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Let $\rho:G \to GL(V)$ a irreducible representation where $|G|=p^3$ and $\dim(V)\neq 1$ over $\mathbb{C}$, then $\rho$ is injective.

I managed to reach the following relationship

$$|G|=|\ker\rho|\dim(V)^2+\sum_{g\notin\ker\rho}|\chi(g)|^2$$

where $\chi$ is the character of $\rho$.

I think this can help to get that the kernel is trivial, but I couldn't get anywhere. I am also wondering about the importance of the order of the group being $p^3$.

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If $\rho$ not injective, then it induces an irreducible representation of $G/{\rm ker}(\rho)$. However this group must have order $1$ or $p$ or $p^2$, so will be abelian and all irreducible representations will have dimension $1$.

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If $|\ker \rho| > 1$ and $\dim(V) > 1$, then both of these numbers must be at least $p$, because they must be divisors of $|G|$. Use this to get a contradiction from the formula you have shown.

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    $\begingroup$ I thought about it, but if $|\ker\rho|=p$ and $\dim(V)=p$, how to ensure that the sum is different from zero? $\endgroup$
    – Mrcrg
    Jul 26 '20 at 6:13
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    $\begingroup$ @Mrcrg Take the inner product of such a character with the trivial character. Or alternatively, what is the definition of the regular character? $\endgroup$ Jul 26 '20 at 14:22

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