Injectiveness of unfaithful representation

Question

I took symmetry course and was wondering about unfaithful group represntations.

I know that representation is unfaithful if group homomorphism is not injective. I also know that if kernel is nontrivial more than one element is mapped to identity. That break injectiveness so representation is unfaithful. If I can find elements that map to same element hence break injectiveness that are not in kernel that means I found an unfaithful representation with trivial kernel. That is impossible according to Wikipedia.

Why is that so?

I also read that elements not in the kernel can map to the same element if they are in the same coset. Is that true? Why?

Answer 1

why is that so?

If $\phi(g)=\phi(h)$, then $\phi(gh^{-1})=I$. If you assumed $g\neq h$, then $gh^{-1}\neq e$ is a second element mapping to the identity.

I also read that elements not in the kernel can map to the same element if they are in the same coset. Is that true?

Trivially, yes. $g\ker\phi=h\ker\phi\iff gh^{-1}\in \ker\phi\iff \phi(g)=\phi(h)$.

Answer 2

Suppose $\rho:G\to H$ is the representation of $G$ (with $H$ being some group, for instance $\operatorname{Aut} V$ if you are talking about linear representations). If $g,g'\in G$ are such that $\rho(g)=\rho(g')$, then $1=\rho(g^{-1}g')$, so $g^{-1}g'\in\ker \rho$, and thus $g'\in g\ker \rho$. This is saying that elements which map to the same thing are in the same coset of the kernel. If $g\neq g'$, this implies the kernel is nontrivial. It might be worth reviewing the isomorphism theorems for groups.