There is a section in my lecturer notes for Group Theory that I am finding difficult to understand. We let the row vector $\hat{\phi}=(\phi_1, \phi_2,...,\phi_l)$. For a normal $l$-fold degeneracy, we can write $\hat{S_a}\hat{\phi}=\hat{\phi}\Gamma (\hat{S_a})$, where $\hat{S_a}$ is the similarity transformation of the Hamiltonian and $\Gamma (\hat{S_a})$ is an $l$-dimensional matrix representation of the Group of the Hamiltonian.
The goal now, and the part I do not understand, is to show that this representation in terms of $\Gamma (\hat{S_a})$ matrices is irreducible. The notes start by replacing the row vector $\hat{\phi}$ by a linear combination $\hat{\phi}U$, where $U$ is an $l$-dimensional square matrix. Then we operate on $\hat{\psi}$ with $\hat{S_a}$:
$$\hat{S_a}\hat{\psi}=...=\hat{\psi}U^{-1}\Gamma{(\hat{S_a})}U$$
where I have skipped some of the intermediate steps. Clearly, the representation based on $\hat{\psi}$ and $\hat{\phi}$ are equivalent, since they are related by a similarity transformation.
Suppose now that the representation based on $\hat{\psi}$ is reducible. Then there would be a unitary transformation of the $\phi_j$ such that there are two or more subsets of the $\psi_j$ that transform only among one another under the symmetry operations of the Hamiltonian. This implies that $\hat{S_a}$ applied to any eigenfunction generates eigenfunctions only in the same subset. The degeneracy with the eigenfunctions in the other subset is therefore accidental, in contrast to our assertion that the degeneracy is normal. Hence the representation obtained for a normal degeneracy is irreducible.
The sentence in bold is what I do not understand. I did not know that reducible representations had this property. Please help me understand this step.