- (a)
- If
*x*∈*X*, then*o*(*g*) is a power of*p*, and since*o*(*g*) =*o*(*xgx*^{-1}), we see that*o*(*g*·*x*)∈*X*. Also*g*·(*h*·*x*) =*g*·(*hxh*^{-1}) =*ghxh*^{-1}*g*^{-1}= (*gh*)·*x*for*g*,*h*∈*P*. Finally 1·*x*=*x*, and so we have an action. - (b)
- {
*z*} is an orbit of size 1 if and only if*g*·*z*=*z*for all*g*∈*P*, if and only if*z*is in the center of*P*. - (c)
- The size of the orbits divide |
*P*| and therefore are powers of*p*. Let*Z*denote the center of*G*. By (b), the number of orbits of size 1 is |*Z*|. Since*p*| |*G*|, we see that*p*| |*P*| and hence*p*| |*Z*|, because the center of a nontrivial*p*-group is nontrivial. The result follows.

- We prove the result by induction on |
*G*|. We may assume that*G*≠1, because if*G*is the trivial group, then*G*has no maximal subgroups. Let*Z*denote the center of*G*and first suppose*H*⊇*Z*. By subgroup correspondence theorem,*H*/*Z*is a maximal subgroup of*G*/*Z*. By induction,*H*/*Z*⊲*G*/*Z*and |(*G*/*Z*)/(*H*/*Z*)| =*p*. Therefore*H*⊲*G*and |*G*/*H*| =*p*.Now assume that

*H*⊉*Z*. Since*HZ*≤*G*and*HZ*≠*H*, we see that*HZ*=*G*. Since the normalizer of*H*in*G*contains*H*and*Z*, we see that*H*⊲*G*. Since*G*/*H*is a nontrivial*p*-group, its center*Y*/*H*is nontrivial and we see that*Y*=*G*, by maximality of*H*. Therefore*G*/*H*is abelian. But then*G*/*H*has a subgroup*K*/*H*of order*p*, and we must have*K*=*G*, again by maximality of*H*, and the result is proven. - Let
*I*⊲*R*. Since*R*is noetherian, there exist*x*_{1},...,*x*_{n}∈*R*such that*I*= (*x*_{1},...,*x*_{n}). Let*g*denote the greatest common divisor of {*x*_{1},...,*x*_{n}}. Since*g*|*x*_{i}for all*i*, there exist*r*_{i}∈*R*such that*x*_{i}=*gr*_{i}and we see that*I*⊆(*g*). Also*x*_{i}/*g*∈*R*for all*i*and no prime divides all the*x*_{i}. Therefore (*x*_{1}/*g*,...,*x*_{n}/*g*) =*R*, in particular there exist*s*_{i}∈*R*such that*x*_{1}*s*_{1}/*g*+ ... +*x*_{n}*s*_{n}/*g*= 1 and hence*g*=*x*_{1}*s*_{1}+ ...*x*_{n}*s*_{n}. Therefore*g*∈*I*, consequently*I*= (*g*) and it follows that*R*is a PID, as required. - Since
*M*is an injective ℤ-module over the PID*I*and*q*≠ 0, we see that*qM*=*M*. Now let*m*⊗*z*be a simple tensor in*M*⊗_{ℤ}ℤ. Since*qM*=*M*, there exists*n*∈*M*such that*qn*=*m*and the*m*⊗*z*=*qn*⊗*z*=*n*⊗*qz*=*n*⊗0 = 0.*M*⊗_{ℤ}ℤ/*q*ℤ = 0. - Since
*M*is a finitely generated module over the PID ℂ[*x*], we may write*M*=*F*⊕*T*, where*F*is a free ℂ[*x*]-module of finite rank and*T*is a finitely generated torsion module. Furthermore we may write*F*= ⊕_{i=1}^{n}ℂ[*x*]/(*x*-*a*_{i})^{bi}for some integers*n*,*b*_{i}and*a*_{i}∈ℂ. First suppose*F*= 0. Note that dim_{ℂ}*T*< ∞, so dim_{ℂ}*M*< ∞, in particular no such*N*can exist ( dim_{ℂ}*M*= dim_{ℂ}*N*⇒*M*≌*N*as ℂ-modules).Therefore we may assume that

