# Irreducible representation is injective

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 $$p^3$$.

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 $$1$$.
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.
• I thought about it, but if $|\ker\rho|=p$ and $\dim(V)=p$, how to ensure that the sum is different from zero? Jul 26 '20 at 6:13