- 1.
- (i)
- By definition
(
*x*) = {*gxg*^{-1}|*g**G*} . Since*x**H**G*, we see that*gxg*^{-1}*H*for all*g**G*and we deduce that (*x*)*H*. Let*C*_{G}(*x*) denote the centralizer of*x*in*G*. Then |(*x*)| = [*G*:*C*_{G}(*x*)]. This last quantity is a power of*p*and it cannot be 1 because*x*is not in the center of*G*. We deduce that |(*x*)| is a nontrivial power of*p*and (i) follows. - (ii)
- It follows from (i) that
*H**Z*is a union of conjugacy classes of the form (*x*) where*x**H**Z*. Since different conjugacy classes intersect trivially, we see from (i) that*p*divides |*H**Z*|. Also |*H*| divides |*G*| and so |*H*| is a nontrivial power of*p*. We deduce that*p*divides |*Z**H*| and (ii) is proven. - (iii)
- Let
*H*be the centralizer of*A*in*G*. Then*H*is a normal subgroup of*G*containing*A*and it will be sufficient to show that*H*=*A*. Let*X*/*A*be the center of*G*/*A*and suppose*H*>*A*. Then we see from (ii) that*X*/*A**H*/*A*> 1 because*H*/*A**G*/*A*, and we deduce that there is a nontrivial cyclic subgroup*Y*/*A*contained in*X*/*A**H*/*A*. Since*Y*/*A**X*/*A*we see that*Y*/*A**G*and we deduce that*Y**G*. Also*A*is contained in the center of*Y*and*Y*/*A*is cyclic, hence*Y*is abelian. This contradicts the fact that*A*is a maximal normal abelian subgroup of*G*and the result follows

- 2.
- (i)
- The number of Sylow 5-subgroups of
*G*divides 36 and is congruent to 1 modulo 5, which means that this number must be 1,6 or 36. It cannot be 1, for then*G*would have a normal Sylow 5-subgroup, which contradicts the fact that*G*is simple. Nor can*G*have 6 Sylow 5-subgroups, for then*G*would be isomorphic to a subgroup of*A*_{6}because*G*is simple. This would mean that*A*_{6}has a subgroup of index 2, and since subgroups of index 2 are always normal, this would contradict the fact that*A*_{6}is simple. We conclude that*G*has 36 Sylow 5-subgroups. - (ii)
- Let
*N*be the normalizer of a Sylow 3-subgroup. Then the number of Sylow 3-subgroups is [*G*:*N*]. Also the number of Sylow 3-subgroups divides 20 and is congruent to 1 modulo 3, so this number is 1,4 or 10. It cannot be 1 because that would mean*G*has a normal Sylow 3-subgroup, which would contradict the fact that*G*is simple. Nor can it be 4, for then*G*would be isomorphic to a subgroup of*A*_{4}because*G*is simple, which is clearly impossible. Therefore the number of Sylow 3-subgroups is 10. We deduce that [*G*:*N*] = 10 and hence*N*has order 18. - (iii)
- A Sylow 3-subgroup of a group of order 18 has order 9. This means the subgroup has index 2, hence the subgroup is normal because in any group, a subgroup of index 2 is normal.
- (d)
- Let
*C*be the centralizer in*G*of*A**B*and suppose*A**B*is not 1. Then*C*contains*A*,*B*and*A**B*is a normal subgroup in*C*, so*C*cannot be the whole of*G*. Therefore the order of*C*is a multiple of 9 and divides 180, and is neither 9 nor 180. Let*d*be the index of*C*in*G*, so |*C*| = |*G*|/*d*. Then*G*is isomorphic to a subgroup of*A*_{d}because*G*is simple. Since the order of*A*_{d}is less than 180 if*d*5, we see that*d*6 and consequently*C*has order 18. From part (iii), the Sylow 3-subgroup of*C*is normal in*C*and therefore*C*has exactly one subgroup of order 9. We now have a contradiction because*C*has two subgroups of order 9, namely*A*and*B*. We conclude that*A**B*= 1. - (e)
- We count the elements in
*G*. Since two Sylow 5-subgroups intersect in 1 and there are 36 Sylow 5-subgroups by (i), we see that*G*has 36*4 = 144 elements of order 5. Also two Sylow 3-subgroups intersect in 1 by (iv) and*G*has 10 Sylow 3-subgroups by (ii). Therefore*G*has 8*10 = 80 elements of order 3 or 9. We conclude that*G*has at least 144 + 80 = 224 elements, which contradicts the fact that*G*has only 180 elements. It follows that no such*G*can exist and thus there is no simple group of order 180.

