Separating the action of Hamiltonian and Symmetry (Block diagonalize Hamiltonian) I heard this statement in one lecture. 

Consider a first quantized Hamiltonian $H$ on a single-particle Hilbert space (of finite dimension $N$). 
  If the Hamiltonian possess symmetries that is unitarily represented, then 
  we can bring the Hamiltonian into a block diagonal form with each block $H^{(\lambda)}$ labelled by the irreducible (unitary) representation $\lambda$ of its symmetry group $G_0$. These irreducible blocks do not exhibit the unitary symmetries.

This seems to be an elementary fact that most papers do not give reference for it. Can anyone point out a proof of it?

Note
The lecturer also gave the following exact statement of his claim:
Suppose we have a Hamiltonian $H$ on single-particle Hilbert space (of
finite dimension $N$). Assume its group of symmetry is $G_0$.  Then
the space $\mathcal{V}$ of single-particle Hilbert space, decomposes
into a direct sum of vector spaces $\mathcal{V}_\lambda$ associated
with the irrep (irreducible representations, labeled by $\lambda$) of
$G_0$.
\begin{equation}
    \mathcal{V} = \oplus_\lambda \mathcal{V}_\lambda
\end{equation}
Let $m_\lambda$ denotes the multiplicity of $\lambda$th irrep.
Denote the dimension of each irrep as $d_\lambda$.
In each vector space $\mathcal{V}_\lambda$, one can choose a
(orthogonal) basis of the form:
\begin{equation}
    |v^{(\lambda)}_\alpha\rangle \otimes |w^{(\lambda)}_k\rangle
\end{equation}
where


*

*$G_0$ acts only only $|w^{(\lambda)}_k\rangle$,
    $k=1,\cdots,d_\lambda$,

*$H$ acts only on $|v^{(\lambda)}_\alpha\rangle$,
    $\alpha=1,\cdots,m_\lambda$.



If you have difficulty understanding @ACuriousMind's answer, please read my comments in that answer for a concrete example.
 A: This seems to be a strange formulation of the fact that $H$ and $G_0$ commute, so there are "joint eigenstates", in particular, each of the $V_\lambda^{(i)}$ (I'm labelling the irreducible representations by $i = 1,\dots,m_\lambda$ here) can be chosen to be an eigenspace of $H$ with energy $E^{(i)}_\lambda$. So, we pick some abstract vector $\lvert E_\lambda^{(i)}\rangle$ and a basis $\lvert v_{\lambda,j}\rangle,j = 1,\dots,d_\lambda$ of $V_\lambda$, and there's an isomorphism from the vector space spanned by $\lvert E_\lambda^{(i)}\rangle\otimes\lvert v_{\lambda,j}\rangle$ for $j = 1,\dots,d_\lambda$ to $V_\lambda^{(i)}$.
Using the tensor product is a bit of a weird notational choice for this - you can do it as there's the isomorphism I've indicated, but usually you'd just pick a basis of the $V^{(i)}_\lambda$ that are eigenvectors of some generators (those in the Cartan subalgebra if we have a Lie group) of $G_0$, and call the resulting basis $\lvert E_\lambda^{(i)},v_{\lambda,j}\rangle$ for whatever eigenvalues $v_{\lambda,j}$ occur in the $V_\lambda$ representation.
