Find bijection between sets of representations (of groups)

Question

I want to show the following:

Let $G$ be group and $N$ a normal subgroup of G.

Show that a bijection exists between two sets of representations, namely:

$$S_1= \{ \rho:G \rightarrow GL(\mathbb{C}^n ); \rho \text{ representation with property: } N \trianglelefteq G \ker \rho \}$$

$$\text{and}$$

$$S_2=\{ \psi:G/N \rightarrow GL(\mathbb{C}^n); \psi \text{ representation } \}$$

My thinking:

I know that if $N$ is a normal subgroup of $G$, then the map $\pi:G \rightarrow G/N$, defined by $\pi(g)=gN$ is a homomorphism (Source: Rotman)

The best way to prove that a bijection exists is to find a bijection. And thats what I was trying to do. My candidate for a bijection: $$\eta:\rho \rightarrow \rho \space\circ\space\pi, \space \rho(.)\rightarrow\rho(\pi(.)) \space .$$

Since $\pi$ is homomorphism, then clearly $\rho(\pi(.))$ will be a representation of $G/N$, because for $g_1,g_2 \in G$, $\rho(\pi(g_1 g_2))=\rho(\pi(g_1)\pi(g_2))=\rho(\pi(g_1))\rho(\pi(g_2))$, the last step follows from the fact that $\rho$ is also a representation. So again $\rho(\pi(.))$ forms a representation of $G/N$ and thus clearly $\eta:S_1 \rightarrow S_2$ . Now I just need to show that $\eta$ is a bijection, and this is where I start to run into trouble. ( I know that $\rho$ cannot be injective if $N \neq \langle e \rangle$ and that $\pi$, on the other hand, is a bijection.)

My question:

Could some please elaborate on my thinking. Am I correct or totally wrong? Could someone provide a different bijection or type of proof. Where does the property $N \trianglelefteq G \ker \rho$ come into play?

Answer 1

The map $\eta:S_2 \to S_1$ defined by $\eta(\rho) = \rho \circ \pi$ is indeed a bijection of sets. To show that this holds, it suffices to use the following fact:

Let $\phi:G \to G'$ be a homomorphism. If $H \subseteq \ker (\phi)$, there is a unique homomorphism $\tilde{\phi}:G/H \to G'$ such that $\phi=\tilde{\phi} \circ \pi$.

A proof of this fact is given in these notes for instance.