normal degeneracy and the “span” of a symmetry group

Question

In Tinkham's "Group Theory and Quantum Mechanics", Tinkham defines normal degeneracy so that the span of the action of the Hamiltonian's symmetry group on any energy eigenstate yields all possible eigenstates with that same energy.

He then goes on to say that under the condition of normal degeneracy, the degenerate subspaces form irreducible representations of the symmetry group.

I'm having trouble connecting the dots. By the definition of the symmetry group it is clear that the span of the action of the symmetry group on some energy eigenstate $v$, i.e.:

$$W_v \equiv span(\{\rho(g)v : \forall g \in G \})$$
(where $\rho(g)$ is the operator associated with the group element $g$ of the symmetry group $G$) does contain only eigenstates of the same energy eigenvalue as $v$, and indeed does form a representation, since, for any $w \in W_v$ and $g_j \in G$ ($G$ being finite):
$$\rho(g_j) w = \rho(g_j) \sum_i w_i \rho(g_i)v = \sum_i w_i \rho(g_j) \rho(g_i) v = \sum_i w_i \rho(g_j g_i) v \in W_v$$
It is only left to show that this representation is irreducible. I'm stuck here.