- Let
0
*a**R*. We must prove that*a*is invertible, so suppose to the contrary that*a*is not invertible. Then*a*^{2}*R*is an ideal of*R*and since*a*is not invertible, we see that*aR**R*and consequently*a*^{2}*R**R*. By hypothesis*a*^{2}*R*is a prime ideal of*R*and since*a*^{2}*a*^{2}*R*, we deduce that*a**a*^{2}*R*. Therefore*a*=*a*^{2}*r*for some*r**R*. Since 0 is a prime ideal of*R*, we see that*R*is an integral domain and we deduce that 1 =*ar*. Thus*a*is invertible and we have a contradiction. This completes the proof. - By hypothesis
*G*has a normal subgroup of*K*of order*p*^{3}. Then*G*/*K*is a group of order*q*^{3}. A nontrivial*q*-group (where*q*is a prime) has a normal subgroup of order*q*, so*G*/*K*has a normal subgroup*H*/*K*of order*q*. Then*H*is a normal subgroup of order*p*^{3}*q*, as required. - Suppose
*R*has an element*a*which is neither a zero divisor nor a unit. Then*aR*is a proper submodule of*R*, because*a*is not a unit. Also the map*r*| - >*ar*is an*R*-map from*R*onto*aR*which has kernel 0, because*a*is a nonzero divisor. This shows that*R*is isomorphic to the proper*R*-submodule*aR*.Conversely suppose

*R*is isomorphic to the proper*R*-sumodule*M*. Then there is an*R*-isomorphism q :*R*- >*M*. Set*a*= q1. Then*aR*= (q1)*R*= q(1*R*) = q*R*=*M*, so*aR**R*and we see that*a*is not a unit. Finally if*ar*= 0, then q*r*= q(1*r*) = (q1)*r*=*ar*= 0 and we deduce that*r*= 0, because q is an isomorphism. Therefore*r*is a nonzero divisor and the result follows. - Since
a satisfies
*X*^{2}- a^{2}*K*(a^{2})[*X*], we see that [*K*(a) :*K*(a^{2})] = 1 or 2. Also [*K*(a) :*K*] = [*K*(a) :*K*(a^{2})][*K*(a^{2}) :*K*]. Since [*K*(a) :*K*] is odd, we deduce that [*K*(a) :*K*(a^{2})] = 1 and the result follows. - By the fundamental structure theorem for finitely generated
abelian groups, we know that
*G*is a direct product of nontrivial cyclic*p*-groups. Since {*x**G*|*x*^{p}= 1} has order*p*^{2}, we see that*G*is a direct product of exactly two nontrivial cyclic*p*-groups. It now follows that*G*/*p*^{5}X /*p*or /*p*^{4}X /*p*^{2}or /*p*^{3}X /*p*^{3}(so there are three possible groups up to isomorphism). - Since
*K*is a splitting over*k*, it can be written as*k*(*a*_{1},...,*a*_{n}) where*a*_{1},...,*a*_{n}are all the roots of some polynomial*f**k*[*X*]. If s Gal(*L*/*k*), then s*a*_{i}also satisfies*f*, because s fixes all the coefficients of*f*, and so s permutes the*a*_{i}. It follows that s*K*=*k*(s*a*_{1},...,s*a*_{n}) =*K*. - Let
*G*be a simple group of order 280. The number of Sylow 7-subgroups is congruent to 1 modulo 7 and divides 40, so there are 1 or 8 Sylow 7-subgroups. There cannot be 1 Sylow 7-subgroup, because then the Sylow 7-subgroup would be normal which contradicts the hypothesis that*G*is simple. Therefore there are 8 Sylow 7-subgroups. Since two distinct Sylow 7-subgroups must have trivial intersection, we see that there are at least 8*6 = 48 elements of order 7. The number of Sylow 5-subgroups is congruent to 1 modulo 5 and divides 56. There cannot be 1 Sylow 5-subgroup, for then it would be normal which would contradict the hypothesis that*G*is simple. Therefore there are 56 Sylow 5-subgroups. Since two distinct Sylow 5-subgroups must intersect in the identity, we see that there are at least 56*4 = 224 elements of order 5. Finally since the Sylow 2-subgroup is not normal, there must be at least 9 elements whose order is a power of 2. We now count elements: we find that*G*has at least 48 + 224 + 9 = 281 elements, which is impossible because*G*has only 280 elements. Therefore no such*G*can exist and we deduce that there is no simple group of order 280. - Let
*K*be a splitting field over*F*which contains*E*, let*G*= Gal(*K*/*F*), and let*H*= Gal(*K*/*E*). Since we are in characteristic zero, everything is separable and hence*K*is a Galois extension of*F*. Therefore by the fundamental theorem of Galois theory, we see that the number of fields between*F*and*E*is equal to the number of subgroups between*G*and*H*. Also [*G*:*H*] =*n*. By considering the permutation representation of*G*on the left cosets of*H*in*G*, we see that there is a normal subgroup*N*of*G*contained in*H*such that |*G*/*N*|*n*!. The number of subgroups between*G*and*H*is at most the number of subgroups between*G*and*N*, which is at most the number of subsets of*G*/*N*. Since the number of subsets of*G*/*N*is 2^{| G/N|}, we deduce that the number of subfields between*F*and*E*is at most 2^{n!}, as required.