quantum mechanics - Collective angular momentum , Dicke states and indistinguishable particles

During course of quantum mechanics we dealt with addition of angular momenta. If we have two particles with spin $j_1$ and $j_2$ we can introduce total spin operator:

$$\mathbf{J} = \mathbf{j}^{(1)} + \mathbf{j}^{(2)},$$ and new basis: $$\mathbf{J}^{2}|J,M\rangle = J(J+1)|J,M\rangle,$$ $$\mathbf{J}_{z}|J,M\rangle = M|J,M\rangle,$$ where $J = |j_1 - j_2|, \ldots, j_1 + j_2$ and $M = -J, \ldots, J$.

The same thing can be done for $N$ particles with e.q. spin $1/2$. One can introduce collective spin operator:

$$\mathbf{J} = \sum\limits_{i=1}^{N}\mathbf{j}^{(i)},$$ and new basis in the Hilbert space $\mathcal{H} = \mathcal{H}_{1} \otimes \ldots \otimes \mathcal{H}_{N}$, $$\mathbf{J}^{2}|J,M\rangle = J(J+1)|J,M\rangle,$$ $$\mathbf{J}_{z}|J,M\rangle = M|J,M\rangle.$$ States with highest momentum $J = N/2$ are known as the Dicke states: $$|N/2, M\rangle, \ \ M = -N/2, \ldots, N/2$$ I always thought that these states are not degenerate with respect to a quantum number $M$, but here they claim that only symmetric Dicke states are uniquely defined. I can't see this cause I construct states with $M$ less than $N/2$ by applying the lowering operator to $|N/2, N/2\rangle$ state (this seems to give me only symmetric Dicke states).

Another issue is the Hilbert space itself. I think it does not take into account that particles are indistinguishable. We simply deal with tensor product of single particle spaces. No symmetrization or antisymmetrization. Am I right?