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If $S_i$ are the simple modules of a split group algebra and $P_i$ are their projective covers, then

Let $g$ be a p-regular element. Its eigenvalues on $M$ are roots of unity of order prime to $p$. If ${\hat{\lambda }}_1,\dots ,{\hat{\lambda }}_d$ are their Teichmuller lifts to characteristic-zero roots of unity, the Brauer character isIt depends only on the conjugacy class of $g$, is additive in short exact sequences, and equals the restriction of an ordinary character whenever the representation lifts.

For linear independence, choose a splitting p-modular system and let $P_i$ be the projective cover of the simple module $S_i$. A projective lattice lifting $P_i$ has an ordinary character ${\Phi }_i$ that vanishes on p-singular elements. Reduction and ordinary character orthogonality givebecause $S_i$ is the head of $P_i$. Pairing a relation ${\sum }_j{c}_j{\chi }_{S_j}=0$ with every ${\Phi }_i$ yields $c_i=0$ for every $i$. HenceThis is the linear-independence part of the Brauer–Nesbitt theorem.

Decompose the projective module aswhere $T$ ranges over the simple modules. Since $P_T/J(P_T)\cong T$, the multiplicity of $S$ in the head of a module $P/J(P)$ is $m_S$. Part (b) gives $\mathrm{Soc}(P_T)\cong T$, so the multiplicity in $\mathrm{Soc}(P)$ is the same $m_S$.

The invariant submodule and coinvariant module satisfyThus $\dim P^G$ is the multiplicity of the trivial module in $\mathrm{Soc}(P)$, while $\dim P_G=\dim {\mathrm{Hom}}_{kG}(P,k)$ is its multiplicity in the head. Applying the same argument to the projective module $P^{\ast }$ yields

The dual $P_S^{\ast }$ is indecomposable projective. Its head is dual to $\mathrm{Soc}(P_S)\cong S$, hence is $S^{\ast }$. Uniqueness of projective covers proves the dual of a projective cover over a group algebra: