- We first factor 480 as 32*3*5. Note that since
*G*is simple, it cannot have a nontrivial subgroup of index 7, because that would mean that*G*is isomorphic to a subgroup of*A*_{7}, which is not possible by Lagrange's theorem.- Let
*A*=*P**Q*and suppose*A*> 1. Since*P*,*Q*C_{G}(*A*), we see that*P*< C_{G}(*A*). Using Lagrange's theorem, we deduce that | C_{G}(*A*)| = 96, 160 or 480. We cannot have 480 because then*A*would be a central and hence a normal subgroup of*G*, which would contradict the hypothesis that*G*is simple. Also we cannot have | C_{G}(*A*)| = 96 or 160, because that would mean that*G*has a subgroup of index 5 or 3. We now have a contradiction, and we conclude that*A*= 1. - The number of Sylow 2-subgroups is congruent to 1 modulo 2 and
divides 15. It cannot be 1 because that would mean that
*G*has a normal Sylow 2-subgroup. Nor can it be 3 or 5, because then*G*would have a subgroup of index 3 or 5. Therefore*G*has 15 Sylow 2-subgroups. Next the number of Sylow 3-subgroups is congruent to 1 modulo 3 and divides 96. This number cannot be 1, because that would mean that the Sylow 3-subgroup is normal. Nor can it be 4, because that would yield a subgroup of index 4. Therefore*G*has at least 10 Sylow 3-subgroups. Finally the number of Sylow 7-subgroups is congruent to 1 modulo 7 and divides 160. This number cannot be 1, because that would mean that the Sylow 7-subgroup is normal. Therefore*G*has at least 8 Sylow 7-subgroups.We now count elements. Since by (a) any two Sylow 2-subgroups intersect trivially, there are 15*31 nontrivial elements whose order is a power of 2. Next any two Sylow 3-subgroups intersect trivially, because a Sylow 3-subgroup has prime order 3, and we see that there are at least 10*2 elements of order 3. Finally any two Sylow 5-subgroups intersect trivially, because a Sylow 5-subgroup has prime order 5, and we deduce that

*G*has at least 6*4 elements of order 5. We now count elements: we find that*G*has at least 15*31 + 10*2 + 6*4 = 509 nontrivial elements. Since*G*has only 480 elements altogether, we have now arrived at a contradiction. We conclude that there is no such group*G*.

- Let
- Since
*R*is a domain and 0*S*, we see that*S*^{-1}*R*is a domain. Also for*a*,*b**R*and*s*,*t**S*, we have*a*/*s*=*b*/*t*if and only if*at*=*bs*. Suppose*p*/1 divides (*a*/*s*)(*b*/*t*) in*S*^{-1}*R*. This means that there exists*c*/*u**S*^{-1}*R*such that (*p*/1)(*c*/*u*) = (*a*/*s*)(*b*/*t*), which means that*pstc*=*abu*. Since*p*is prime, we see that*p*divides at least one of*a*,*b*,*u*. If*p*divides*u*, then*p*/1 is a unit in*S*^{-1}*R*because*u*/1 is a unit in*S*^{-1}*R*. Therefore we may assume that*p*does not divide*u*; without loss of generality, we may assume that*p*divides*a*, say*pq*=*a*. Then (*p*/1)(*q*/*s*) =*a*/*s*and we see that*p*/1 divides*a*/*s*. Therefore if*p*/1 is not a unit, it is prime and the result follows. - Let
