Abstract
We show that Thompson’s \(A\times B\)-lemma can be obtained as a consequence of Brauer’s third main theorem.
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Let k be an algebraically closed field of prime characteristic p. Brauer’s third main theorem [2, Theorem 3], rephrased using Brauer pairs (cf. [1, Theorem 3.13] or [7, Theorem 6.3.14], or also [5, Theorem 7] for a different proof), states that if b is the principal block idempotent of a finite group algebra kG, then \(\mathrm {Br}_Q(b)\) is the principal block idempotent of \(kC_G(Q)\) for any p-subgroup Q of G. Here \(\mathrm {Br}_Q : (kG)^Q\rightarrow kC_G(Q)\) denotes the Brauer homomorphism (cf. [3, §1.2] or [6, Theorem 5.4.1]). In particular, if kG has a unique block, then \(kC_G(Q)\) has a unique block. Moreover, if kG has a unique block, then \(O_{p'}(G)=1\) because otherwise \(\frac{1}{|O_{p'}(G)|} \sum _{g \in O_{p'}(G)} g\) would be a central idempotent in kG different from 1. Also, it is well-known that if G has a self-centralising normal p-subgroup, then kG has a unique block (cf. [7, Corollary 6.2.8]). We use these facts to give a proof of the following result.
(Thompson’s \(A \times B\)-lemma; cf. [4, Chapter 5, Theorem 3.4]). Let \(A \times B\) be a subgroup of the automorphism group of a finite p-group P, with A a \(p'\)-group and B a p-group. If A acts trivially on \(C_P(B)\), then \(A=1\).
FormalPara ProofConsider the group \(G=P \rtimes (A\times B)\), where the notation is as in the statement, and suppose that A acts trivially on \(C_P(B)\). Note that \(S=P\rtimes B\) is the unique Sylow p-subgroup of G, and that A can be regarded as a \(p'\)-subgroup of the automorphism group of S. Thus S is self-centralising and normal in G, and hence kG has a unique block.
By the assumptions, A acts trivially on the group \(Q = C_P(B) \times B\). That is, we have \(A \le C_G(Q)\), and hence \(C_G(Q) = C_S(Q) \rtimes A\). Since kG has a unique block, \(kC_G(Q)\) has a unique block. We now show that Q is self-centralising in S. Let \(x\in C_S(Q)\). Write \(x = yu\) for some \(y\in P\) and \(u\in B\). Since \(u\in B\), it follows that conjugation by u preserves the decomposition \(Q = C_P(B) \times B\). Thus conjugation by y preserves this decomposition as well. In particular, y normalises B. By elementary group theory, it follows that y centralises B. Indeed, if \(u\in \) B, then \(yuy^{-1}u^{-1} \in P \cap B =1 \). This shows that \(C_S(Q) \le Q \). Since \(Q \le C_G(A)\), it follows that \(C_G(Q) = C_S(Q) \times A\). By the above, \(kC_G(Q)\) has a single block, and hence \(A=O_{p'}(C_G(Q))=1\). \(\square \)
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Acknowledgements
The authors would like to thank G.R. Robinson for helpful comments on an earlier version of this note. The first author acknowledges support from EPSRC Grant EP/T004592/1.
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Kessar, R., Linckelmann, M. A block theoretic proof of Thompson’s \(A\times B\)-lemma. Arch. Math. 117, 481–483 (2021). https://doi.org/10.1007/s00013-021-01638-5
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DOI: https://doi.org/10.1007/s00013-021-01638-5