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Let $G$ be any group and $k$ be any field. While studying representation theory, I saw some theorems which are special cases of the following theorem:

$\textbf{Theorem. }$ Let $\rho_1,\rho_2:G\to \text{GL}_n(k)$ be two semisimple representations of $G$ over $k$. If characteristic polynomial of $\rho_1(g)$ and $\rho_2(g)$ are the same for all $g\in G$, then $\rho_1\simeq \rho_2$.

$\textbf{Question}$: Is this theorem true?

Since I couldn't find proof of this theorem with full generality, I tried to prove it myself. After a bit of work, I could prove the theorem if $G$ is finite or $k$ is perfect, and I posted it here. Now I wonder whether this theorem is true or false if $G$ is infinite and $k$ is imperfect. I suspect there is a counterexample since an important lemma in my proof (Theorem 4 of the link) is no longer true. However, I cannot find any counterexample yet.

Daebeom Choi
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    The theorem you state is false for every field $k$. Do you want to assume that $\rho_1$ and $\rho_2$ are semisimple? – spin May 08 '22 at 13:52
  • Oh, that is a BIG typo... thank you. I fixed it! – Daebeom Choi May 08 '22 at 13:54
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    See question here for short proof. Previously here, refers to Section 5.2, Theorem 5.10 in this MSc thesis. – spin May 08 '22 at 14:19
  • Thank you! In fact, I saw the second question before, but the link to the thesis in that question was broken, so I ignored it...

    I think the most important thing I missed is that for any semisimple $A$-module $M$, the image of $A$ in $\text{End}_k(M)$ is a semisimple algebra. This makes many things a lot simpler, in both approach. Anyway, thank you again!

     – Daebeom Choi May 08 '22 at 14:41

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Adding an answer so this is not unanswered.

The theorem is true, more generally for finite-dimensional semisimple representations of any $k$-algebra $A$. Short proof is given in an answer by David Speyer here.

The key observation is that you can reduce to the case where $A$ is a finite-dimensional semisimple $k$-algebra.

spin
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