- 1.
- The order of
*G*is 5^{3}7^{3}. The number of Sylow 5-subgroups of*G*divides 7^{3}and is congruent to 1 modulo 5; the only possibility is 1. Therefore*G*has a normal Sylow 5-subgroup*A*. The number of Sylow 7-subgroups of*G*divides 5^{3}and is congruent to 1 modulo 7; the only possibility is 1. Therefore*G*has a normal Sylow 7-subgroup*B*. Since (|*A*|,|*B*|) = 1, we see that*A**B*= 1. We next show that every element of*A*commutes with every element of*B*. Suppose*a**A*and*b**B*. Then*aba*^{-1}*b*^{-1}=*a*(*ba*^{-1}*b*^{-1}) and since*A**G*, we see that*ba*^{-1}*b*^{-1}*A*and consequently*aba*^{-1}*b*^{-1}*A*. Similarly*aba*^{-1}*b*^{-1}*B*and we deduce that*aba*^{-1}*b*^{-1}*A**B*= 1. Therefore*aba*^{-1}*b*^{-1}= 1 and we conclude that*ab*=*ba*, in other words every element of*A*commutes with every element of*B*.Since a group of prime power order has normal subgroups of order

*m*for all*m*dividing the order of the group, we see that*A*has a normal subgroup*H*of order 25. From the previous paragraph,*B*centralizes*H*and so certainly normalizes*H*. Thus*A*and*B*normalize*H*, hence |*A*| and |*B*| divide the order of the normalizer of*H*in*G*and we conclude that*H**G*. This completes the solution. - 2.
- First we write
*G*as a direct product of cyclic groups of prime power order:*G*/4/9/9/5 . Any subgroup of*G*is isomorphic to a product of subgroups, where one subgroup is taken from each factor. Thus*H*/9/3 or /3/9 . These last two groups are isomorphic, so we conclude that*H*/3/9 . - 3.
- (a)
- Since
*D*is a conjugacy class in*f*^{-1}(*C*), we may write*D*= {*bdb*^{-1}|*b**f*^{-1}(*C*)} for some fixed*d**D*. Then*f*(*D*) = {*f*(*bdb*^{-1}) =*f*(*b*)*f*(*d*)*f*(*b*)^{-1}|*b**f*^{-1}(*C*)}.*f*(*D*) = {*cf*(*d*)*c*^{-1}|*c**C*} , and (a) follows. - (b)
- Let
*D*(*g*) denote the conjugacy class of*g*in*f*^{-1}(*C*). Since*f*(*g*) is centralized by*C*, the conjugacy class containing*f*(*g*) is precisely {*f*(*g*)} . Since*f*(*D*(*g*)) is by (a) the conjugacy class containing*f*(*g*), we see that |*f*(*D*(*g*))| = 1. Therefore all elements of*D*(*g*) are in the same coset of ker*f*and we conclude that |*D*(*g*)|| ker*f*| as required. - (c)
- Let
*K*denote the centralizer of*g*in*f*^{-1}(*C*). Then the order of the centralizer of*g*in*G*is at least |*K*|. Now the order of the conjugacy class of*g*in*f*^{-1}(*C*) is [*f*^{-1}(*C*) :*K*], and by (b) this order is at most | ker*f*|. Therefore [*f*^{-1}(*C*) :*K*]| ker*f*|, consequently|

because*K*||*f*^{-1}(*C*)/ker*f*| = |*C*|*f*^{-1}(*C*)/ker*f**C*. The result follows.

