- (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*A*_{6}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.

- (a)
- Since each
*I*_{i}is principal, there exist*a*_{i}∈*I*_{i}such that*I*_{i}= (*a*_{i}) for all*i*∈ℕ. Write*a*_{1}=*up*_{1}...*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 {*p*_{1},...,*p*_{d}} than*a*_{i}. The result follows. - (b)
- Note that if
*I*is an ideal generated by finitely many elements*a*_{1},...,*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*= (*a*_{1},...,*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*a*_{1},*a*_{2},...∈*I*such that*a*_{n+1}∉(*a*_{n}). Set*I*_{n}= (*a*_{1},...,*a*_{n}). Then*I*_{n}is a principal ideal for all*n*because it is finitely generated. This contradicts (a).

- (a)
- This is true. Since
*P*is projective, we may write*P*⊕*Q*=*F*, where*F*is a free*S*-module. Then(Now*R*⊗_{S}*P*)⊕(*R*⊗_{S}*Q*)≌*R*⊗_{S}(*P*⊕*Q*)≌*R*⊗_{S}*F*.*R*⊗_{S}*S*≌*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*⊗_{S}*F*is free on 1⊗*x*, so*R*⊗_{S}*P*is a direct summand of the free*R*-module*R*⊗_{S}*F*. This proves that*R*⊗_{S}*P*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*⊗_{S}*S*≌*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.

- 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})^{di}*N*≌ *R*^{e}⊕⊕_{i}(*R*/*Rq*_{i})^{ei}

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*M*^{3}≌*N*^{2}, we deduce from uniqueness that 3*d*= 2*e*and 3*d*_{i}= 2*e*_{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})^{di}. - (a)
- Since
*A*is similar to*A*^{2}, there exists an invertible matrix*X*such that*XAX*^{-1}=*A*^{2}and we see that*XA*^{m}*X*^{-1}=*A*^{2m}for all*m*∈ℕ. Then for*n*∈ℕ, we have*X*^{n}*Ax*^{-n}=*X*^{n-1}*A*^{2}*X*^{-n+1}=*X*^{n-2}*A*^{4}*X*^{2-n}= ... =*A*^{2n},*A*is similar to*A*^{2n}. - (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*A*^{2}, we see that*J*is similar to*J*^{2}and by part (a), we deduce that*J*is similar to*J*^{2n}for all*n*∈ℕ. Choose*n*greater than the size of the matrix*A*and set*e*= 2^{n}. 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*K*^{2}=*a*^{2}*I*+*N*^{2}, and repeating this*n*times, we obtain*K*^{e}=*a*^{e}*I*, a diagonal matrix, and the result follows.

- 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 (*x*^{2}- 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. - Write
ψ = Ind
_{H}^{G}(χ). For*h*∈*H*, we have|because*H*|ψ(*h*) = ∑_{g∈G}χ(*ghg*^{-1}) = |*G*|χ(*h*)*ghg*^{-1}=*h*for all*g*∈*G*. Therefore ψ|_{H}= |*G*/*H*|χ. By Frobenius reciprocity, we now see that(ψ,ψ)which proves that ψ is not irreducible when |_{G}= (χ,ψ|_{H})_{H}= |*G*/*H*|(χ,χ)_{H},*G*/*H*| > 1.

Peter Linnell 2016-08-25