Let $G$ be a finite group, and let $k$ be a field whose characteristic divides $\left|G\right|$. Let $\rho:G\to \mathrm{End} V$ be a (finite-dimensional) representation of $G$ over $k$. Prove or disprove that


$\sum\limits\_{g\in G} \frac{1}{\det\left(\mathrm{id}-T\rho\left(g\right)\right)} = 0$ as an equality between power series in $k\left[T\right]$.




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**Motivation:**


Let $G$ be a finite group, and $k$ be any field. Let $\rho:G\to \mathrm{End} V$ be a (finite-dimensional) representation of $G$ over $k$. For every $d\geq 0$, let $n\_d$ denote the dimension of the space $\left(\mathrm{Sym}^d V^{\ast}\right)^G$ of the $G$-invariant symmetric $d$-ary forms on $V$. [Molien's formula](http://en.wikipedia.org/wiki/Molien_series) states that


$\left|G\right|\cdot \sum\limits\_{n=0}^{\infty} n\_dT^d = \sum\limits\_{g\in G} \frac{1}{\det\left(\mathrm{id}-T\rho\left(g\right)\right)}$ as an equality between power series in $k\left[T\right]$


if $\mathrm{char} k$ does not divide $\left|G\right|$. I am wondering whether this formula still holds if $\mathrm{char} k\mid \left|G\right|$. The standard proof, using the Maschke projection, does not make much sense in this case...



Molien for modular representations?
A simple-minded argument:


Pick any element $g\in G$ of order $p$, the characteristic of $k$, and let $C(g)$ be the centralizer of $g$.


Now $G\setminus C(g)$ is the disjoint union of orbits under the action of the inner automorphism $\iota\_g:h\in G\mapsto ghg^{-1}\in G$, and those orbits are of size $p$. The sum of the terms in your sum corresponding to the elements of $g$ in one of those orbits is then zero, for those terms are all equal.


If $h\in C(g)$, then the terms in your sum corresponding to the elements $h$, $gh$, $g^2h$, $\dots$, $g^{p-1}h$ are all equal---because $g$ and $h$ commute, you can take them simutaneously to Jordan canonical form and $g$ has only $1$ as an eigenvalue---so that their sum is also zero.


We have thus partitioned $G$ into, on one hand, the orbits of $\iota\_g$ in $G\setminus C(g)$, and, on the other, the cosets of $\langle g\rangle$ in $C(g)$, and checked that the sum of the terms in each part of this partition is zero. Therefore your sum is zero.


**NB:** Notice that the specific form of the terms in your sum does not really matter, as long as it only depend on the eigenvalues of the $g\in G$. Thus, for example, exactly the same reasoning shows that the "other" Molien formula $$\sum\_{g\in G}\det(I-t\rho(g))$$ also vanishes.