- 3.
- Since
*M*is a cyclic*R*-module, we know that*M**R*/*Rs*for some*s**R*. By the uniqueness part of the fundamental structure theorem for finitely generated modules over a PID, we cannot write*R*=*A**B*where*A*,*B*are nonzero*R*-modules. Therefore*s*0. Since*sM*= 0, we may take*r*=*s*.Since

*rM*= 0, we see that*rR**sR*and hence*s*divides*r*. Suppose there do not exist distinct primes*p*,*q*dividing*r*. Then the same is true for*s*because*s*divides*r*, and we deduce that*s*is a prime power, say*p*^{e}for some prime*p*. From the uniqueness statement in the fundamental structure theorem for finitely generated modules over a PID, we cannot write*R*/*Rp*^{e}=*A**B*where*A*,*B*are nonzero*R*-modules and we have a contradiction. This finishes the proof. - 4.
- (i)
- Choose integers
*r*,*s*such that*qr*+ 2*s*= 1. Then in [[*X*]]/(*X*- 2) we have*qr*= 1 -*sX*mod (*X*- 2). Since 1 -*sX*is invertible in [[*X*]] (with inverse 1 +*sX*+*s*^{2}*X*^{2}+ ...), it follows that*q*is invertible in [[*X*]]/(*X*- 2). - (ii)
- The general element of
*R*is*a*_{i}*X*^{i}mod (*X*- 2), where*a*_{i}for all*i*. Now each*a*_{i}is of the form*p*/*q*where*p*,*q*and*q*is odd. Using (i), we may now write*a*_{i}=*b*_{i}mod (*X*- 2) where*b*_{i}[[*X*]], and then we may write the general element of*R*in the form*b*_{i}*X*^{i}mod (*X*- 2). This proves that is surjective. We now determine the kernel of . Obviously (*X*- 2) ker . Conversely suppose*f*ker . Then we may write*f*= (*X*- 2)*g*where*g*[[*X*]]. We want to show that*g*[[*X*]]. Write*g*=*g*_{i}*X*^{i}where*g*_{i}. Then the coefficient of*X*^{n}in*g*(*X*- 2) is*g*_{n - 1}- 2*g*_{n}for*n*> 0, and the constant coefficient is -2*g*_{0}. By induction on*n*we see that 2*g*_{n}and since*g*_{n}, we conclude that*g*_{n}for all*n*. This proves that ker = (*X*- 2) and it now follows from the fundamental isomorphism theorem that*R*[[*X*]]/(*X*- 2). - (iii)
- By considering the homomorphism
[
*X*] determined by sending*X*to 2, we see that [*X*]/(*X*- 2) . Since 3 is not invertible in , we see that 3 is not invertible in [*X*]/(*X*- 2). But 3 is invertible in*R*and the result follows.

- 5.
- (i)
- Obviously
*K*()*K*(). Also is separable over*K*() and satisfies the polynomial*X*^{p}- . Since is the only root of*X*^{p}- , it follows that*K*() and hence*K*() =*K*(). Now we consider the minimum polynomial of over*K*. This has degree*p*because [*K*() :*K*] =*p*, and must be a polynomial in*X*^{p}because is not separable over*K*. Thus the minimum polynomial must be of the form*X*^{p}-*b*for some*b**K*and it follows that =*b**K*. - (ii)
- Since we are in characteristic
*p*, we have ( + )^{p}= + . But*K*by (i), hence*K*( + ). Therefore*K*()*K*( + ) and it now follows from (i) that*K*()*K*( + ) as required. - (iii)
- Since
*K*( + )*K*() by (ii), we have [*K*( + ) :*K*] = [*K*( + ) :*K*()][*K*() :*K*] and since [*K*() :*K*] =*d*, it remains to prove that [*K*( + ) :*K*()] =*p*. Now ( + )^{p}= +*K*() which shows that [*K*( + ) :*K*()] =*p*or 1, because we are in characteristic*p*. It remains to prove that [*K*( + ) :*K*()] 1, or equivalently that*K*(). But is separable over*K*, hence every element of*K*() is separable over*K*which shows that*K*() because is not separable over*K*. This completes the proof.

- 6.
- (i)
- It will be sufficient to prove that
*X*^{p}-*t*is irreducible in [*t*,*X*], or equivalently that*t*-*X*^{p}is irreducible in [*X*,*t*]. Let*R*= [*X*] and let*F*= (*X*), the field of fractions of*R*. Then*t*-*X*^{p}is a monic polynomial in*R*[*t*] and is irreducible in*F*[*t*], hence it is irreducible in*R*[*t*] = [*X*,*t*] and the result follows. - (ii)
- Let
*y*be one of the roots of*X*^{p}-*t*in*L*. Since*X*^{p}-*t*is irreducible in*K*[*X*], we see that [*K*(*y*) :*K*] =*p*. Also the roots of*X*^{p}-*t*are*e*^{2ni/p}*y*where*n*= 0,...,*p*- 1, and since*e*^{2ni/p}for all*n*, we deduce that all the roots of*X*^{p}-*t*are in*K*(*y*). It follows that*K*(*y*) =*L*and we conclude that [*L*:*K*] =*p*. Therefore the Galois group of*L*over*K*has order*p*(note that*L*/*K*is a separable extension because we are in characteristic zero). Since groups of order*p*are cyclic, we conclude that the Galois group of*L*over*K*is cyclic of order*p*and hence isomorphic to /*p*.