훾ℎ)(푓) = 훾푔(ℎ푓ℎ ) = 푔(ℎ푓ℎ )푔 = (푔ℎ)(푓)(푔ℎ) = 훾푔ℎ(푓) . MATH 893 : 2020 page 1 of 2 We deduce that 훾푔 ∘ 훾푔−1 = 훾푒 = id퐺 = 훾푔−1 ∘ 훾푔, so the map 훾푔 is a bijection. (ii) The equation (‡) implies that Γ(푔ℎ) = 훾푔ℎ = 훾푔 ∘ 훾ℎ = Γ(푔) Γ(ℎ), so the map Γ is a group homomorphism. (iii) Since we have the following sequence of equivalences −1 푔 ∈ 푍(퐺) ⇔ 푔푓푔 = 푓 ∀푓 ∈ 퐺 ⇔ 훾푔(푓) = 푓 ∀푓 ∈ 퐺 ⇔ 훾푔 = id퐺 ⇔ Γ(푔) = 푒Aut(퐺) , we conclude that that Ker(Γ) = 푍(퐺). (iv) When 휑 ∈ Aut(퐺) and 훾푔 ∈ Inn(퐺), it follow that, for all 푓 ∈ 퐺, we have −1 −1 −1 −1 (휑 ∘ 훾푔 ∘ 휑 )(푓) = 휑(푔 휑 (푓) 푔 ) = 휑(푔) 푓 휑(푔) = 훾휑(푔)(푓) . −1 Hence, we obtain 휑∘훾푔 ∘휑 = 훾휑(푔) ∈ Inn(퐺), so the image Inn(퐺) is a normal subgroup of Aut(퐺). □ 3. Fix 푛 ∈ ℕ. Two permutations 휍, 휏 ∈ 픖푛 have the same cycle structure if, for all 푘 ∈ ℕ, their factorizations into disjoint cycles have the same number of cycles of length 푘. The cycle type of a permutation is the list 휆 of cycles lengths from its factorization into disjoint cycles arranged in non-increasing order. −1 (i) For all permutations 휍, 휏 ∈ 픖푛, prove that the conjugate permutation 휍 휏 휍 has the same cycle structure as 휏 and is obtained by applying 휍 to the entries in the cycles of 휏. (ii) Prove that permutations are conjugate if and only if they have the same cycle type. Solution. (i) Let 휛 the permutation with the same cycle structure as 휏 obtained by applying 휍 to the entries in the cycles of 휏. We consider two cases. • Suppose that 휏(푖) = 푖. The permutation 휛 also fixes 휍(푖) because 휍(푖) lies in a cycle of length 1. Since (휍 휏 휍−1)(휍(푖)) = (휍 휏)(푖) = 휍(푖) = 휛(휍(푖)), the permutation 휍 휏 휍−1 also fixes 휍(푖). • Suppose that 휏(푖) = 푗 ≠ 푖. By definition, we have 휛(휍(푖)) = 휍(푗). On the other hand, we see that (휍 휏 휍−1)(휍(푖)) = (휍 휏)(푖) = 휍(푗) = 휛(휍(푖)). Since 휛 and 휍 휏 휍−1 agree on all 푖 ∈ [푛], we conclude that 휛 = 휍 휏 휍−1. (ii) Part (i) shows that conjugate permutations have the same cycle type. For the converse, suppose that permutations 휛 and 휏 have the same cycle type. We −1 need to produce a permutation 휍 ∈ 픖푛 such that 휛 = 휍 휏 휍 . To define 휍, place the factorization of 휏 into disjoint cycles over that of 휛 so that the cycles of the same length correspond. Let 휍 be the function sending the top row to the bottom. For example, if 휏 = (3 1 2)(5 4)(6) and 휛 = (3 1)(4)(6 2 5), then we have the array 3 1 2 5 4 6 1 2 3 4 5 6 ( ) ≡ ( ) 6 2 5 3 1 4 2 5 6 1 3 4 so 휍 = (6 4 1 2 5 3). Notice that 휍 is not uniquely determined by this procedure (it implicitly depends on the choice of bijection between the cycles of the same lengths in 휏 and 휛). Nevertheless, the part (i) implies that 휛 = 휍 휏 휍−1, so the permutations 휛 and 휏 are conjugate. □ MATH 893 : 2020 page 2 of 2.