- (a)
*H*acts on the set of conjugates of*Q*according to the formula*gQg*^{-1}| - >*hgQg*^{-1}*h*^{-1}for*h**H*and*g**G*. Note that*gQg*^{-1}__<__*G*for all*g*, so O_{Q}is a set of subgroups. Let*S*= {*h**H*|*hQh*^{-1}=*Q*}, the stabilizer of*Q*in*H*. Then |O_{Q}||*S*| = |*H*|. Now*h**S*if and only if*h**H*/\ N_{G}(*Q*) = 1, hence |*S*| = 1 and the result follows.- (b)
- Let
*P*be a Sylow*p*-subgroup of G. We apply the above with*H*=*P*. Since*P*/\ N_{G}(*Q*) =*P*/\*Q*= 1, we see that O_{Q}has |*P*| =*p*^{m}subgroups. Furthermore all the subgroups of O_{Q}have prime order*q*, so any two of them must intersect in 1. Now each nonidentity element of a subgroup in O_{Q}has order*q*, consequently each subgroup of O_{Q}yields*q*- 1 elements of order*q*and we deduce that*G*has at least (*q*- 1)*p*^{m}elements of order*q*. Therefore*G*has at most*pq*^{m}- (*q*- 1)*p*^{m}=*p*^{m}elements of order a power of*p*. Since |*P*| =*p*^{m}and every element of a Sylow*p*-subgroup has order a power of*p*, we conclude that*P*is the only Sylow*p*-subgroup of*G*. Therefore*P*is normal in*G*and we are finished.

- Let
*F*=**Q**(,) and*N*= Gal(*K*/*F*). Then*F**K*and is the splitting field over**Q**for (*x*^{2}- 2)(*x*^{2}- 3). Therefore*N*is a normal subgroup in Gal(*K*/**Q**) of index [*F*:**Q**]. Now satisfies*x*^{2}- 2 and**Q**. Therefore [**Q**() :**Q**] = 2.Next we show that

**Q**(). Suppose**Q**(). Then we could write =*a*+*b*with*a*,*b***Q**, because every element of**Q**() can be written in this form. Squaring, we obtain 3 =*a*^{2}+ 2*b*^{2}+ 2*ab*. Since**Q**, we deduce that*a*or*b*= 0. But*a*= 0 yields**Q**, while*b*= 0 yields**Q**, neither of which is true. We conclude that**Q**(). Since satisfies*x*^{2}- 3, we deduce that [*F*:**Q**()] = 2. Therefore[Thus*F*:**Q**] = [**Q**(,) :**Q**()][**Q**() :**Q**] = 2*2 = 4.*N*is a normal subgroup of index 4 in Gal(*K*/**Q**).Suppose

*K*. Since satisfies*x*^{8}- 2 and*x*^{8}- 2 is irreducible over**Q**by Eisenstein's criterion for the prime 2, we see that 8 divides [*K*:**Q**], consequently 8 divides | Gal(*K*/**Q**)|. But | Gal(*K*/**Q**)| = 4|*N*|, so this is not possible if |*N*| is odd. - (a)
- If
*X*and*Y*are right*R*-modules and q :*X*- >*Y*is an*R*-module homomorphism, then there is a unique group homomorphism q 1 :*X**C*- >*Y**C*such that q(*x**c*) = (q*x*)*c*for all*x**X*and*c**C*.First we apply this to the maps

( *a*,*b*) | - >*a*:*A**B*- - >*A*( *a*,*b*) | - >*b*:*A**B*- - >*B*.

We obtain a group homomorphismq : (such that q((*A**B*)*C*- - >*A**C**B**C**a*,*b*)*c*) = (*a**c*,*b**c*). Next we apply it to the maps*a*| - > (*a*, 0)*c*:*A**C*- - > (*A**B*)*C**a*| - > (0,*b*)*c*:*B**C*- - > (*A**B*)*C*.

We obtain a group homomorphismf : (such that f(*A**C*) (*B**C*) - - > (*A**B*)*C**a**c*,*b**d*)= (*a*, 0)*c*+ (0,*b*)*d*.Finally we show that fq is the identity on (

*A**B*)*C*, and that qf is the identity on (*A**C*) (*B**C*). We havefq(Since (*a*,*b*)*c*= f(*a**c*,*b**c*) = (*a*,*b*)*c*.*A**B*)*C*is generated as an abelian group by elements of the form (*a*,*b*)*c*, we see that fq is the identity. Alsoqf(Since (*a**c*,*b**d*)= q(*a*, 0)*c*+ q(0,*b*)*d*= (*a**c*,*b**d*).*A**C*) (*B**C*) is generated as an abelian group by elements of the form (*a**c*,*b**d*), we deduce that qf is the identity. It now follows that (*A**B*)*C*@ (*A**C*) (*B**C*). - (b)
- Since
*M*is a finitely generated**Z**-module, we may express it as a finite direct sum of cyclic**Z**-modules, say*M*=**Z**/*a*_{i}**Z**, where we may assume that*a*_{i}=/= ±1 for all*i*. Then by the first part, we see that*M**M*@**Z**/*a*_{i}**Z****Z**/*a*_{j}**Z**.**Z**/*a*_{i}**Z****Z**/*a*_{i}**Z**=/= 0 for all*i*(of course even for just one*i*will be sufficient). However we can define a bilinear mapq :by q(**Z**/*a*_{i}**Z**X**Z**/*a*_{i}**Z**- >**Z**/*a*_{i}**Z***x*,*y*) =*xy*. This induces a**Z**-module homomorphism**Z**/*a*_{i}**Z****Z**/*a*_{i}**Z**- >**Z**/*a*_{i}**Z**, which is obviously onto. We conclude that**Z**/*a*_{i}**Z****Z**/*a*_{i}**Z**=/= 0, as required.

