- We use without further comment the property that a
*p*-subgroup of a group is a Sylow*p*-subgroup if and only if it has index in the group prime to*p*.First suppose

*P*is a Sylow*p*-subgroup of*G*. Then*P*∩*H*is a subgroup of*P*, so*P*∩*H*is a*p*-subgroup of*H*. Also*P*/*P*∩*H*≌*PH*/*H*, hence |*H*|/|*P*∩*H*| = |*PH*|/|*P*|. Furthermore |*G*|/|*P*| = |*G*|/|*PH*|·|*PH*|/|*P*| and we deduce that |*H*|/|*P*∩*H*| divides |*G*|/|*P*|. Since |*G*|/|*P*| is prime to*p*, it follows that |*H*|/|*P*∩*H*| is also prime to*p*, which proves that*P*∩*H*is a Sylow*p*-subgroup of*H*.Next,

*PH*/*H*≌*P*/*P*∩*H*, so*PH*/*H*is a*p*-subgroup of*G*/*H*. Furthermore |*G*|/|*P*| = |*G*|/|*PH*|·|*PH*|/|*P*|, and we see that |*G*|/|*PH*| is prime to*p*. Therefore |*G*/*H*|/|*PH*/*H*| is also prime to*p*and it follows that*PH*/*H*is a Sylow*p*-subgroup of*G*/*H*.Now suppose

*P*∩*H*and*PH*/*H*are Sylow*p*-subgroups. Since*P*/*P*∩*H*≌*PH*/*H*and |*P*| = |*P*/*P*∩*H*|·|*P*∩*H*|, we see that*P*is a*p*-group. Finally if |*H*| =*p*^{a}*x*and |*G*/*H*| =*p*^{b}*y*, where*x*and*y*are prime to*p*, then |*G*| =*p*^{a+b}*xy*and*xy*is prime to*p*. This means that a Sylow*p*-subgroup of*G*has order*p*^{a+b}. But since |*P*∩*H*| =*p*^{a}and |*PH*/*H*| =*p*^{b}, we see that |*P*| =*p*^{a+b}and hence*P*is a Sylow*p*-subgroup of*G*, as required. - Suppose
*G*is simple group of order 576. The number of Sylow 2-subgroups is congruent to 1 mod 2 and divides 9, so has to be 1, 3 or 9. It cannot be 1, because then*G*would have a normal Sylow 2-subgroup. Nor can it be 3, otherwise*G*would be isomorphic to a subgroup of*A*_{3}. Finally suppose it is 9. Then*G*is isomorphic to a subgroup of*A*_{9}; unfortunately at first sight this seems possible, because 576 divides |*A*_{9}|. However the isomorphism is induced by the representation of*G*on the 9 left cosets of a Sylow 2-subgroup*P*in*G*. Thus*g*∈*G*gives the permutation*xP*↦*gxP*. Then*g*stabilizes some*xP*if and only if*g*is in some Sylow 2-subgroup. So if*g*has order 6, it can be considered as an element of*A*_{9}which fixes no points; a quick check shows that this is not possible and therefore*G*has no element of order 6.Now consider the Sylow 3-subgroups. If

*P*and*Q*are distinct Sylow 3-subgroups and 1≠*x*∈*P*∩*Q*, then C_{G}(*x*) contains*P*and*Q*and hence contains an element of order 2. It follows that*G*has an element of order 6, which is not possible by the previous paragraph, so distinct Sylow 3-subgroups intersect trivially.Next the number of Sylow 3-subgroups is 16 or 64. If it is 16, we consider the representation of

*G*on the left cosets of a Sylow 3-subgroup*P*. If*Q*is another Sylow 3-subgroup, then there is an orbit under*Q*which has order 3, which shows that there exists 1≠*q*∈*Q*such that*q*∈*P*∩*Q*, contradicting the previous paragraph. Finally if there are 64 Sylow 3-subgroups, then since two distinct Sylow 3-subgroups intersect in the identity, we can count elements to show that*G*has a normal Sylow 2-subgroup. - Consider
*p*^{m}+*q*^{n}. If this is not a unit, then there exists some prime which divides it, which without loss of generality we may assume is*p*. Thus*p*divides*p*^{m}+*q*^{n}, hence*p*divides*q*^{n}, which is not possible.Now let

