Algebra Prelim Solutions, August 2020

1. Let G be a group of order 63 = 3²·7. The number n₇ of Sylow 7-subgroups of G divides 9 and is congruent to 1 mod 7. Hence n₇ = 1 and G has a normal Sylow 7-subgroup Q≌ℤ₇. We have G≌P⋉Q for any Sylow 3-subgroup P < G.

   The group Aut(Q)≌ℤ₇^× is cyclic of order 6. Fix a generator b of Q and let σ∈Aut(Q) be the element of order 3 such that σ(b) = b². Since P≌ℤ₉ or P≌ℤ₃×ℤ₃, any homomorphism P→Aut(Q) has image in ⟨σ⟩.

   We deduce that there exist exactly 4 non-isomorphic groups of order 63: the abelian groups ℤ₉×ℤ₇ and ℤ₃×ℤ₃×ℤ₇, and two non-abelian semi-direct products ℤ₉⋉ℤ₇ and (ℤ₃×ℤ₃)⋉ℤ₇. The various possibilities for the latter are seen to be isomorphic by suitably changing generators. (For example, if ℤ₉⋉ℤ₇ is generated by a, b such that a⁹ = b⁷ = e and aba^-1 = b² = σ(b), then in terms of the generators a², b we have a²ba^-2 = b⁴ = σ²(b), corresponding to the other non-trivial homomorphism P→Aut(Q) in the case P≌ℤ₉.)

2. We know that A is similar to matrix in Jordan canonical form, with nonzero eigenvalues (since A is invertible). Without loss of generality, we may assume that A is an n×n Jordan block matrix with eigenvalue μ≠ 0.

   Consider a single n×n Jordan block J with eigenvalue λ≠ 0 on the diagonal and 1 on the superdiagonal. Its square J² has λ² on the diagonal, 2λ on the superdiagonal, 1 on the next diagonal, and all other entries 0. In particular, one has (J² - λ²I)ⁿ = 0 and no smaller power of x - λ² annihilates J². Thus the Jordan canonical form of J² also consists of a single block, i.e., J² is similar to a n×n Jordan block with eigenvalue λ².

   Now pick λ∈ℂ so that λ² = μ. By the previous paragraph, we have PJ²P^-1 = A for some invertible matrix P. Then A = B² for B = PJP^-1.

3. If A and B are both free, then so is A⊗_RB. If either A or B is 0, then A⊗_RB is also 0 and hence free.

   Suppose A and B are both nonzero, but not both free. Without loss of generality, we may assume that A has an elementary divisor pⁱ. If the free part of B is nonzero, then A⊗_RB contains a (torsion) submodule isomorphic to R/(pⁱ)⊗_RR≌R/(pⁱ) and therefore cannot be free. Thus, in order for A⊗_RB to be free, B must be a torsion module. But then, for the same reason, A must also be a torsion module.

   Assuming A and B are nonzero torsion modules, the elementary divisors of A⊗_RB are determined from those of A and B by using the bilinearity of the tensor product and the isomorphism R/I⊗_RR/J≌R/(I + J) (or by direct arguments), which gives

   R/(pi)⊗RR/(pj)	≌R/(pmin{i, j})
   R/(pi)⊗RR/(qj)	= 0

   for distinct primes p, q∈R and positive integers i, j. Thus, in order for A⊗_RB to be free, A and B cannot have elementary divisors for a common prime.

   By the classification theorem, we conclude that A⊗_RB is free if and only if one of the following holds:

   • one of A, B is 0
   • both A and B are free
   • both A and B are nonzero torsion modules and they do not having elementary divisors for a common prime

4. The classification of finite abelian groups gives that the Sylow p-subgroup of A is A(p) = {a∈A : pⁿa = 0}. A homomorphism f : ℤ/pⁿℤ→A is uniquely determined by f (1̄), which must belong to A(p). Moreover, for each a∈A(p), since pⁿa = 0, there exists unique such f with f (1̄) = a, by the First Isomorphism Theorem. In other words, the map sending f∈Hom(ℤ/pⁿℤ, A) to f (1̄)∈A(p) is a bijection. This map is a group isomorphism, since f₁ + f₂↦(f₁ + f₂)(1̄) = f₁(1̄) + f₂(1̄).

5. (a)

   Let α =  ³√4∈ℝ, β = αω, γ = αω² be the roots of x³ - 4, where ω = e^2πi/3 is a primitive 3^rd root of unity. We have E = ℚ(α,ω).

   We claim that E/ℚ has degree 6. We know that [E : ℚ]≤3! = 6. The polynomial x³ - 4 is irreducible over ℚ, since it has degree 3 and no rational roots (since ±1,±2,±4 are not roots). The minimal polynomial of ω is x² + x + 1. Since E contains α and ω, its degree [E : ℚ] must be divisible by 2 and 3, hence by 6. Therefore [E : ℚ] = 6.

   The action of G = Gal(E/ℚ) on α,β,γ gives an injective homomorphism G→S₃. Since | G| = 6, it must be an isomorphism G≌S₃.

   By the Galois correspondence, E/ℚ has one intermediate field for each subgroup of S₃, of which there are 6. Thus

   ℚ,ℚ(ω),ℚ(α),ℚ(β),ℚ(γ), E

   are all intermediate fields of E/ℚ. (These are the fixed fields of the subgroups S₃,⟨(123)⟩,⟨(12)⟩,⟨(13)⟩,⟨(23)⟩,⟨e⟩, respectively, where we identify α with 1, β with 2, and γ with 3.)

   (b)

   It suffices to find θ∈E whose stabilizer in G is trivial. One checks that θ = α + ω has this property:

   α + ω	--(12)→β + ω-1 = αω - ω - 1
   α + ω	--(13)→γ + β/γ = αω2 + ω-1 = α(-ω -1) - ω - 1
   α + ω	--(23)→α + γ/α = α - ω - 1
   α + ω	--(123)→β + ω = αω + ω
   α + ω	--(132)→γ + ω = αω2 + ω

   Here we use that {1,α,α²,ω,αω,α²ω} is a ℚ-basis of E to see that none of these images are equal to α + ω.

6. (a)

   We know 𝓢 is not empty, because the zero ideal belongs to 𝓢. Let I₁⊂I₂⊂... be a chain in 𝓢. The union I = ⋃_n=0^∞I_n is an ideal of R and a^k∉I for all k≥ 0. Hence I belongs to 𝓢 and is an upper bound for the chain. We have verified the conditions of Zorn's lemma.

   (b)

   Let P be a maximal element of 𝓢. Then a∉P. We will show that P is a prime ideal of R. Suppose xy∈P for some x, y∈R, but x, y∉P. Then the ideals (x) + P and (y) + P strictly contain P, and therefore do not belong to 𝓢. Hence a^k∈(x) + P and a^l∈(y) + P for some k, l≥ 0. But then a^k+l∈(xy) + P = P, which is a contradiction, since P∈𝓢.

7. (a)

   We apply the orbit-stabilizer theorem. We have hxk^-1 = x if and only if h = xkx^-1∈H∩xKx^-1. Hence the stabilizer of x has | H∩xKx^-1| elements, and the orbit of x has | H×K|/| H∩xKx^-1| elements.

   (b)

   The orbit through the identity is HeH = H, which has | H| = q(q - 1)² elements. Taking any s∉H, e.g.,

   s =

   (

   	0	1
   	1	0

   ),

   we find that | H∩sHs^-1| = (q - 1)² and hence, by (a), the orbit through s has size q²(q - 1)⁴/(q - 1)² = q²(q - 1)². Since (q² + q)(q - 1)² = q(q + 1)(q - 1)² = | G| we are done.