Showing 1-dim representations factor through $G/G'$ I have a question that is as follows:
Show that the 1-dim complex representations of $G$ are those that factor through $G/G'$.
Now I am a bit confused by this question, what exactly does it mean factor through?
 A: So, elaborating on bartogian's comment, a representation of $G$ on a vector space $V$ is a group homomorphism $\rho$ from $G$ to $GL(V)$. Then, we say that the representation factors through $G/N$ for some normal subgroup $N$ if there exists a group homomorphism $\rho' : G/N \rightarrow GL(V)$ such that $\rho = \rho' \circ \pi$ where $\pi$ is the canonical projection.
But by the universal property of quotients, $\rho$ factors through $G/N$ if and only if $N$ is contained in the kernel of $\rho$. So, in your problem, we need to show that if $\rho$ is a homomorphism from $G$ to $GL(1)$, then $[G, G]$ is in the kernel of $\rho$. But this follows from the fact that $G/\ker(\rho)$ is abelian because $GL(1)$ is abelian.
A: $\newcommand{\GL}{\operatorname{GL}}$I assume that $G' = \langle ghg^{-1}h^{-1} \mid g,h\in G\rangle$ is the derived subgroup. That is, $G$ is the group generated by all the commutators.
Let $\pi: G \to \GL(\mathbb{C}) = \mathbb{C}^\times$ be a one-dimensional representation. Then note that for $ghg^{-1}h^{-1} \in G'$ you have
$$
\pi(ghg^{-1}h^{-1}) = 1.
$$
That means that $\pi$ is trivial on $G'$ and it factors through $G / G'$, $\pi: G / G' \to \mathbb{C}^\times$ via:
$$
\pi(gG') = \pi(g).
$$
This is well defined exactly because $\pi$ is trivial on $G'$.
