In Wikipedia's article on the Brauer-Nesbitt theorem, they state that given a group $G$ and a field $E$, two semisimple representations $\rho_1,\rho_2 : G\longrightarrow \operatorname{GL}_n(E)$ are isomorphic if and only if $\rho_1(g),\rho_2(g)$ have the same characteristic polynomials for all $g\in G$.

The references they provide however only seem to apply to finite groups. Does anyone have a reference for the version stated above?

In characteristic 0, this appears in Lang's Algebra as Corollary 3.8 (p650)

In the general case, this is treated in Theorem 5.7 of Eggermont's masters thesis, but it feels sketchy to cite an unpublished masters thesis.

(EDIT: I forgot to include the stipulation that $\rho_1,\rho_2$ are semisimple)


1 Answer 1


What you have asked for, is false. If $G$ is a unipotent group such that $\rho _1$ and $\rho _2$ both have unipotent images and have the same dimension $n$, then the characteristic polynomials are $(X-1)^n$ for both (for any element of the group $G$).

The Wikipedia article to which you have sent a link, states this only for semi-simple representations.

If the characteristic of the field is zero ( or is sufficiently large), then this condition on the char polynomials is equivalent to the traces of the two representations being the same, and that is a well known result. One can deduce it from Weddernburn's theorem on semi-simple algebras.

  • $\begingroup$ Of course, I apologize I forgot to include semisimple. I've also noted that the characteristic 0 version is in Lang's Algebra. I'm still looking for a good reference for the positive characteristic case. $\endgroup$ Oct 27, 2021 at 17:58

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