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 AB = 1 and AB = H, it follows that HA×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 Ii is principal, there exist aiIi such that Ii = (ai) for all i∈ℕ. Write a1 = up1...pd where u is a unit and the pi are primes. Since (ai)⊆(ai+1), we see that ai+1 divides ai for all i. But R is a UFD, so either ai and ai+1 are associates in which case (ai) = (ai+1), or ai+1 is divisible by at least one fewer prime of the primes in {p1,..., pd} than ai. The result follows.

(b)
Note that if I is an ideal generated by finitely many elements a1,..., ad where d≥2, then (ad-1, ad) is principal, so is equal to (b) for some bR and then I = (a1,..., ad-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 an+1∉(an). Set In = (a1,..., an). Then In 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 PQ = F, where F is a free S-module. Then

(RSP)⊕(RSQ)≌RS(PQ)≌RSF.

Now RSSR as left R-modules (via the map induced by rsrs, which has inverse rr⊗1). If F is free on X, then RSF is free on 1⊗x, so RSP is a direct summand of the free R-module RSF. This proves that RSP 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 RSSR (see above). This is not injective; consider the F[x]-submodule xF[x] of F[x]. The map fxf 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≠kK. Then 0≠xkxF[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 ≌Rd⊕⊕i(R/Rqi)di N ≌Re⊕⊕i(R/Rqi)ei

where d, e, di, ei≥ 0, the qi are distinct prime powers in R, and uniquely so apart from the possibility that some of the d, e, di, ei = 0. Since M3N2, we deduce from uniqueness that 3d = 2e and 3di = 2ei for all i. Therefore d and all di are divisible by 2 and we may set P = Rd/2i(R/Rqi)di.

5. (a)
Since A is similar to A2, there exists an invertible matrix X such that XAX-1 = A2 and we see that XAmX-1 = A2m for all m∈ℕ. Then for n∈ℕ, we have

XnAx-n = Xn-1A2X-n+1 = Xn-2A4X2-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 Je 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 Nd-1 = 0, in particular Ne = 0. By Freshman's dream, we get K2 = a2I + N2, and repeating this n times, we obtain Ke = aeI, a diagonal matrix, and the result follows.

6. Let ω = e2πi/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 = IndHG(χ). For hH, we have

| H|ψ(h) = ∑g∈Gχ(ghg-1) = | G|χ(h)

because ghg-1 = h for all gG. 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.

Peter Linnell 2016-08-25