- 4.
- (a)
- If
*R*has no prime elements, then*R*is a field and so certainly a PID. Therefore we may suppose that*R*has exactly one prime*p*(up to a multiple of a unit), and we need to prove that*R*is a PID. Let*I*be a nonzero ideal of*R*. Then each nonzero element of*I*can be written in the form*up*^{n}for some nonnegative integer*n*and some unit*u*, because*p*is the only prime (up to a multiple of a unit) of*R*. Let*N*be the smallest nonnegative integer such that*up*^{N}*I*for some unit*u*. We now show that*I*=*p*^{N}*R*; clearly*p*^{N}*R**I*. If*x**I*\ 0, then we may write*x*=*vp*^{n}for some unit*v*and some integer*n**N*. Thus*x*=*p*^{N}*vp*^{n - N}which shows that*x**p*^{N}*R*, and the result follows. - (b)
- Suppose every maximal ideal of
*R*is principal. Then each maximal ideal of*R*is of the form*pR*where*p*is a prime of*R*. Suppose by way of contradiction that*I*is a nonprincipal ideal of*R*. Clearly 0*I**R*. Choose a nonzero element*x**I*, and write*x*=*p*_{1}^{d1}...*p*_{n}^{dn}, where the*p*_{i}are nonassociate primes and the*d*_{i}are positive integers.For each prime

*p*, let*e*(*p*) denote the largest integer such that*p*^{e(p)}*R**I*. If*p*is not an associate of one of the*p*_{i}, then*e*(*p*) = 0. Set*y*=*p*_{1}^{e1}...*p*_{n}^{en}. We claim that*I*=*yR*.First we show that

*I**yR*. If*z**I*, then by unique factorization we may write*z*=*qp*_{1}^{f1}...*p*_{n}^{fn}, where the*f*_{i}are nonnegative integers and*q*is a product of primes which are not associate to any of the*p*_{i}. Again using unique factorization, we must have*f*_{i}*e*_{i}for all*i*and we deduce that*z**yR*.Finally we show that

*yR**I*. Set*J*= {*r**R*|*yr**I*} (so*J*=*y*^{-1}*I*). Clearly*J*is an ideal of*R*and*yJ*=*I*. If*J**R*, then by Zorn's lemma*J*is contained in a maximal ideal of*R*, which we may assume is of the form*pR*where*p*is a prime of*R*. It would follow that*ypR**I*, which contradicts the maximality of the*e*(*p*). Therefore*J*=*R*and we deduce that*yR**I*. Thus*I*=*yR*and the proof is complete.

- 5.
- Let
:
*N**X**N*denote the projection onto*N*, so (*n*,*x*) =*n*for all*n**N*and*x**X*, and let denote the identity map on*N*. Then (*f*)*i*= = . This shows that*i*has a left inverse, consequently the sequence0

splits (the map*N**M**M*/*N*0*M**M*/*N*above is of course the natural epimorphism). This shows that*M**N**M*/*N*as required. - 6.
- Let
*G*denote the Galois group of*E*over*F*. Since*E*is a Galois extension of*F*with [*E*:*F*] =*p*^{n}, we see that |*G*| =*p*^{n}. Since*G*is a*p*-group, it has a series*G*=*G*_{0}*G*_{1}...*G*_{n - 1}*G*_{n}= 1, with*G*_{i}*G*and |*G*_{i}/*G*_{i - 1}| =*p*for all*i*. Set*K*_{i}= {*e**E*|*ge*=*e*for all*g**G*_{i}} . Then by the Galois correspondence,*K*_{i}is normal over*F*and [*K*_{i}:*K*_{i - 1}] = [*G*_{i}:*G*_{i - 1}] =*p*for all*i*, as required. - 7.
- Suppose
*K*is a finite field which is algebraically closed. Let*n*be a positive integer which is prime to the characteristic of*K*, and consider the polynomial*X*^{n}- 1. The derivative of*X*^{n}- 1 is*nX*^{n - 1}which is prime to*X*^{n}- 1 in*K*[*X*], because*n*is prime to the characteristic of*K*. This tells us that the roots of*X*^{n}- 1 in a splitting field for*K*are distinct. If*K*is algebraically closed, then all these roots would be in*K*, and we would deduce that |*K*|*n*. Since*n*can be arbitrarily large, this would contradict the assumption that*K*is finite, and the result is proven.