Let $\rho : G \rightarrow GL_n(\mathbb{C})$ be a representation of $G$.
If $G$ is finitely generated as a group, does that mean that $im(\rho) \leq GL_n(\mathbb{C}) $ is finitely generated? Because, if $g$ is a generator for $G$, this does not necessarily imply that $\rho(g)$ is a generator for $im(\rho)$ does it?
For example, does $G = \langle x_1, ..., x_n \rangle$ imply that $im(\rho) = \langle \rho(x_1), ... , \rho(x_n) \rangle $?
Proof. "$\supseteq$" since $f(S)\subseteq f(G)$ then by definition of the generated subgroup we get $\langle f(S)\rangle\subseteq f(G)$.
"$\subseteq$" Assume that $H'$ is a subgroup of $H$ such that $f(S)\subseteq H'$. It follows that $S\subseteq f^{-1}(H')$ and since $f^{-1}(H')$ is a subgroup of $G$ and $G=\langle S\rangle$ then $f^{-1}(H')=G$. Therefore $f(G)= ff^{-1}(H')\subseteq H'$. So we've proved that every subgroup containing $f(S)$ has to contain $f(G)$. In other words $f(G)\subseteq \langle f(S)\rangle$. $\Box$
Note that this is pure group theory. No notion of matrices, complex numbers or representations is need.
Corollary. If $G$ is finitely generated then so is $f(G)$.
Proof. Simply because an image of a finite set via any function is finite. Apply previous lemma. $\Box$
Note however that the image of an infinitely generated group need not be infinitely generated as trivial homomorphism shows.
Yes, although note that $\rho$ need not be injective. I.e. it may happen that $G=\langle x, y\rangle$ and $im(\rho)=\langle\rho(x)\rangle$ if $\rho(x)=\rho(y)$.