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Let $V$ a $\mathbb KG$-module such that the character of $V$ is such that $\chi_V(g)=0$ for all $g\in G\setminus \{1\}$. Show that there is an $m$ s.t. $$V=\underbrace{\mathbb KG\oplus\cdots\oplus \mathbb KG}_{m\text{ times}}.$$

To me, $\chi_V$ is the regular representation. I know that $$\mathbb KG=U_1\oplus\cdots\oplus U_n$$ where $U_i$ are irreducible submodule of $\mathbb KG$. So, $$\underbrace{\mathbb KG\oplus\cdots\oplus \mathbb KG}_{m\text{ times}}=U_1^{\oplus m}\oplus \cdots \oplus U_n^{\oplus m},$$ but how can I prove that it's $V$ ?

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  • $\begingroup$ Are you familiar with the usual inner product on the characters? $\endgroup$ – Tobias Kildetoft Jun 20 '16 at 19:41
  • $\begingroup$ @TobiasKildetoft That approach only works if $\mathbb{K}$ is of characteristic zero. The OP does not say anything about that (although I highly doubt that the statement will hold in positive characteristics). $\endgroup$ – Batominovski Jun 20 '16 at 21:28
  • $\begingroup$ @Batominovski Indeed, even making sense of it becomes less trivial (what should be meant by character for example). $\endgroup$ – Tobias Kildetoft Jun 21 '16 at 4:59
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Edit As @Batominovski and @TobiasKildetoft remarked, one should assume characteristic zero here. In positive characteristic $p$ and non-trivial $G$, the statement is indeed not true, at least if one takes the naive definition of characters: For example, the $p$-fold sum of the trivial representation satisfies the assumption but is not of the form $KG^{\oplus m}$ since the latter is not trivial.

For any irreducible character $\chi_U$ you have ${\mathbb Z}\ni \langle\chi_V,\chi_U\rangle = \frac{\chi_V(1)\chi_U(1)}{|G|}=\frac{\dim(V)}{|G|}\dim(U)$. In particular, taking $U$ to be the trivial representation shows $\frac{\dim(V)}{|G|}=:m\in {\mathbb N}$, and so any irreducible representation $U$ of $G$ occurs exactly $m \dim(U)$ times in $V$. Since $KG=\bigoplus_U U^{\oplus\dim U}$, it follows that $V\cong KG^{\oplus m}$.

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  • $\begingroup$ In positive characteristic, one would need to specify what is even meant by a character (just taking traces loses too much information to be something one usually does). $\endgroup$ – Tobias Kildetoft Jun 21 '16 at 7:41
  • $\begingroup$ @TobiasKildetoft Thanks, I made a comment. $\endgroup$ – Hanno Jun 21 '16 at 10:37

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