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 An where n = 2, 4 or 8. Thus | G| divides | An| = 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 IiIi+1 (strict inclusion) for all i. Write Ii = (ai) where aiR (assume that ai = 1 if Ii = R). Since (ai)⊂(ai+1), we see that ai+1 divides ai and that ai is not an associate of ai+1 for all i. Write a1 = p1...pd where the pi are primes. From the previous sentence and the fact that R is a UFD, we see that ai 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 = Pqd for all n∈ℕ. Let 0≠pP. Then pF and it will have a nonzero coordinate; let p1∈ℤ be the value of this nonzero coordinate. Choose d∈ℕ such that qd | p1. Then there is no fF such that fqd = 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 = UV where V is a subspace of P/Pq. Then the formula (u, v) |--> u yields an epimorphism P/PqU. Composing this with the natural epimorphism PP/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 PP -> ℤ/qℤ, which is clearly onto. This shows that PP≠ 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 mM, 0≠r∈ℤ and mr = 0, then m)r = θ(mr) = θ0 = 0, which shows that θST. Similarly φTS. Since S| = | S| and T| = | T|, we deduce that θS = T and φT = S, in particular ST. Let π : NN/T denote the natural epimorphism. Since θS = T, we see that ker(πθ) = S and hence πθ induces a monomorphism θ1 : M/SN/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≌ℤa and N/T≌ℤb for some nonnegative integers a, b, and then the previous sentence shows that ab. Similarly ba and we deduce that M/SN/T. Since MM/SS and NN/TT, we conclude that MN 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) = xr for some positive integer rn. Since rd, we see that Ad = 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) = xd.

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 ∀ hH}. 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,..., fN).

Now let sS. Then fn(s)≠ 0 for some positive integer n. Also fn = f1h1 + ... + fNhN for some hik[x1,..., xd]. Therefore fn(s) = f1(s)h1(s) + ... + fN(s)hN(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 χi⊗ψj. 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

Peter Linnell 2012-08-12