Algebra Prelim Solutions, August 2012

1. (a)

   The number of Sylow 11-subgroups is congruent to 1 mod 11 and divides 2³·7². Since G is simple, this number cannot be 1. The only remaining possibility is 56. This means that the index of the normalizer of a Sylow 11-subgroup is 56. It follows that G has a subgroup of order 4312/56 = 77.

   (b)

   Let H be a subgroup of order 77 of G, which exists by (a). The number of Sylow 7-subgroups of H is congruent to 1 mod 7 and divides 11, hence must be 1. Therefore H has a normal subgroup P of order 7. Now P is contained in a Sylow 7-subgroup K, and K will have order 49. Since groups of prime squared order are abelian, we see that P <| K. It follows that the normalizer N of P contains subgroups of order 49 and 77, namely K and H.

   (c)

   Since N contains subgroups of order 49 and 77, we see that [G : N] divides 8, and cannot be 1 because G is not simple. It follows that G is isomorphic to a subgroups of A_n where n = 2, 4 or 8. Thus | G| divides | A_n| = n!/2, where 2≤n≤8. This is not possible because 7² divides | G|, and the result follows.

2. Suppose the result is not true. Then we may assume that I_i⊂I_i+1 (strict inclusion) for all i. Write I_i = (a_i) where a_i∈R (assume that a_i = 1 if I_i = R). Since (a_i)⊂(a_i+1), we see that a_i+1 divides a_i and that a_i is not an associate of a_i+1 for all i. Write a₁ = p₁...p_d where the p_i are primes. From the previous sentence and the fact that R is a UFD, we see that a_i is a product of at most d + 1 -i primes, which is not possible if i > d + 1. This finishes the proof.

3. (a)

   Since P is projective, it is a submodule of a free ℤ-module F. Suppose P = Pq. Then P = Pq^d for all n∈ℕ. Let 0≠p∈P. Then p∈F and it will have a nonzero coordinate; let p₁∈ℤ be the value of this nonzero coordinate. Choose d∈ℕ such that q^d | p₁. Then there is no f∈F such that fq^d = p and we have a contradiction.

   (b)

   We may view P/Pq as a vector space over the field ℤ/qℤ. Let U≌ℤ/qℤ be a one-dimensional subspace of P/Pq and write P/Pq = U⊕V where V is a subspace of P/Pq. Then the formula (u, v) |--> u yields an epimorphism P/Pq↠U. Composing this with the natural epimorphism P↠P/Pq, we obtain an epimorphism π : P↠ℤ/qℤ.

   (c)

   Define θ : P×P -> ℤ/qℤ by θ(p, r) = πpπr. It is easily checked that θ is a ℤ-balanced map and hence induces a ℤ-homomorphism P⊗_ℤP -> ℤ/qℤ, which is clearly onto. This shows that P⊗_ℤP≠ 0.

4. Let S and T denote the torsion submodules of M and N respectively. Since M and N are finitely generated, we see that S and T are finitely generated torsion modules and hence finite. By hypothesis, we have monomorphisms θ : M -> N and φ : N -> M. If m∈M, 0≠r∈ℤ and mr = 0, then (θm)r = θ(mr) = θ0 = 0, which shows that θS⊆T. Similarly φT⊆S. Since |θS| = | S| and |φT| = | T|, we deduce that θS = T and φT = S, in particular S≌T. Let π : N↠N/T denote the natural epimorphism. Since θS = T, we see that ker(πθ) = S and hence πθ induces a monomorphism θ₁ : M/S↪N/T, so M/S is isomorphic to a submodule of N/T. Now M/S and N/T are finitely generated torsion-free modules, so M/S≌ℤ^a and N/T≌ℤ^b for some nonnegative integers a, b, and then the previous sentence shows that a≤b. Similarly b≤a and we deduce that M/S≌N/T. Since M≌M/S⊕S and N≌N/T⊕T, we conclude that M≌N as required.

5. Let c(x) denote the characteristic polynomial of A and let m(x) denote the minimum polynomial of A. Since Aⁿ = 0, we see that m(x) | xⁿ, so m(x) = x^r for some positive integer r≤n. Since r≤d, we see that A^d = 0. Also the irreducible factors of c(x) divide m(x), hence the only irreducible factor of c(x) is x, and it follows that c(x) = x^d.

6. ℚ(ζ) is the splitting field of the irreducible polynomial (x¹³ - 1)/(x - 1) over ℚ. It follows that ℚ(ζ) is a Galois extension of ℚ of degree 12. Also Gal(ℚ(ζ)/ℚ) is cyclic of order 12. Now the subfields of ℚ(ζ) are in a one-to-one correspondence with the subgroups of Gal(ℚ(ζ)/ℚ), and a subfield of degree d over ℚ corresponds to a subgroup of index d in Gal(ℚ(ζ)/ℚ). Since a cyclic group of order 12 has exactly one subgroup of order 2, we see that Gal(ℚ(ζ)/ℚ) has a unique subgroup H of order 2. It follows that ℚ(ζ) has exactly one subfield K such that [K : ℚ] = 6. Specifically K = Fix(H) : = {z∈ℚ(ζ) | hz = z ∀ h∈H}. Also H <| Gal(ℚ(ζ)/ℚ), because all subgroups of a cyclic group are normal. It follows that K is Galois over ℚ. Finally let γ denote complex conjugation. Then γ defines a nontrivial element of Gal(ℚ(ζ)/ℚ). Since γ has order 2, it follows that H = < γ >. Therefore complex conjugation fixes K elementwise, and we deduce that K⊂ℝ.

7. By Hilbert's basis theorem, there exists a positive integer N such that
   (f₁, f₂,...) = (f₁, f₂,..., f_N).
   Now let s∈S. Then f_n(s)≠ 0 for some positive integer n. Also f_n = f₁h₁ + ... + f_Nh_N for some h_i∈k[x₁,..., x_d]. Therefore f_n(s) = f₁(s)h₁(s) + ... + f_N(s)h_N(s), and the result follows.

8. The character table for S₃ is

   Class Size	1	3	2
   Class Rep	1	(1 2)	(1 2 3)
   χ1	1	1	1
   χ2	1	-1	1
   χ3	2	0	-1

   while the character table for ℤ/2ℤ is

   Class Size	1	1
   Class Rep	0	1
   ψ1	1	1
   ψ2	1	-1

   The conjugacy classes for S₃×ℤ/2ℤ are of the form 𝓢×𝓣, where 𝓢 is a conjugacy class for S₃ and 𝓣 is a conjugacy class for ℤ/2ℤ. Thus in particular S₃×ℤ/2ℤ has 3*2 = 6 conjugacy classes, and hence it has 6 irreducible representations. We get the six irreducible representations from taking the tensor product of irreducible representations of S₃ and ℤ/2ℤ, namely the representations χ_i⊗ψ_j. Thus the character table of S₃×ℤ/2ℤ is

   Class Size	1	1	3	3	2	2
   Class Rep	((1), 0)	((1), 1)	((1 2), 0)	((1 2), 1)	((1 2 3), 0)	((1 2 3), 1)
   χ1⊗ψ1	1	1	1	1	1	1
   χ1⊗ψ2	1	-1	1	-1	1	-1
   χ2⊗ψ1	1	1	-1	-1	1	1
   χ2⊗ψ2	1	-1	-1	1	1	-1
   χ3⊗ψ1	2	2	0	0	-1	-1
   χ3⊗ψ2	2	-2	0	0	-1	1