Algebra Prelim Solutions, August 2018

  1. The number of Sylow 5-subgroups is congruent to 1 mod 5 and divides 99, so if G does not have a normal Sylow 5-subgroup, it has 11 Sylow 5-subgroups and hence 44 elements of order 5. The number of Sylow 11-subgroups is congruent to 1 mod 11 and divides 45, so if G does not have a normal subgroup of order 11, it has 45 subgroups of order 11 and hence 450 elements of order 11. If G does not have a normal Sylow 3-subgroup, then there are at least 10 elements of order a power of 3. We now see that G has at least 44 + 450 + 10 = 504 > 495, too many elements, and it now follows that G has normal Sylow p-subgroup for p = 3, or 5, or 11, as required.

    If G does not have a normal Sylow 3-subgroup, then it has either a normal Sylow 5-subgroup or a normal Sylow 11-subgroup. Suppose G has a normal Sylow 5-subgroup A. Then G/A is a group of order 99, and therefore G/A has a normal subgroup B/A of order 9, because the number of Sylow 3-subgroups in a group of order 99 is 1. Since | B| = 45, the number of Sylow 3-subgroups of B is 1 and we deduce that B has a characteristic subgroup C of order 9. We conclude that C⊲G as required. On the other hand if G has a normal Sylow 11-subgroup D, then G/D is a group of order 45 which has a normal subgroup E/D of order 9. Then E is a group of order 99 and the number of Sylow 3-subgroups is 1, consequently E has a characteristic subgroup F of order 9. It follows that G has a normal subgroup F of order 9 as required and the proof is complete.

  2. Let f∈ℤ[x] be a monic polynomial of degree n. We will show that I = (f ). If this is not the case, then we may choose g∈I∖(f ) with smallest possible degree. Clearly g≠ 0 and therefore deg(g)≥n, so we may write g = amxm + am-1xm-1 + ... + a0 where m = deg(g) and am≠ 0. Then m≥n and g -amxm-nf∈I and has degree strictly less than m, so we have a contradiction and the result is proven.

  3. To show that ℚ/ℤ⊗I = 0, it will be sufficient to show that every simple tensor x⊗y = 0. Choose n∈ℕ such that nx = 0. Since I is injective, we know that nI = I, so there exists z∈I such that nz = y. Then

    x⊗y = x⊗nz = nx⊗z = 0

    as required.

  4. By the structure theorem for finitely generated modules over a PID, invariant factor form, we may write

    M = Rd⊕R/Rt1⊕...⊕R/Rtn

    where t1| t2|...| tn≠ 0. Here T ≔R/Rt1⊕...⊕R/Rtn is the torsion submodule of M. First suppose C≌R. Then M is not a torsion module, so d≥1 and we may write M = R⊕S, where S = Rd-1⊕T. Then M/S≌R and it follows that we have an epimorphism M↠C.

    Now suppose C≇R. Then C is a torsion module, so C⊆T. Since tnT = 0, we see that tnC = 0 and we deduce that C≌R/sR where s| tn. Since there exists an R-epimorphism R/tnR↠R/sR and R/tnR is a direct summand of M, we deduce that there exists an R-epimorphism M↠C, which completes the proof.

  5. (a)
    Let G = Gal(K/ℚ). Note that G has a subgroup H of order 10, for example the normalizer of a Sylow 5-subgroup, because the number of Sylow 5-subgroups is 6. Let F denote the fixed field of H. Then [F : ℚ] = 6 and by the primitive element theorem, F = ℚ(p) for some p∈F. Let f denote the minimal polynomial of p over ℚ. Then f is irreducible, deg f = 6, and the splitting field for f is a Galois extension of ℚ contained in K. Since A5 is simple, the only Galois extensions of ℚ contained in K are ℚ and K, and we deduce that K is the splitting field of f.

    (b)
    Complex conjugation is an element γ of G, because K is a Galois extension of ℚ. The order of γ is 2, because K is not contained in the real numbers, and the fixed field of γ is R. Therefore [K : R] = 2 and we deduce that [R : ℚ] = 30.

    (c)
    By considering its action on the roots of f, we get an embedding of G into S6. If γ is an odd permutation, then the even permutations yield a subgroup of index 2 in G, which is not possible because G is simple. It follows that γ is an even permutation of order 2, so it must be a product of two 2-cycles. We deduce that f has exactly two real roots.

    (d)
    From (c), let a, b be the two real roots of f. Then ℚ(a, b)⊆R. Since 6|[ℚ(a, b) : ℚ] and [R : ℚ] = 30, we see that either [ℚ(a, b) : ℚ] = 6 or [ℚ(a, b) : ℚ] = 30. If the latter is true, then we must have R = ℚ(a, b) as required. On the other hand if [ℚ(a, b) : ℚ] = 6, then the corresponding subgroup for ℚ(a, b) has order 10 in G and therefore must contain a 5-cycle σ. But then σ can only fix one root and we have a contradiction.

  6. Let A be the given matrix. The characteristic polynomial f of A must be x3 + x2 + ax + 1, where a = 0 or 1. First suppose a = 0. Then f (x)≠ 0 for x = 0 or 1 and we see that f has no linear factor. It follows that f is irreducible and it we deduce that the rational canonical form for A is the companion matrix for x3 + x2 + 1, that is

     (
     0 0 1
     1 0 0
     0 1 1
    		)  

    Now suppose a = 1. Then f (x) = (x + 1)3 and we see that there are three possibilities for the invariant factors, namely {(x + 1)3}, {(x + 1)2,(x + 1)} and {x + 1, x + 1, x + 1}. The corresponding rational canonical forms are

     (
     0 0 1
     1 0 1
     0 1 1
    		)

     (
     0 1 0
     1 0 0
     0 0 1
    		)

     (
     1 0 0
     0 1 0
     0 0 1
    		)

    Thus there are four possible rational canonical forms, as described above.

  7. Representatives for the conjugacy classes for A4 are (1), (1 2 3), (1 3 2) and (1 2)(3 4). The sizes of the conjugacy classes are 1, 4, 4 and 3 respectively. Let V denote the Sylow 2-subgroup of A4, a normal subgroup of order 4 consisting of 1 and the fourth conjugacy class above. Then A4/V is a group of order 3, so it has 3 one-dimensional representations, and hence A4 has 3 one-dimensional representations. Since A4 has 4 conjugacy classes, it has 4 irreducible representations. Thus A4 has one more irreducible representation, which will have degree 3, because the sum of the squares of the degrees of the irreducible representations of A4 is | A4| = 12. Let ι denote the trivial character, let α,β denote the two other degree 1 characters, and let χ denote the irreducible degree 3 character. Let ω = e2πi/3 = (-1 + i√3)/2 denote a primitive cube root of 1. Then the character table is

    Class Size 1 4 4 3
    Class Rep 1 (1 2 3) (1 3 2) (1 2)(3 4)
    ι 1 1 1 1
    α 1 ω ω2 1
    β 1 ω2 ω 1
    χ 3 0 0 -1

    The character χ is derived from the rest of the character table and the orthogonality relations.