- Let
a be a root of
*f*in*K*. Then [*F*(a) :*F*]1 because*f*has no roots in*F*, is less than 6 because*f*has degree 6, and divides 21. It follows that [*F*(a) :*F*] = 3. Let*g*be the minimum polynomial of a over*F*. Then*g*is an irreducible polynomial of degree 3 which divides*f*in*F*[*X*], so we may write*f*=*gh*in*F*[*X*] where*h*has degree 3. Since*h*has no root in*F*, we see that*h*is irreducible in*F*[*X*], so*f*=*gh*is the factorization of*f*into irreducible polynomials in*F*[*X*].Since

*f*has at most two roots in*K*, we see that*g*and*h*have at most one root in*K*. It follows that we may write*g*=*g*_{1}*g*_{2}and*h*=*h*_{1}*h*_{2}where*g*_{1},*g*_{2},*h*_{1},*h*_{2}are irreducible in*K*[*X*],*g*_{1}and*h*_{1}have degree 1, and*g*_{2}and*h*_{2}have degree 2. Then*f*=*g*_{1}*g*_{2}*h*_{1}*h*_{2}is the factorization of*f*into irreducible polynomials in*K*[*X*]. - The number of Sylow 59-subgroups divides 33 and is congruent to
1 modulo 59. Therefore there is only one Sylow 59-subgroup which
means that
*G*has a normal subgroup*H*of order 59.Now

*G*/*H*is a group of order 33 and so the number of Sylow 3-subgroups of*G*/*H*is congruent to 1 modulo 3 and divides 11. Therefore*G*/*H*has a normal Sylow 3-subgroup, which we may write as*A*/*H*where*A*is a normal subgroup of*G*. Then*A*is a group of order 3*59 and the number of Sylow 3-subgroups of*A*is congruent to 1 modulo 3 and divides 59. Therefore*A*has a normal subgroup*K*of order 3. Observe that if*g**G*, then*gKg*^{-1}is a subgroup of order 3 contained in*gAg*^{-1}. Since*A*is normal,*gAg*^{-1}=*A*, so*gKg*^{-1}is a subgroup of order 3 in*A*and hence*gKg*^{-1}=*K*, because*A*has exactly one subgroup of order 3. Therefore*K*is a normal subgroup of order 3 in*G*.Using exactly the same argument as above with the primes 3 and 11 interchanged, we see that

*G*has a normal subgroup*L*of order 11. We have now proved that all the Sylow subgroups of*G*are normal, so*G*is isomorphic to a direct product of its Sylow subgroups. Also each nontrivial Sylow subgroup has prime order and is therefore cyclic. It follows that*G*abelian, and then by using the structure theorem for finitely generated abelian groups, we conclude that*G*is cyclic. - Since
*G*is a nontrivial*p*-group, its center is nontrivial and therefore it has a central subgroup*Z*of order*p*. Then*G*/*Z*is a group of order*p*^{n - 1}and since*n*2, we see that*G*/*Z*is nontrivial*p*-group and hence it has a central subgroup of order*p*. We may write this subgroup as*A*/*Z*where*A*is a normal subgroup of*G*. Then*A*has order*p*^{2}and since groups of order*p*^{2}are abelian, it follows that*A*is a normal abelian subgroup of order*p*^{2}as required. - The roots of
*X*^{3}- 2 are , w and w^{2}and it follows easily that*K*is the splitting field for*X*^{3}- 2. Therefore*K*/ is a Galois extension of . Also [() : ] = 3 and ()*K*. Since the splitting field of a polynomial of degree 3 has degree dividing 6 and the Galois group is isomorphic to a subgroup of*S*_{3}, we conclude that [*K*: ] = 6 and the Galois group of*K*over is isomorphic to*S*_{3}. - Obviously
0
*A**R*. Let*a*,*b**A**R*. Then*a*-*b**A*and*a*-*b**R*, so*a*-*b**A**R*. Finally let*r**R*. Then*ar**A*because*A*is an ideal of*S*, and*ar**R*. Thus*ar**A**R*and we have proved that*A**R*is an ideal of*R*. - Let
*x**A*. Since*F*is the field of fractions of the PID*R*, we may write*x*=*ab*^{-1}with*a*,*b**R*and (*a*,*b*) = 1. Then there exist*p*,*q**R*such that*ap*+*bq*= 1, so*px*+*q*=*b*^{-1}. Since*p*,*q*,*x**S*, we see that*b*^{-1}*S*. Now*A*is an ideal of*S*, so*xb*=*a**A**R*=*Rd*, so there exists*r**R*such that*xb*=*rd*. Then we have*x*=*b*^{-1}*rd**Sd*and the result follows.

- Obviously
0
- We have
[
*G*/*M*:*PM*/*M*] = [*G*:*PM*] and [*G*:*P*] = [*G*:*PM*][*PM*:*P*], so [*G*/*M*:*PM*/*M*] divides [*G*:*P*]. Since*P*is a Sylow*p*-subgroup of*G*, we see that [*G*:*P*] is prime to*p*and hence [*G*/*M*:*PM*/*M*] is prime to*p*. This shows that*PM*/*M*is a Sylow*p*-subgroup of*G*. - Let
*n**N*. Then*nPn*^{-1}=*P*and*nMn*^{-1}=*M*, hence*nPMn*^{-1}=*PM*. This shows that*Mn*is in the normalizer of*PM*/*M*in*G*/*M*and we conclude that*n**H*. The result follows. - The number of Sylow
*p*-subgroups of*G*/*M*is [*G*/*M*:*H*/*M*] = [*G*:*H*], and the number of Sylow*p*-subgroups of*G*is [*G*:*N*]. Since*N**H*, we see that [*G*:*H*] divides [*G*:*N*] and the result follows.

- We have
[
*A*_{5}X /59.- /2 X /2 X /885.

- We shall use the structure theorem for finitely generated
abelian groups. We may write
*G*^{a}/*p*_{i}^{ai}*H*^{b}/*p*_{i}^{bi}

for certain integers*a*,*b*,*a*_{i},*b*_{i},*n*, and the*p*_{i}are distinct primes. Since*G**G**H**H*, we see that^{2a}/*p*_{i}^{2ai}^{2b}/*p*_{i}^{2bi}*a*= 2*b*and 2*a*_{i}= 2*b*_{i}for all*i*. Therefore*a*=*b*and*a*_{i}=*b*_{i}for all*i*, which proves that*G**H*.