Let $(V,\rho)$ be a finite, irreducible representation of a finite group $G$, where $V$ is a finite-dimensional vector space. So:
$$ \rho: G \to \textrm{GL}(V) $$
is a homomorphism.
Let $Z(V)$ denote the centre of $\textrm{GL}(V)$, and define the projective general linear group as the quotient:
$$ \textrm{PGL}(V) = \textrm{GL}(V)/Z(V) $$
and the corresponding canonical projection:
$$ \pi: \textrm{GL}(V) \to \textrm{PGL}(V). $$
I want to show that, if $Z(\rho(G))$ is the centre of $\rho(G) < \textrm{GL}(V)$, then:
$$ \pi(\rho(G)) \cong \rho(G)/Z(\rho(G)). $$
I have been told that this follows immediately from Schur's lemma, but I do not see how.
I suppose that by arguing by induction on the order of $G$, we could assume that $\rho$ is faithful, so that we have $\rho(G)/Z(\rho(G)) \cong G/Z(G)$.
As the centre of the group is Abelian, it would follow from Schur's lemma that the restriction of $\rho$ to $Z(G)$ is the sum of irreducible degree 1 characters, but I do not see how this applies to the above.
All help would be highly appreciated.
I'll make a try. From first isom. theorem we have $$\pi(\rho(G))\cong \rho(G)/ker\pi|_{\rho(G)}$$
It suffies to show that $ker\pi|_{\rho(G)}=Z(\rho(G))$. It is $$ker\pi|_{\rho(G)}=\{\rho(g)\in \rho(G)|\ \rho(g)\in Z(V)\}=\\ =\{\rho(g)\in \rho(G)|\ \rho(g)=k\cdot Id\}=\\=Z(\rho(G))$$
where the second equality folllows from Schur's Lemma.