- Let
*s*∈*S*and let*H*denote the stabilizer of*s*in*G*. Since*G*acts transitively on*S*, we have |*G*| =*p*^{n}|*H*|, hence*p*^{n}| |*G*|/|*P*∩*H*| and we deduce that*p*^{n}| |*P*|/|*P*∩*H*, because*p*~~|~~|*G*|/|*P*|. Therefore*p*^{n}divides the size of the orbit of*s*under*P*, because*P*∩*H*is the stabilizer of*s*in*P*. Thus we must have the orbit of*s*under*P*is the whole of*S*and the result is proven. - Let
*G*be a simple group of order 448. The number of Sylow 2-subgroups of*G*is congruent to 1 mod 2 and divides 7, and cannot be 1 because*G*is not simple. Therefore*G*has exactly 7 Sylow 2-subgroups and because*G*is simple, we deduce that*G*is isomorphic to a subgroup of*A*_{7}. This is not possible because 448 does not divide |*A*_{7}|, so the result is proven. - (a)
- If
*x*^{2}+ 1 was not irreducible, then it would have a root in**Z**/3**Z**. This is not the case, because*x*^{2}= 0 or 1 mod 3. - (b)
- We have an epimorphism
**Z**/3**Z**[*x*] --> >**Z**[*i*]/3**Z**[*i*] determined by*x*|-->*i*whose kernel contains*x*^{2}+ 1. Thus from part (a), we see that**Z**[*i*]/3**Z**[*i*] is a field and hence 3 is a prime in**Z**[*i*]. We can now apply Eisenstein's criterion for the prime 3. Since 3 divides 3 and -9, but 3^{2}does not divide 12 in**Z**[*i*], the result is proven.

- By the structure theorem for finitely generated modules over a PID,
there is an
*R*-submodule*K*of*M*containing*N*such that*M*/*K*is a torsion module and*K*/*N*is a free module, so there exists 0≠*r*∈*R*such that*Mr*⊆*K*. Since*K*/*N*is free, there exists a submodule*L*of*K*such that*L*+*N*=*K*and*L*∩*N*= 0. The result follows. - Let
*b*∈*B*. Since*f*is onto, there exists*a*∈*A*such that*f*(*a*) =*b*. Now set*k*(*b*) =*g*(*a*). If we had instead chosen*a'*∈*A*such that*f*(*a'*) =*b*, then*jg*(*a'*) =*hf*(*a'*) =*h*(*b*) =*hf*(*a*) =*jg*(*a*)*g*(*a'*) =*g*(*a*) because*j*is one-to-one; in other words, the definition of*k*does not depend on the choice of*a*. Next we need to show that*k*is an*R*-module homomorphism. Suppose*b*,*b'*∈*B*and choose*a*,*a'*∈*A*such that*f*(*a*) =*b*and*f*(*a'*) =*b'*. Then*f*(*a*+*a'*) =*b*+*b'*. Thus*k*(*b*+*b'*) =*g*(*a*+*a'*) =*g*(*a*) +*g*(*a'*) =*k*(*b*) +*k*(*b'*). Also if*r*∈*R*, then*f*(*ar*) =*br*, consequently*k*(*br*) =*g*(*ar*) =*g*(*a*)*r*=*k*(*b*)*r*and we have shown that*k*is an*R*-module homomorphism. Clearly*kf*=*g*. Furthermore*jkf*=*jg*=*hf*and since*f*is onto, we deduce that*jk*=*h*. Finally*k*is unique because*j*is one-to-one. - Solving
*x*^{4}-2*x*^{2}+ 9 = 0, we find that*x*^{2}= 1±2√2*i*and we deduce that the roots of*x*^{4}-2*x*^{2}+ 9 are ±√2±*i*. It follows that the splitting field is**Q**[*i*,√2]. Since this has degree 4 over**Q**, we see that the Galois group has order 4. The automorphisms induced by*i*|--> -*i*, √2 |--> √2 and*i*|-->*i*, √2 |--> - √2 both have order 2 and we conclude that the Galois group is isomorphic to**Z**/2**Z**×**Z**/2**Z**. - We can define an
*R*-bilinear map*R*/*I*×*R*/*J*->*R*/(*I*+*J*) by (*r*+*I*,*s*+*J*) |-->*rs*. This induces an*R*-module map θ :*R*/*I*⊗_{R}*R*/*J*->*R*/(*I*+*J*) satisfying θ((*r*+*I*)⊗(*s*+*J*)) =*rs*+*I*+*J*. Now define φ :*R*->*R*/*I*⊗_{R}*R*/*J*by φ(*r*) = (*r*+*I*)⊗_{R}(1 +*J*). Then φ is an*R*-module map and clearly*I*⊆kerφ. Also if*j*∈*J*, then φ(*j*) = (*j*+*I*)⊗(1 +*J*) = (1 +*I*)⊗(*j*+*J*) = 0. It follows that*I*+*J*⊆kerφ and we deduce that φ induces an*R*-module map ψ :*R*/(*I*+*J*) ->*R*/*I*⊗_{R}*R*/*J*such that ψ(*r*+*I*+*J*) = (*r*+*I*)⊗(1 +*J*). Note that θψ(*r*+*I*+*J*) = θ((*r*+*I*)⊗(1 +*J*)) =*r*+*I*+*J*so θψ is the identity map. Finally ψθ(*r*+*I*)⊗(*s*+*J*) = ψ(*rs*+*I*+*J*) = (*rs*+*I*)⊗(1 +*J*) = (*r*+*I*)⊗(*s*+*J*) and we conclude that ψθ is also the identity map. This shows that θ and ψ are isomorphisms, and the result is proven.

Peter Linnell 2011-07-21