Algebra Prelim Solutions, August 2016

1. (a)

   The number of Sylow 5-subgroups is congruent to 1 mod 5 and divides 48, so the possibilities are 1, 6 and 16. However G is simple, so 1 is not possible, nor is 6 because then G would be isomorphic to a subgroup of A₆ which has order 360, but | G| does not divide 360. Therefore G has exactly 16 Sylow 5-subgroups, which means that G has a subgroup of order 240/16 = 15. Let H be a group of order 15. The number of Sylow 3-subgroups is congruent to 1 mod 3 and divides 5, so there is a unique Sylow 3-subgroup A which must be normal. Similarly the number of Sylow 5-subgroups is congruent to 1 mod 5 and divides 3, so there is a unique Sylow 5-subgroup B which must be normal. Since A∩B = 1 and AB = H, it follows that H≌A×B, so H is an abelian group of order 15 and it follows from the structure theorem for finitely generated abelian groups that H is cyclic.

   (b)

   From (a), we see that the normalizer of a Sylow 3-subgroup has a subgroup of order 15, and we deduce that the number of Sylow 3-subgroups divides 240/15 = 16. Therefore number of Sylow 3-subgroups is congruent to 1 mod 3 and divides 16, so the possibilities are 1, 4 and 16. However 1 and 4 are not possible because G is simple. Therefore G has exactly 16 Sylow 3-subgroups. Since two distinct Sylow 3-subgroups intersect in the identity, we conclude that G has exactly 32 elements of order 3.

2. (a)

   Since each I_i is principal, there exist a_i∈I_i such that I_i = (a_i) for all i∈ℕ. Write a₁ = up₁...p_d where u is a unit and the p_i are primes. Since (a_i)⊆(a_i+1), we see that a_i+1 divides a_i for all i. But R is a UFD, so either a_i and a_i+1 are associates in which case (a_i) = (a_i+1), or a_i+1 is divisible by at least one fewer prime of the primes in {p₁,..., p_d} than a_i. The result follows.

   (b)

   Note that if I is an ideal generated by finitely many elements a₁,..., a_d where d≥2, then (a_d-1, a_d) is principal, so is equal to (b) for some b∈R and then I = (a₁,..., a_d-2, b). Thus I can be generated by d - 1 elements and it follows by induction on d that I is principal. Now let I be an arbitrary ideal. If I is not finitely generated, then we can find an infinite sequence a₁, a₂,...∈I such that a_n+1∉(a_n). Set I_n = (a₁,..., a_n). Then I_n is a principal ideal for all n because it is finitely generated. This contradicts (a).

3. (a)

   This is true. Since P is projective, we may write P⊕Q = F, where F is a free S-module. Then

   (R⊗_SP)⊕(R⊗_SQ)≌R⊗_S(P⊕Q)≌R⊗_SF.

   Now R⊗_SS≌R as left R-modules (via the map induced by r⊗s↦rs, which has inverse r↦r⊗1). If F is free on X, then R⊗_SF is free on 1⊗x, so R⊗_SP is a direct summand of the free R-module R⊗_SF. This proves that R⊗_SP is a projective R-module.

   (b)

   Let F be a field (e.g. ℚ), let S = F and let R = F[x]. Then S is an injective S-module (over a field all modules are both injective and projective; this is just a consequence of the fact that every subspace has a direct complement). On the other hand R⊗_SS≌R (see above). This is not injective; consider the F[x]-submodule xF[x] of F[x]. The map f↦xf show that F[x]≌xF[x], so if xF[x] was injective, then xF[x] would be also and we would conclude that xF[x] is a direct summand of F[x], say F[x] = xF[x]⊕K, where K is an ideal of F[x]. Since xF[x]≠F[x] (because xF[x] consists of polynomials of degree at least 1), we see that K≠ 0. Let 0≠k∈K. Then 0≠xk∈xF[x]∩K, which contradicts the direct sum property. Therefore F[x] is not an injective F[x]-module, as required.

4. We use the structure theorem for finitely generated modules over a PID, elementary divisor form. We may write

   M ≌R^d⊕⊕_i(R/Rq_i)^d_i

   N ≌R^e⊕⊕_i(R/Rq_i)^e_i

   
   
   where d, e, d_i, e_i≥ 0, the q_i are distinct prime powers in R, and uniquely so apart from the possibility that some of the d, e, d_i, e_i = 0. Since M³≌N², we deduce from uniqueness that 3d = 2e and 3d_i = 2e_i for all i. Therefore d and all d_i are divisible by 2 and we may set P = R^d/2⊕_i(R/Rq_i)^d_i.

5. (a)

   Since A is similar to A², there exists an invertible matrix X such that XAX^-1 = A² and we see that XA^mX^-1 = A^2m for all m∈ℕ. Then for n∈ℕ, we have

   XⁿAx^-n = X^n-1A²X^-n+1 = X^n-2A⁴X^2-n = ... = A²ⁿ,

   which proves that A is similar to A²ⁿ.

   (b)

   We show that the Jordan canonical form J for A is a diagonal matrix, which will prove the result because J is similar to A. Since A is similar to A², we see that J is similar to J² and by part (a), we deduce that J is similar to J²ⁿ for all n∈ℕ. Choose n greater than the size of the matrix A and set e = 2ⁿ. We show that J^e is a diagonal matrix. Let K = J(d, a) be a Jordan block of A, that is a d×d matrix with a's on the main diagonal and 1's on the superdiagonal. Then we may write K = aI + N where I is the identity matrix. Note that aI and N commute, because everything commutes with the identity matrix, and N^d-1 = 0, in particular N^e = 0. By Freshman's dream, we get K² = a²I + N², and repeating this n times, we obtain K^e = a^eI, a diagonal matrix, and the result follows.

6. Let ω = e^2πi/7, a primitive 7th root of 1. Then ℚ(ω) is a Galois extension of ℚ with degree 6 and abelian Galois group G of degree 6. Complex conjugation γ is an element of order 2 of G, and its fixed field F will be a Galois extension of ℚ of degree 3 (Galois because all subgroups of G are normal). Since ω + γ(ω)∈F - ℚ, we see that ℚ(ω + γ(ω)) is a Galois extension of degree 3 over ℚ. We conclude that ℚ(cos(2π/7)) is a Galois extension of ℚ. If K is any such field, then K is the splitting field of some polynomial f∈ℚ[x]. Then K(√2) is the splitting field for (x² - 2)f and we see that K(√2) is a Galois extension of ℚ. We cannot have √2∈K, because [ℚ(√2) : ℚ] = 2 and [K : ℚ] = 3. Therefore K(√2) is a Galois extension of degree 6 over ℚ; let G denote the Galois group. Since K is a Galois extension of degree 3 over ℚ, we see that G has a normal subgroup of index 3, i.e. of order 2. Also ℚ(√2) is a Galois extension of degree 2 over ℚ, so G also has a normal subgroup of index 2, i.e. of order 3. It follows that G is abelian and hence isomorphic to ℤ/6ℤ.

7. Write ψ = Ind_H^G(χ). For h∈H, we have
   | H|ψ(h) = ∑_g∈Gχ(ghg^-1) = | G|χ(h)
   because ghg^-1 = h for all g∈G. Therefore ψ|_H = | G/H|χ. By Frobenius reciprocity, we now see that
   (ψ,ψ)_G = (χ,ψ|_H)_H = | G/H|(χ,χ)_H,
   which proves that ψ is not irreducible when | G/H| > 1.