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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 representations.Going to the quotient group G/<a> <b>, we have a surjection QG → Q<b> Q × Q(ω), where ω is as in (1). This gives two simpleQG-modules: Q with the trivial G-action, and Q(ω) with a acting trivially andb acting by multiplication by ω. The surjection π gives Q and Q(ω) as two simple components of QG.Our next step is to construct a 6-dimensional simple QG-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 QG-module by letting b actvia the field automorphism σ Gal(Q(ζ)/Q) of order 3 sending ζ→ ζ2. This gives a well-defined G-actionsince, for any k K, (a2b)k= ζ2σ(k), while (ba)k =b(ζk) = σ(ζk) = ζ2σ(k).
Thus, K is a simple QG-module.Let E = End(QGK). 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 + σ2ζ = ζ + ζ2 + ζ4 F hasminimal equation α2 + α+2= 0, so F = Q(α) = Q(√−7 ).) We have
now seen that E F. Since dimEK = 3, a simple component of QG isgiven by End(KE) M3(E) M3(Q(√−7 )).
This has Q-dimension 18, so we musthave QG Q × Q(ω) ×M3(Q(√−7)), and Q, Q(ω), and K are the three simple QG-modules.

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