# Answer to Question #23879 in Abstract Algebra for Irvin

Question #23879

Let G be the group of order 21 generated by two elements a, b with the relations a^7 = 1, b^3 = 1, and bab^−1 = a^2. Construct the irreducible rational representations of G and determine the Wedderburn decomposition of the group algebra QG.

Expert's answer

We turn to the case of rational

Thus,

now seen that

This has Q-dimension 18, so we musthave Q

*representations.Going to the quotient group**G/<a>**∼**<b>*, we have a surjection Q*G →*Q*<b>**∼*Q*×*Q(*ω*), where*ω*is as in (1). This gives two simpleQ*G*-modules: Q with the trivial*G*-action, and Q(*ω*) with*a*acting trivially and*b*acting by multiplication by*ω*. The surjection*π*gives Q and Q(*ω*) as two simple components of Q*G*.Our next step is to construct a 6-dimensional simple Q*G*-module. We startwith the simple Q*<a>*-module*K*= Q(*ζ*) (where*ζ*is as before), on which*a*actsvia multiplication by*ζ*. We can make*K*into a Q*G*-module by letting*b*actvia the field automorphism*σ**∈**Gal(Q(**ζ*)*/*Q) of order 3 sending*ζ**→ ζ*^{2}. This gives a well-defined*G*-actionsince, for any*k**∈**K*, (*a*^{2}*b*)*k*=*ζ*^{2}*σ*(*k*), while (*ba*)*k*=*b*(*ζ**k*) =*σ*(*ζ**k*) =*ζ*^{2}*σ*(*k*)*.*Thus,

*K*is a simple Q*G*-module.Let*E*= End(Q*GK*). Any*f**∈**E*is right multiplication by some element*l**∈**K*, and we must have (*bk*)*f*=*b*(*kf*)(for every*k**∈**K*); that is,*σ*(*k*)*l*=*σ*(*kl*) =*σ*(*k*)*σ*(*l*)*,*which amounts to*l**∈**Q(**ζ*)*. Now*^{σ}*F*:= Q(*ζ*)*is the unique quadratic extension ofQ in*^{σ}*K*, which is well-known to be Q(*√−*7 ). (More explicitly,*α*=*ζ*+*σζ*+*σ*2*ζ*=*ζ*+*ζ*2 +*ζ*4*∈**F*hasminimal equation*α*^{2}+*α*+2= 0, so*F*= Q(*α*) = Q(*√−*7 ).) We havenow seen that

*E**∼**F*. Since dim*= 3, a simple component of Q*_{E}K*G*isgiven by End(*K*)_{E}*∼*M_{3}(*E*)*∼*M_{3}(Q(*√−*7 ))*.*This has Q-dimension 18, so we musthave Q

*G**∼*Q*×*Q(*ω*)*×*M_{3}(Q(*√−*7))*,*and Q, Q(*ω*), and*K*are the three simple Q*G*-modules.Need a fast expert's response?

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