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* *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*)^{σ}. Now *F *:= Q(*ζ*)^{σ} is the unique quadratic extension ofQ in *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 have

now seen that*E **∼* *F*. Since dim_{E}K = 3, a simple component of Q*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.

Thus,

now seen that

This has Q-dimension 18, so we musthave Q

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