*I*⊲*R*. We want to prove*I*is a principal ideal, and since 0 is clearly a principal ideal, we may assume that*I*≠ 0. Each nonzero element of*I*has a factorization*up*^{i}*q*^{j}, where*u*is a unit and*i*,*j*are nonnegative integers. Choose 0≠*x*∈*I*such that*x*=*p*^{i}*q*^{j}, with*i*as small as possible, and then choose 0≠*y*∈*I*such that*y*=*p*^{k}*q*^{l}, with*l*as small as possible. We will show that*I*= (*p*^{i}*q*^{l}). Clearly*I*⊆(*p*^{i}*q*^{l}). On the other hand*p*^{k-i}+*q*^{j-l}is a unit by the above, hence*p*^{i}*q*^{l}is an associate of*y*+*x*and we see that*p*^{i}*q*^{l}∈*I*. This proves that*I*= (*p*^{i}*q*^{l}). - Note that
*k*[*x*] is a PID. By the structure theorem for finitely generated modules over a PID, we may write*M*≌*k*[*x*]^{d}⊕*k*[*x*]/(*f*_{1})⊕...⊕*k*[*x*]/(*f*_{n}), where the*f*_{i}are monic polynomials, say of degree*a*_{i}, and*f*_{i}|*f*_{i+1}for all*i*. Suppose*d*= 0. Then dim_{k}*M*= ∑_{i=1}^{n}*a*_{i}, and then it is clear that if*N*is a proper*k*[*x*]-submodule of*M*, then*N*≇*M*, because dim_{k}*N*< dim_{k}*M*. Therefore*d*> 0, in particular there is an epimorphism*M*↠*k*[*x*]. Since*C*is a cyclic*k*[*x*]-module, there exists an epimorphism*k*[*x*]↠*C*. By composing these two epimorphisms, we obtain a*k*[*x*]-module epimorphism*M*↠*C*. - Let
*p*denote the characteristic of*K*. Then we may write |*K*| =*p*^{n}where*n*∈ℕ. Set*M*=*K*^{+}⊗_{ℤ}*L*^{×}. Then |*K*^{+}|*M*= 0 and |*L*^{×}|*M*= 0, so if (|*K*|,|*L*|-- 1) = 1, we see that |*M*| = 0. Now suppose (|*K*|,|*L*|-- 1)≠1. Then*p*divides |*L*|-- 1. Also*K*^{+}≌(ℤ/*p*ℤ)^{n}and*L*^{×}≌ℤ/(|*L*|-- 1)ℤ.Now we have well defined homomorphisms θ : ℤ/

*p*ℤ⊗_{ℤ}ℤ/(|*L*|-- 1)ℤ and φ : ℤ/*p*ℤ→ℤ/*p*ℤ⊗_{ℤ}ℤ/(|*L*|-- 1)ℤ determined by θ(⊗) = and φ() = ⊗, and θφ and φθ are the identity maps. This shows that ℤ/*p*ℤ⊗_{ℤ}ℤ/(|*L*|-- 1)ℤ≌ℤ/*p*ℤ and we deduce that*M*≌(ℤ/*p*ℤ)^{n}. Therefore if (|*K*|,|*L*|-- 1)≠1, it follows that |*M*| = |*K*|. - Write
*K*∩*L*= ℚ(α_{1},...,α_{n}), let*f*_{i}denote the minimum polynomial of α_{i}over ℚ, and set*f*=*f*_{1}...*f*_{n}. Let*F*denote the splitting field of*f*over ℚ, a subfield of ℂ. Since*K*and*L*are Galois extensions of ℚ, all the roots of all*f*_{i}lie in both*K*and*L*and hence the splitting field of*f*is contained in*K*∩*L*. Therefore*K*∩*L*is the splitting field of*f*and it follows that*K*∩*L*is a Galois extension of ℚ. - We can split up the given exact sequence into two short exact
sequences, namely
0→ℤ→
*P*→*Y*→ 0 and 0→*Y*→*Q*→ℤ→ 0. Then using the long exact sequence for Ext in the first variable, we obtain*H*^{1}(*G*,*X*) = Ext_{ℤG}^{1}(ℤ,*X*)≌Ext_{ℤG}^{2}(*Y*,*X*)≌Ext_{ℤG}^{3}(ℤ,*X*) =*H*^{3}(*G*,*X*),

Peter Linnell 2013-08-23