Symmetric and antisymmetric states of spin singlet/triplet I'm studying the book "Quarks and Leptons: An Introductory Course In Modern Particle Physics". When I reach chapter 2 (page 33), I encounter this exercise problem that I couldn't understand:

EXERCISE 2.1 Justify the decomposition shown in (2.1) by either (1)
  considering the symmetry of the states under interchange of the nucleons or
  (2) using the angular momentum "step-down" operator.


Here's the solution in the book:

My questions are: Why do these "justify" the decomposition in (2.1)? Why the state $|S=1,M_S=0>$ is said to be "also" symmetric? What does it mean that the state $|S=0,M_S=0>$ be antisymmetric due to orthogonality? I'm really confused...
I'm really new to the subject, so it'd be great if someone could help me. Thank you!
 A: If you interchange the spins, then the states are either symmetric or antisymmetric under permutation: this is more explicit if you write
$$
\sqrt{2}\vert S=1,M=0\rangle=\vert\uparrow\rangle_1\vert \downarrow\rangle_2 + \vert\downarrow\rangle_1\vert\uparrow\rangle_2\, ,
\tag{1}
$$
where $\vert \uparrow\rangle_1$ denotes particle 1 in the spin-up state etc.  
Clearly (1) is symmetric under permutation of the particle indices, i.e. it comes back to $+1$ times itself under permutation.  On the other hand
$$
\sqrt{2}\vert S=0,M=0\rangle=\vert\uparrow\rangle_1\vert \downarrow\rangle_2 - \vert\downarrow\rangle_1\vert\uparrow\rangle_2\, ,
\tag{2}
$$
is clearly antisymmetric under permutation of particles indices, i.e. this states comes back to $-1$ times itself under permutation.
Finally, note that all $S=1$ states are symmetric in the sense above irrespective of the value of $M$.
Since $L_x=L_x^{(1)}+L_x^{(2)}$, i.e. the total $L_x$ is the sum of the projection for particles $1$ and $2$, with $L_x^{(1)}$ action on states for particle 1 etc, is unchanged under permutation of particles, the operator $L_\pm = L_\pm^{(1)}+L_\pm^{(1)}$ will not change the permutation symmetry of the state $\vert S,M\rangle$ when stepping up or down since, if $\vert S,M\rangle$ has a definite permutation symmetry, 
$$
L_\pm \vert S,M\rangle
$$
will have the same permutation symmetry as $\vert S,M\rangle$.