*P*be a finitely generated projective*k*[*X*]/(*X*^{3}+*X*)-module. Then there is a*k*[*X*]/(*X*^{3}+*X*)-module*Q*and an integer*e*such that*P**Q*(*k*[*X*]/(*X*^{3}+*X*))^{e}. Note that*X*^{3}+*X*=*X*(*X*+ 1)^{2}and*k*[*X*]/(*X*^{3}+*X*)*k*[*X*]/(*X*)*k*[*X*]/(*X*+ 1)^{2}, so*P**Q*(*k*[*X*]/(*X*))^{e}(*k*[*X*]/(*X*+ 1)^{2})^{e}. We may view this as an isomorphism of finitely generated*k*[*X*]-modules. We use repeatedly without comment that a map between*k*[*X*]/(*X*^{3}+*X*)-modules is an isomorphism as*k*[*X*]/(*X*^{3}+*X*)-modules if and only if it is an isomorphism as*k*[*X*]-modules. Since*k*is a field,*k*[*X*] is a PID, so the structure theorem for finitely generated modules over a PID tells us that*P**k*[*X*]/(*f*_{i}) and*Q**k*[*X*]/(*g*_{i})*f*_{i},*g*_{i}are either 0 or positive powers of monic irreducible polynomials. Then we have*k*[*X*]/(*f*_{i})*k*[*X*]/(*g*_{i}) (*k*[*X*]/(*X*))^{e}(*k*[*X*]/(*X*+ 1)^{2})^{e}.*f*_{i}=*X*or (*X*+ 1)^{2}for all*i*. It follows that a finitely generated*k*[*X*]/(*X*^{3}+*X*)-module is isomorphic to a finite direct sum of modules of the form*k*[*X*]/(*X*) or*k*[*X*]/(*X*^{2}+ 1). - Suppose there is a positive integer
*n*such that*MJ*^{n}=*MJ*^{n + 1}0. Then*MJ*^{n}is finitely generated because*M*is Noetherian, and (*MJ*^{n})*J*=*MJ*^{n + 1}=*MJ*^{n}. By Nakayama's lemma we deduce that*MJ*^{n}= 0, as required. - Since
*R*is a right Artinian ring with no nonzero nilpotent ideals, the Wedderburn structure theorem tells us that*R**R*_{1}...*R*_{n}, where*n*is a positive integer, and the*R*_{i}are matrix rings over division rings. If*n*> 1, then (1, 0,..., 0) is a nontrivial idempotent, so*n*= 1 which means that*R*is a matrix ring over a division ring. If this matrix ring has degree > 1, then the matrix with 1 in the (1, 1) position and zeros elsewhere is a nontrivial idempotent. Therefore*R*is isomorphic to a matrix ring over a division ring, and the result follows. - The character table for
*S*_{4}is given below; the irreducible characters are c_{1},...,c_{5}. c_{1}is the principal character, c_{2}is the character coming from the sign of the permutation, r is the permutation character (not irreducible), c_{4}= r - c_{1}, and c_{5}= c_{2}c_{4}. The remaining row, the character of c_{3}, can easily be filled in using the orthogonality relations.Class Size 1 6 8 6 3 Class Rep (1) (12) (123) (1234) (12)(34) c _{1}1 1 1 1 1 c _{2}1 -1 1 -1 1 c _{3}2 0 -1 0 2 c _{4}3 -1 0 1 -1 c _{5}3 1 0 -1 -1 r 4 2 1 0 0 - Since splitting fields are determined up to isomorphism, we may
as well assume that
*K*. Since the roots of*X*^{4}- 2 are ±, ±*i*, we see that*K*= ([],*i*). Now*X*^{4}- 2 is irreducible over by Eisenstein's criterion for the prime 2, so [() : ] = 4. Also*i*() and*i*satisfies*X*^{2}+ 1 = 0, consequently [*K*: ()] = 2. We deduce that [*K*: ] = 8 and therefore | Gal(*K*/)| = 8. Now a group of order 8 has a normal subgroup of order 2 (a*p*-group has normal subgroups of any order dividing the order of the group); let*H*be a normal subgroup of order 2 in*G*, and let*L*be the fixed field of*H*. Then*H*has index 4 in*G*, so by the fundamental theorem of Galois theory, we see that*L*is a normal extension of degree 4 over , as required.