Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 138 / 2

The roots-of-unity criterion for a splitting field for finite group representations says that the assumed $m$th roots of unity make $k$ a splitting field for $G$. Fix a multiplicative identification of the group ${\mu }_m(k)$ with the complex $m$th roots of unity. For a p-regular element $g$, its order divides $m$; the operator $\rho (g)$ is diagonalizable because $X^{|g|}-1$ has distinct roots in $k$. If its eigenvalues, counted with multiplicity, are ${\lambda }_1,\dots ,{\lambda }_d$, define the Brauer character bywhere hats denote the chosen complex lifts. This defines a class function on the p-regular elements, not on arbitrary elements of $G$.

A short exact sequence can be represented by block triangular matrices, so the eigenvalue multiset is the union of those on its submodule and quotient. Thus Brauer characters are additive on short exact sequences. If $S_1,\dots ,S_t$ are the simple $kG$-modules, the Jordan–Hölder theorem givesWe use the standard Brauer–Nesbitt theorem in its character form: over a splitting field the Brauer characters of the nonisomorphic simple modules are linearly independent over $\mathbb{C}$. ThereforeBoth the modular splitting-field criterion and this independence theorem are the representation-theoretic results used here.

Let $g$ be a p-regular element, with eigenvalues ${\lambda }_i$ on $V$. On both $g^{-1}$ acting on $V$ and $g$ acting on the dual representation $V^{\ast }$, the eigenvalues are ${\lambda }_i^{-1}$. Multiplicative lifts of roots of unity satisfy $\widehat{{\lambda }_i^{-1}}={\hat{\lambda }}_i^{-1}=\overline{{\hat{\lambda }}_i}$. HenceAll these identities concern the domain of a Brauer character, namely the p-regular elements. If necessary, extend the p-modular system to contain the relevant roots of unity and use one consistent choice of lifts.

For a p-regular element $g$, diagonalize its actions on $V$ and $V^{\prime }$. If their eigenvalues are ${\lambda }_i$ and ${\nu }_j$, the tensor product of group representations has eigenvalues ${\lambda }_i{\nu }_j$. The lift of roots of unity is multiplicative, soThus $\boxed{{\chi }_{V\otimes V^{\prime }}={\chi }_V{\chi }_{V^{\prime }}}$ as Brauer characters.

For a nonzero group representation $V$, there is a $G$-equivariant identificationwhere $G$ acts on endomorphisms by conjugation. The identity endomorphism is a nonzero fixed vector, so its span is a trivial submodule. It follows that the trivial simple module is a Jordan–Hölder factor of $V\otimes V^{\ast }$, and character additivity gives $\boxed{1_G\text{ is a constituent of }\chi \overline{\chi }}$.

Here “constituent” means a composition factor; the trivial submodule need not be a direct summand. Also the assertion needs $\chi \ne 0$: the zero-dimensional module has the zero Brauer character, whose product with its conjugate has no constituent. This is an implicit nonzero-character qualification in the printed statement.

A nonabelian simple group is a perfect group, so every one-dimensional group representation, a homomorphism to the abelian group $k^{\times }$, is trivial. Thus a nontrivial irreducible Brauer character cannot have degree $1$.

Suppose instead that its representation has dimension $2$, working over a splitting extension if necessary. Its determinant is a one-dimensional representation and is therefore trivial. Its kernel is a normal subgroup, and the representation is nontrivial, so simplicity makes it faithful. Consequently $G$ embeds in ${\mathrm{SL}}_2(k)$.

We use the Feit–Thompson theorem: every finite group of odd order is solvable. Therefore this nonabelian simple group has even order, and Cauchy's theorem for finite groups supplies an involution $t$. In odd characteristic, its representing matrix $A$ satisfies $A^2=I$ and is diagonalizable with eigenvalues in $\{1,-1\}$. Since $\det A=1$, it is $I$ or $-I$. Faithfulness excludes $I$, so $t$ acts as the scalar matrix $-I$. It commutes with the whole image; faithfulness then makes $t$ central in $G$, contradicting nonabelian simplicity. Hence $\boxed{\chi (1)>2}$.

The odd order theorem is the deep standard group-theoretic input in this proof; its use is explicit rather than hidden in an unsupported assertion that $G$ has an involution.