Mathematics W4041x Introduction to Modern Algebra Answers to Final Exam December 16, 2010 1. Any finite group is isomorphic to a subgroup of a symmetric group. 2. (a) 1 C A7 C Σ7 ; (b) 1 C {±1} C {±1, ±i} C Q; (c) 1 C A6 × 1 × 1 C A6 × A7 × 1 C A6 × A7 × A8 ; (d) 1 C 3Z15 C Z15 C D30 . 3. Yes, (12345); yes, (12345)(67); no, since the order is the least common multiple of the cycle lengths in a disjoint cycle decomposition, and cycle types 5 + 3 = 8 or 15 are too long for Σ7 ; (1234)(567) has order 12. 4. #A5 = 5!/2 = 60 = 22 · 3 · 5. The Sylow 3-subgroups and 5-subgroups have prime orders 3 and 5, hence are isomorphic to Z3 and Z5 . In fact (123) and (12345) generate them. The Sylow 2-subgroups have order 4. One of them is {e, (12)(34), (13)(24), (14)(23)}, which is ∼ = Z2 × Z2 : any map taking e to e is an isomorphism. 5. By the Chinese remainder theorem, all but Z15 × Z6 are isomorphic to Z9 × Z5 × Z2 . But Z15 × Z6 ∼ = Z3 × Z3 × Z5 × Z2 is not isomorphic to the others, as it contains no elements of order 9. 6. True. Let σ = (123 · · · `) and τ = (` + 1 · · · n). Then στ = τ σ, |σ| = `, |τ | = m, and τ |{1,...,`} = id{1,...,`} as well as σ|{`+1,...,n} = id{`+1,...,n} . Let φ : Z` × Zm → Σn be given by φ([i], [j]) = σ i τ j . 0 0 0 0 Then φ([i + i0 ], [j + j 0 ]) = σ i+i τ j+j = σ i τ j σ i τ j = φ([i], [j])φ([i0 ], [j 0 ]), so φ is a homomorphism. And φ([i], [j]) = id implies φ|{1,...,`} = id implies `|i implies [i] = [0], and similarly [j] = [0], so φ is injective. 7. False: Z3 × Z5 < A8 , as in the previous problem. Or Z2 × Z2 ∼ = {e, (12)(34), (13)(24), (14)(23)} < A5 . 8. Say N C G but N = {e, n} for n 6= e. Since normal, for all g ∈ G, we have gng −1 ∈ N . But gng −1 6= e, for otherwise n = g −1 eg = e. So gng −1 = n. Hence gn = ng and n ∈ ZG. So N < ZG. 9. ∆ C G × G ⇔ for all g ∈ G and (h1 , h2 ) ∈ G × G, (h1 , h2 )(g, g)(h1 , h2 )−1 ∈ ∆ ⇔ for all (h1 , h2 ) ∈ G × G, −1 h1 gh−1 = h2 gh−1 = g, hence hg = gh, so G is 1 2 . Taking h1 = h and h2 = e, the latter implies hgh −1 −1 abelian. Conversely, if G is abelian, then h1 gh1 = g = h2 gh2 . 10. Note 99 = 32 · 11. By Sylow Thm 3, the no. of Sylow 3’s divides 11 and is ≡ 1 mod 3, hence is 1. Also, the no. of Sylow 11’s divides 9 and is ≡ 1 mod 11, hence is 1. So both are normal, and by the Main Thm on Direct Products, G ∼ = S3 × S11 . Moreover, S11 is of prime order, hence ∼ = Z11 , while S3 is of order p2 ∼ where p = 3 is prime, hence = Z9 or Z3 × Z3 . Hence the 2 possibilities are G ∼ = Z9 × Z11 or Z3 × Z3 × Z11 . (The first has an element of order 9, the second doesn’t.) 11. Let φ : N o K → K be φ(n, k) = k. Then φ((n1 , k1 )(n2 , k2 )) = φ(n1 (k1 · n2 ), k1 k2 ) = k1 k2 = φ(n1 , k1 )φ(n2 , k2 ), so φ is a homomorphism. It is clearly surjective since φ(e, k) = k, and its kernel is exactly N o 1, so the result follows from the first isomorphism theorem. 12. Let S be a Sylow p-subgroup; if p divides #G, then S is nontrivial, say of order pn . Let g be any nontrivial m−1 element; by Lagrange, g has order pm where 0 < m ≤ n. Let h = g p ; then h has order p. 13. Because the arrows point clockwise, there is no reflection symmetry, so the symmetry group is only rotations Zp . It acts on the set S of colorings, that is, functions Zp → {1, . . . P , n}. The number of different types of 1 g e necklace is the number of orbits. By Burnside’s lemma, this is #Z g∈Zp S . For g = e, we have S = S p e p and #S = n . Any other g acts transitively on the beads, so the only fixed necklaces are those in a solid color, and then #S g = n. Hence the total number is p1 (np + (p − 1)n) = n(1 + (np−1 − 1)/p). (Notice, by the way, that we already know p | (np−1 − 1) by Fermat’s little theorem.) 14. (a) Proof 1: If g1 , g2 ∈ N (H), then g1 Hg1−1 = g2 Hg2−1 = H, so (g1 g2 )H(g1 g2 )−1 = g1 (g2 Hg2−1 )g1−1 = g1 Hg1−1 = H, so N (H) is closed under multiplication. Also, multiplying g1 Hg1−1 = H on the left by g1−1 and on the right by g1 yields H = g1−1 Hg1 , so g1−1 ∈ N (H), so N (H) is closed under inversion. Finally, eHe−1 = H, so e ∈ N (H). Proof 2: (a) follows from (c), since N (H) is a stabilizer. (b) This is trivial, since for all g ∈ N (H), gHg −1 = H, which we know implies N C N (H). (c) Let G act by conjugation on the set of all subgroups of G. Then c is the size of the orbit through H. On the other hand, since g ∈ N (H) ⇔ gHg −1 = H, it follows that N (H) is the stabilizer group of H. By the counting formula (aka orbit-stabilizer theorem), c = [G : N (H)]. (d) The orbit of H consists of all subgroups generated by a 3-cycle (abc). Since (abc) and (cba) (and all recyclings) generate the same subgroup, such subgroups are in bijective correspondence with 3-element n! subsets of {1, . . . , n}, which number c = 3!(n−3)! . Hence #N (H) = #Σn /c, that is, #N (H) = 3!(n − 3)!. ∼ (One can also show directly that N (H) = Σ3 × Σn−3 , but this takes more effort.)