- 1.
- Suppose
*G*is a simple group with exactly three elements of order two. Consider the conjugation action of*G*on the three elements of order two: specifically if*g**G*and*x*is an element of order two, then we define*g**x*=*gxg*^{-1}. This action yields a homomorphism :*G**S*_{3}. Suppose ker =*G*. Then*gxg*^{-1}=*x*for all*g**G*and for all elements*x*of order two. Thus if*x*is an element of order two, we see that*x*is in the center of*G*and hence*G*is not simple. On the other hand if ker*G*, then since*G*is simple we must have ker = 1 and it follows that*G*is isomorphic to a subgroup of*S*_{3}. The only subgroup of*S*_{3}which has exactly three elements of order two is*S*_{3}itself. But*S*_{3}is not simple (because*A*_{3}is a nontrivial normal subgroup), hence*G*is not simple and the result is proven. - 2.
- Let
*F*denote the free group on generators*x*,*y*, and define a homomorphism*f*:*F**S*_{3}by*f*(*x*) = (123) and*f*(*y*) = (12). Since*f*(*x*^{6}) = (123)^{6}=*e*,*f*(*y*^{4}) = (12)^{4}=*e*, and*f*(*yxy*^{-1}) = (213) =*x*^{-1}, we see that*f*induces a homomorphism from*G*to*S*_{3}. This homomorphism is onto because its image contains*f*(*x*) = (123) and*f*(*y*) = (12), and the elements (123), (12) generate*G*. Thus*G*has a homomorphic image isomorphic to*S*_{3}.We prove that

*G*is not isomorphic to*S*_{3}by showing it has an element whose order is a multiple of 4 or , which will establish the result because the orders of elements in*S*_{3}are 1,2 and 3. We shall use bars to denote the image of a number in /4 . Thus is the identity of /4 under the operation of addition. Define*h*:*F*/4 by*h*(*x*) = and*h*(*y*) = . Since*h*(*x*^{6}) = 6* = ,*h*(*y*^{4}) = 4* , and*h*(*yxy*^{-1}) = + - = =*h*(*x*^{-1}),*h*induces a homomorphism from*G*to /4 , which has in its image. Since has order 4, it follows that*G*has an element whose order is either a multiple of 4 or infinity. This completes the proof. - 3.
- (i)
- Since
*R*is not a field, we may choose 0*s**R*such that*s*is not a unit, equivalently*sR**R*. Define*f*:*R**R*by*f*(*r*) =*sr*. Then*f*is an*R*-module homomorphism which is injective, because*R*is a PID and*s*0, and is not onto because*s*is not a unit. This proves that*R*is isomorphic to the proper submodule*sR*of*R*. - (ii)
- Using the fundamental structure theorem for finitely generated
modules over a PID, we may write
*M*as a direct sum of cyclic*R*-modules. Since*M*is not a torsion module, at least one of these summands must be*R*; in other words we may write*M**R**N*for some*R*-submodule*N*of*M*. Then*M**sR**N*and since*sR**N*is a proper submodule of*R**N*, we have proven that*M*is isomorphic to a proper submodule of itself.

- 4.
- (i)
- We will write mappings on the left.
Let
:
*B**B**C*, :*C**B**C*denote the natural injections (so*b*= (*b*, 0)), and let :*B**C**B*, :*B**C**C*denote the natural epimorphisms (so (*b*,*c*) =*b*). Define: Hom

by (_{R}(*A*,*B**C*)Hom_{R}(*A*,*B*) Hom_{R}(*A*,*C*)*f*)= (*f*,*f*), and: Hom

by (_{R}(*A*,*B*) Hom_{R}(*A*,*C*)Hom_{R}(*A*,*B**C*)*f*,*g*) =*f*+*g*. It is easily checked that and are*R*-module homomorphisms, so will suffice to prove that and are the identity maps. We have(

