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Tuesday, July 30, 2013

Group Generation Through Exterior Algebras (11.5.9)

Dummit and Foote Abstract Algebra, section 11.5, exercise 9:

MathJax TeX Test Page Let R=ZG be the group ring over the group G={1,σ}. Letting M=Z×Z and defining σ(e1)=e1+2e2,σ(e2)=e2, show M is an R-module and 2M is a group of order 2 generated by e1e2.

Proof: Since there is a unique representation for rR and mM we may define the action of R on M from their mutual bases' actions to obtain rm=(z1σ+z2)(a,b)=(z1a+z2a,2z1az1b+z2b). Note that σ(σ(a,b))=(a,b) so that it satisfies action associativity.

Observe(a,b)(c,d)=((a,0)+(0,b))((c,0)+(0,d))=(ad+bc)(e1e2)so that 2M is generated by e1e2 as an R-module. As well, we have σ(e1e2)=(σe1)e2=(e1+2e2)e2=e1e2 and σ(e1e2)=e1(σe2)=e1e2=(e1e2) so that (z1σ+z2)(e1e2)=(¯z1+z2)(e1e2) where the overline denotes reduction modulo 2, implying 2M is either of order 1 or 2. To prove the latter, we shall construct a bilinear alternating form φ on M such that φ(e1,e2)0.

First, define ψ:RZ/2Z by ψ(z1σ+z2)=¯z1+z2 and prove it is a ring homomorphismψ(z1σ+z2+z1σ+z2)=¯z1+z2+z1+z2=ψ(z1σ+z2)+ψ(z1σ+z2)ψ((z1σ+z2)(z1σ+z2))=ψ((z1z2+z2z1)σ+(z1z1+z2z2))=¯z1z2+z2z1+z1z1+z2z2=¯(z1+z2)(z1+z2)=ψ(z1σ+z2)ψ(z1σ+z2)Now define φ:M×MZ/2Z by φ((a,b),(c,d))=ψ(adbc). It alternates and is 1 on (e1,e2) so all that remains to show is that it is bilinear by being additive in its componentsφ((a,b)+(a,b),(c,d))=ψ((a+a)d(b+b)c))=ψ(adbc)+ψ(adbc)=φ((a,b),(c,d))+φ((a,b),(c,d))and linear over Rφ((z1σ+z2)(a,b),(c,d))=φ((z1a+z2a,2z1a+z2bz1b),(c,d))=¯z1ad+z1bc+z2adz2bc=¯z1adz1bc+z2adz2bc=¯z1(adbc)+z2(adbc)=(z1σ+z2)φ((a,b),(c,d)) 

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