I keep seeing this following fact stated;
If $G$ is a simple group, then the only $1$ dimensional representation of $G$ is the trivial representation.
But I’ve not seen a proof and I can’t seem to find one online.
Could anyone please show me why this is true or provide me a reference to a proof?
Thank you!
A 1-dimensional representation of $G$ is a morphism from $G$ to the abelian group $\mathbb C^*$; hence, it must factor through the quotient $G/[G;G]$ by the derived subgroup. Because $G$ is simple, the derived subgroup $[G,G]$ is either $G$ (so the morphism is trivial) or $(0)$ (so the group is simple abelian, and $G=\mathbb Z/p\mathbb Z$).
As Lord Shark the Unknown stated in the comment, this is all one can ask for, since $\mathbb Z/p\mathbb Z$ admits non trivial 1-dimensional representations.