*F*≠ 0. Now choose any*c*∈ℂ with*c*≠*a*_{i}for all*i*. By the Chinese remainder theorem (*x*-*c*)*T*=*T*. Also (*x*-*c*)*F*≌*F*and hence (*x*-*c*)*M*≌*M*. Finally*F*= ℂ[*x*]^{m}for some*m*∈ℕ, consequently (*x*-*c*)*F*= (*x*-*c*)ℂ[*x*]^{m}and we deduce that (*x*-*c*)*F*≠*F*. Therefore (*x*-*c*)*M*≠*M*and the result follows (in fact there exist infinitely many such*c*). - (a)
- A polynomial has a degree 1 factor if and only if it has a root.
Therefore a degree 2 polynomial
*f*∈𝔽_{2}[*x*] is irreducible if and only if*f*(0) =*f*(1) = 1. There are only 4 degree 2 polynomials, and it is easy to see that only*x*^{2}+*x*+ 1 satisfies this criterion. - (b)
- If
*g*≔*x*^{5}+*x*^{3}+ 1 is not irreducible, it has a factor of degree 1 or 2. But*g*(0) =*g*(1) = 1 and*x*^{2}+*x*+ 1 does not divide*g*. Therefore*g*is irreducible and it follows that [𝔽_{2}(α) : 𝔽] = 5. If*h*≔*x*^{4}+*x*+ 1 is not irreducible, it has a factor of degree 1 or 2. But*h*(0) =*h*(1) = 1 and*x*^{2}+*x*+ 1 does not divide*h*. Therefore*h*is also irreducible and it follows that [𝔽_{2}(β) : 𝔽_{2}] = 4. Since 4 and 5 are coprime, we deduce that [𝔽_{2}(α,β) : 𝔽_{2}] = 4·5 = 20. - (c)
- Since all field extensions involving finite fields are Galois
extensions,
*K*= 𝔽_{2}(α,β). Also we know that the Galois group is cyclic with order the degree of the extension. Therefore Gal(*K*/𝔽_{2})≌ℤ/20ℤ.

- First we compute the character table for
*S*_{3}. There are three conjugacy classes in*S*_{3}, and representatives are 1, (1 2) and (1 2 3). There is the trivial representation with character χ_{1}defined by χ_{1}(*x*) = 1 for all*x*∈*S*_{3}. Then there is the character χ_{2}which is defined by the sign of a permutation, so χ_{2}(1) = χ_{2}(1 2 3) = 1 and χ_{2}(1 2) = -1. The number of irreducible characters equals the number of conjugacy classes, so there are exactly three irreducible characters. The final character χ_{3}can be determined by the orthogonality relations. We have χ_{3}(1) = 2. Since χ an irreducible character if and only if (complex conjugate) is an irreducible character, we see that χ_{3}(1 2) and χ_{3}(1 2 3) are real numbers. Taking the inner product of the first two columns of the character table, we obtain χ_{3}(1 2) = 0, and then it follows easily that χ_{3}(1 2 3) = -1. Thus the character table of*S*_{3}isClass Size 1 3 2 Class Rep 1 (1 2) (1 2 3) χ _{1}1 1 1 χ _{2}1 -1 1 χ _{3}1 0 -1 Since ℤ/3ℤ is an abelian group, all its irreducible characters are of degree one and correspond to homomorphisms into the cube roots of 1 in ℂ, because |ℤ/3ℤ| = 3. Let ω =

*e*^{2πi/3}, a primitive cube root of 1. Let 0, 1, 2 represent the conjugacy classes , , respectively. Then the character table for ℤ/3ℤ isClass Size 1 1 1 Class Rep 0 1 2 ψ _{1}1 1 1 ψ _{2}1 ω ω ^{2}ψ _{3}1 ω ^{2}ω Now for a finite group of the form

*G*×*H*, the conjugacy classes of*G*×*H*is*C*×*D*, where*C*is the set of conjugacy classes of*G*and*D*is the set of conjugacy classes of*D*, and then the irreducible characters are of the form χ_{i}ψ_{j}≔χ_{i}(*c*)ψ_{j}(*d*), in particular there are |*C*|·|*D*| irreducible characters. Therefore the character table of*S*_{3}×ℤ/3ℤ isClass Size 1 1 1 3 3 3 2 2 2 Class Rep (1, 0) (1, 1) (1, 2) ((1 2), 0) ((1 2), 1) ((1 2), 2) ((1 2 3), 0) ((1 2 3), 1) ((1 2 3), 2) χ _{1}ψ_{1}1 1 1 1 1 1 1 1 1 χ _{1}ψ_{2}1 ω ω ^{2}1 ω ω ^{2}1 ω ω ^{2}χ _{1}ψ_{3}1 ω ^{2}ω 1 ω ^{2}ω 1 ω ^{2}ω χ _{2}ψ_{1}1 1 1 -1 -1 -1 1 1 1 χ _{2}ψ_{2}1 ω ω ^{2}-1 - ω - ω ^{2}1 ω ω ^{2}χ _{2}ψ_{3}1 ω ^{2}ω -1 - ω ^{2}- ω 1 ω ^{2}ω χ _{3}ψ_{1}2 2 2 0 0 0 -1 -1 -1 χ _{3}ψ_{2}2 2ω 2ω ^{2}0 0 0 -1 - ω - ω ^{2}χ _{3}ψ_{3}2 2ω ^{2}2ω 0 0 0 -1 - ω ^{2}- ω

Peter Linnell 2014-08-12