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

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.