- (a)
- The number of Sylow 11-subgroups is congruent to 1 mod 11 and divides
2
^{3}·7^{2}. 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 7^{2}divides |*G*|, and the result follows.

- 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*a*_{1}=*p*_{1}...*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. - (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*p*_{1}∈ℤ be the value of this nonzero coordinate. Choose*d*∈ℕ such that*q*^{d}~~|~~*p*_{1}. 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/*q*ℤ. Let*U*≌ℤ/*q*ℤ 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/*q*ℤ. - (c)
- Define
θ :
*P*×*P*-> ℤ/*q*ℤ by θ(*p*,*r*) = π*p*π*r*. It is easily checked that θ is a ℤ-balanced map and hence induces a ℤ-homomorphism*P*⊗_{ℤ}*P*-> ℤ/*q*ℤ, which is clearly onto. This shows that*P*⊗_{ℤ}*P*≠ 0.

- 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 (θ*m*)*r*= θ(*mr*) = θ0 = 0, which shows that θ*S*⊆*T*. Similarly φ*T*⊆*S*. Since |θ*S*| = |*S*| and |φ*T*| = |*T*|, we deduce that θ*S*=*T*and φ*T*=*S*, in particular*S*≌*T*. Let π :*N*↠*N*/*T*denote the natural epimorphism. Since θ*S*=*T*, we see that ker(πθ) =*S*and hence πθ induces a monomorphism θ_{1}:*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. - Let
*c*(*x*) denote the characteristic polynomial of*A*and let*m*(*x*) denote the minimum polynomial of*A*. Since*A*^{n}= 0, we see that*m*(*x*) |*x*^{n}, 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}. -
ℚ(ζ) is the splitting field of the irreducible
polynomial
(
*x*^{13}- 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. - By Hilbert's basis theorem, there exists a positive integer
*N*such that(Now let*f*_{1},*f*_{2},...) = (*f*_{1},*f*_{2},...,*f*_{N}).*s*∈*S*. Then*f*_{n}(*s*)≠ 0 for some positive integer*n*. Also*f*_{n}=*f*_{1}*h*_{1}+ ... +*f*_{N}*h*_{N}for some*h*_{i}∈*k*[*x*_{1},...,*x*_{d}]. Therefore*f*_{n}(*s*) =*f*_{1}(*s*)*h*_{1}(*s*) + ... +*f*_{N}(*s*)*h*_{N}(*s*), and the result follows. - The character table for
*S*_{3}isClass Size 1 3 2 Class Rep 1 (1 2) (1 2 3) χ _{1}1 1 1 χ _{2}1 -1 1 χ _{3}2 0 -1 while the character table for ℤ/2ℤ is

Class Size 1 1 Class Rep 0 1 ψ _{1}1 1 ψ _{2}1 -1 The conjugacy classes for

*S*_{3}×ℤ/2ℤ are of the form 𝓢×𝓣, where 𝓢 is a conjugacy class for*S*_{3}and 𝓣 is a conjugacy class for ℤ/2ℤ. Thus in particular*S*_{3}×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*S*_{3}and ℤ/2ℤ, namely the representations χ_{i}⊗ψ_{j}. Thus the character table of*S*_{3}×ℤ/2ℤ isClass 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) χ _{1}⊗ψ_{1}1 1 1 1 1 1 χ _{1}⊗ψ_{2}1 -1 1 -1 1 -1 χ _{2}⊗ψ_{1}1 1 -1 -1 1 1 χ _{2}⊗ψ_{2}1 -1 -1 1 1 -1 χ _{3}⊗ψ_{1}2 2 0 0 -1 -1 χ _{3}⊗ψ_{2}2 -2 0 0 -1 1

Peter Linnell 2012-08-12