- Let
*G*be a group of order 18. The number of Sylow 3-subgroups is congruent to 1 mod 3 and divides 18/9 = 2. Therefore*G*has a unique Sylow 3-subgroup*H*of order 9, and*H*<|*G*. Also*G*has an element*x*of order 2. Then*G*≌*H*X| <*x*>. Since groups of order*p*^{2}are prime,*H*is an abelian group. For a positive integer*n*, let*C*_{n}denote the cyclic group of order*n*. We have three cases to consider.- (a)
- The conjugation action of
*x*on*H*is trivial, that is*xhx*^{-1}=*h*for all*h*∈*H*and we have*G*≌*H*× <*x*>. There are two isomorphism classes for*H*, namely*C*_{9}and*C*_{3}×*C*_{3}. It follows that there are two isomorphism classes for*G*in this case. - (b)
- The conjugation action of
*x*on*H*is nontrivial and*H*≌*C*_{9}. There is exactly one automorphism of*H*of order two, namely*h*|-->*h*^{-1}for*h*∈*H*. It follows that there is exactly one isomorphism class for*G*in this case. - (c)
- The conjugation action of
*x*on*H*is nontrivial and*H*≌*C*_{3}×*C*_{3}. Either*x*acts by inversion on*H*and this gives us one isomorphism class for*G*. Otherwise we may write*H*=*A*×*B*where*A*,*B*≌*C*_{3},*x*centralizes*A*and*x*acts by inversion on*B*. This yields a second isomorphism class for*G*.

- First suppose that
*A*is Noetherian. Then*A*[*X*,*Y*] is Noetherian by Hilbert's basis theorem. Since factor rings of Noetherian rings are Noetherian, we see that*A*[*X*,*Y*]/(*X*^{2}-*Y*^{2}) is also Noetherian.Conversely suppose

*A*[*X*,*Y*]/(*X*^{2}-*Y*^{2}) is Noetherian. Since (*X*,*Y*)⊃(*X*^{2}-*Y*^{2}), we see that*A*[*X*,*Y*]/(*X*,*Y*) is also Noetherian. But*A*[*X*,*Y*]/(*X*,*Y*)≌*A*, because the homomorphism*X*|--> 0,*Y*|--> 0 :*A*[*X*,*Y*] ->*A**X*,*Y*), and the result follows. - Let
0≠
*u*∈*F*^{2}and let*S*denote the stabilizer in GL_{n}(*F*) of the one-dimensional subspace*Fu*. We need to prove that*S*is not simple. Set*D*= {diag(*f*,*f*) | 0≠*f*∈*F*}, where diag(*f*,*f*) indicates the invertible matrix in GL_{2}(*F*) which has*f*'s on the main diagonal and zeros elsewhere. Then*D*is a central subgroup of*S*and since |*F*|≥3, it is not 1. Let*v*be an element of*F*^{2}which is not in*Fu*. Then {*u*,*v*} is a basis of*F*^{2}and so we can define a linear isomorphism of*F*^{2}by*u*|-->*u*,*v*|-->*u*+*v*. This yields an element of*S*\*D*. Thus*D*is a normal subgroup of*S*which is neither 1 nor*D*, and we conclude that*S*is not simple. - Clearly
*M*cannot be free of rank 0. Nor can*M*be free of rank at least 2, because if*a*,*b*∈*M*were part of a free*R*-basis for*M*, we would have 0≠*ab*∈*aR*∩*bR*, which would mean that {*a*,*b*} was not linearly independent over*R*. Therefore the only possibility of*M*being free is that it is free of rank 1. This means we can write 2*R*+*XR*=*cR*for some*c*∈*R*. There are several methods to show that this is not possible; we present one of them.Since 2∈

*cR*, we see that*c*is a polynomial of degree zero and thus*c*= ±1 or ±2. Without loss of generality, we may assume that*c*= 1 or 2. Since*X*∈*cR*, we may write*X*=*cf*for some polynomial*f*∈ℤ[*X*]. By considering the leading coefficient (degree 1) of*f*, we see that*c*= 1 and we deduce that there exist*g*,*h*∈*R*such that 2*g*+*Xh*= 1. This is not possible because the left hand side has constant coefficient ∈2ℤ and in particular cannot be 1. It follows that*M*is not a free*R*-module. - Since
σ
*a*-*a*∈*F*for all σ∈*G*, we see that (∑_{σ∈G}σ*a*) - |*G*|*a*∈*F*. Now τ∑_{σ∈G}σ*a*= ∑_{σ∈G}σ*a*for all τ∈*G*. Since*K*/*F*is a Galois extension with Galois group*G*, it follows that ∑_{σ∈G}σ*a*∈*F*and we deduce that |*G*|*a*∈*F*. We conclude that*a*∈*F*because*F*has characteristic zero. - Set
*m*= √*n*. A finite dimensional simple algebra over an algebraically closed field is isomorphic to a full matrix ring over the field. In this situation, this means*S*is isomorphic to M_{m}(ℂ), the*m*×*m*matrices over ℂ. Let*e*_{ij}( 1≤*i*,*j*__<__*m*) denote the matrix units of M_{m}(ℂ), so*e*_{ij}has 1 in the (*i*,*j*)th position and zeros elsewhere. Then*Se*_{ii}is the*i*th column of*S*and we see that*S*=*Se*_{11}⊕*Se*_{22}⊕...⊕*Se*_{mm}. All that remains to prove is that*Se*_{ii}is irreducible for all*i*. Without loss of generality, we may assume that*i*= 1. Suppose*M*is a nonzero*R*-submodule of*Se*_{11}. The general element α of*Se*_{11}is of the form ∑_{i}*a*_{i}*e*_{i1}. If this is a nonzero element of*M*, then*a*_{i}≠ 0 for some*i*, 1≤*i*≤*m*, and we deduce that*e*_{11}=*a*_{i}^{-1}*e*_{1i}α∈*M*. Thus*M*=*Se*_{11}and the result follows. - The map
*f*|-->*f*⊗1 : -> ⊗_{F}*L*is an algebra monomorphism with image ⊗1. Furthermore, if {λ_{1},...,λ_{n}} is a basis for*L*over*F*, then {1⊗λ_{1},..., 1⊗λ_{n}} is a basis for ⊗_{F}*L*over ⊗1. It follows that ⊗_{F}*L*is a field extension of degree*n*over and since is algebraically closed, we deduce that*n*= 1. Therefore*L*=*F*as required.

Peter Linnell 2012-08-04