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 $p^3$.
Best Answer
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 $1$.