- Note that
Ann(
*m*) is an ideal of*R*. Since*R*is Noetherian, we may choose*w**M*such that Ann(*w*) is maximal (that is Ann(*w*) is as large as possible, but not*R*). Suppose Ann(*w*) is not prime. Then there exists*a*,*b**R*\Ann(*w*) such that*ab*Ann(*w*), i.e.*abw*= 0. But then*a*Ann(*bw*) and Ann(*w*) Ann(*bw*). Furthermore Ann(*bw*) =/=*R*because*bw*=/= 0, so the maximality of Ann(*w*) has been contradicted and the result follows. - Suppose
*R*is a field. Then an*R*-module is the same thing as an*R*-vector space, and since every vector space has a basis this means that every*R*-module is free; in particular every*R*-module is projective.Conversely suppose every

*R*-module is projective. Since*R*is an integral domain, to prove*R*is a field we only need show that every nonzero element of*R*is invertible. Suppose to the contrary that*x*is a nonzero element of*R*which is not invertible. Then*R*/*Rx*is a nonzero*R*-module, so it has a nonzero element*u*. Note that*xu*= 0. Consider the exact sequence0 - - >Since*Rx*- - >*R*- - >*R*/*xR*- - > 0.*R*/*xR*is projective, the sequence splits, in particular*R*/*xR*is isomorphic to a submodule of*R*. Now*R*is an integral domain, so*xv*=/= 0 for all nonzero*v**R*and we deduce that*xu*=/= 0. We now have a contradiction and the result follows. - Since
Z
_{N}= Z_{N + 1}, we see that Z(*G*/Z_{N}) = 1. Therefore*K*_{N}.Now suppose

*L*is a normal subgroup of*G*such that Z(*G*/*L*) = 1 and*L*does not contain Z_{N}. Then there is a nonnegative integer*n*such thatZChoose_{n}*L*, Z_{n + 1}*L*.*x*Z_{n + 1}\*L*. Then*xL*=/= 1 in*G*/*L*. Also*xgx*^{-1}*g*^{-1}Z_{n + 1}for all*g**G*, because*x*Z_{n + 1}Z(*G*/Z_{n + 1}). Therefore*xgx*^{-1}*g*^{-1}*L*and we deduce that*xL*Z(*G*/*L*). This is a contradiction, and so the result is proven. - Let
*I*be the set of matrices in M_{2}(**Q**[*x*]/(*x*^{2}- 1)) of the form( *a*(*x*+ 1) + (*x*^{2}- 1)0 *b*(*x*+ 1) + (*x*^{2}- 1)0 ) with

*a*,*b***Q**. Note that if*f***Q**[*x*], then (*x*- 1) divides*f*(*x*) -*f*(1), consequently*f*(*x*)(*x*+ 1) + (*x*^{2}- 1) =*f*(1)(*x*+ 1) + (*x*^{2}- 1) in**Q**[*x*]/(*x*^{2}- 1).Now we verify that

*I*is a left ideal of**Q**[*x*]/(*x*^{2}- 1). Clearly*I*is an abelian group under addition. Since( *f*(*x*) + (*x*^{2}- 1)*g*(*x*) + (*x*^{2}- 1)*h*(*x*) + (*x*^{2}- 1)*k*(*x*) + (*x*^{2}- 1)) ( *a*(*x*+ 1) + (*x*^{2}- 1)0 *b*(*x*+ 1) + (*x*^{2}- 1)0 ) = ( *af*(1)(*x*+ 1) + (*x*^{2}- 1)0 *bh*(1)(*x*+ 1) + (*x*^{2}- 1)0 ) we see that

*I*is closed under left multiplication by elements of**Q**[*x*]/(*x*^{2}- 1), and it now follows that*I*is a left ideal.Finally we need to show that

*I*is a minimal ideal. Obviously*I*=/= 0 (note*x*+ 1(*x*^{2}- 1)). Suppose*J*is a nonzero left ideal contained in*I*. We need to show that*J*=*I*. By multiplying on the left by the matrix( 0 1 + ( *x*^{2}- 1)1 + ( *x*^{2}- 1)0 ) if necessary, we may assume that

*I*contains a matrix of the form( *a*(*x*+ 1) + (*x*^{2}- 1)0 *b*(*x*+ 1) + (*x*^{2}- 1)0 ) with

*a*=/= 0. Then by multiplying on the left by( *c*/*a*+ (*x*^{2}- 1)0 *d*/*a*+ (*x*^{2}- 1)0 ) we see that

*J*must be the whole of*I*and the result follows.