because , are the zero maps, and , are the identity maps. Therefore is the identity map. Also*f*,*g*) = (*f*+*g*) = ((*f*+*g*),(*f*+*g*)) = (*f*,*g*)(

because + is the identity map. Thus is the identity map and (i) is proven.*h*) = (*h*,*h*) =*h*+*h*= ( + )*h*=*h* - (ii)
- Write
*Hom*_{R}(*A*,*A*) =*X*. If Hom_{R}(*A*,*A**A*) , then by the first part we would have*X**X*. Thus*X*0, and we see that is the direct sum of two nonzero groups. This is not possible and the result follows.

- 5.
- (i)
- Let
*I*be an ideal of*S*^{-1}*R*. We need to prove that*I*is finitely generated. Let*J*= {*r**R*|*r*/1*S*^{-1}*I*} (where we view*S*^{-1}*R*as elements of the form*r*/*s*where*r**R*and*s**S*). Then*J*is an ideal of*R*and since*R*is Noetherian, there exist elements*x*_{1},...,*x*_{n}which generate*J*as an ideal, which means*J*=*x*_{1}*R*+ ... +*x*_{n}*R*. We claim that*I*is generated by {*x*_{1}/1,...,*x*_{n}/1} . Indeed if*r*/*s**I*, then*r*=*r*_{1}*x*_{1}+ ... +*r*_{n}*x*_{n}for some*r*_{i}*R*, and hence*r*/*s*=*r*_{1}/*s**x*_{1}+ ... +*r*_{n}/*s**x*_{n}. This proves (i). - (ii)
- Let
*S*be the multiplicative subset {1,*X*,*X*^{2},...} . Then every element of*S*is invertible in*R*[[*X*,*X*^{-1}]] and hence the identity map*R*[[*X*]]*R*[[*X*]] extends to a homomorphism*S*^{-1}*R*[[*X*]]*R*[[*X*,*X*^{-1}]]. It is easily checked that this map is an isomorphism. Since*R*[[*X*]] is Noetherian, it follows from (i) that*S*^{-1}*R*[[*X*]] is Noetherian and hence*R*[[*X*,*X*^{-1}]] is Noetherian as required.

- 6.
- (i)
- Set
*Y*=*X*- 1. Then

*X*^{4}+*X*^{3}+*X*^{2}+*X*+ 1= ( *X*^{5}- 1)/(*X*- 1) = ((*Y*+ 1)^{5}- 1)/*Y*= ( *Y*^{5}+ 5*Y*^{4}+ 10*Y*^{3}+ 10*Y*^{2}+ 5*Y*)/*Y*= *Y*^{4}+ 5*Y*^{3}+ 10*Y*^{2}+ 10*Y*+ 5.*Y*^{4}+ 5*Y*^{3}+ 10*Y*^{2}+ 10*Y*+ 5 is irreducible in [*Y*]. Since*Y**Y*+ 1 induces an automorphism of [*Y*], we deduce that*X*^{4}+*X*^{3}+*X*^{2}+*X*+ 1 is irreducible. - (ii)
- Let
*c*(*X*) denote the characteristic polynomial of*A*, and let*m*(*X*) denote the minimum polynomial of*A*. Since*A*^{5}=*I*, we see that*m*(*X*) divides*X*^{5}- 1, and since 1 is not an eigenvalue of*A*, we see that*X*- 1 does not divide*m*(*X*). Therefore*m*(*X*) divides*X*^{4}+*X*^{3}+*X*^{2}+*X*+ 1 and using (i), we deduce that the only irreducible factor of*m*(*X*) is*X*^{4}+*X*^{3}+*X*^{2}+*X*+ 1. It follows that the only irreducible factor of*c*(*X*) is*X*^{4}+*X*^{3}+*X*^{2}+*X*+ 1, which shows that the degree of*c*(*X*) is a multiple of 4. This completes the proof, because*n*is the degree of*c*(*X*).