Group structure of QCD‘s chiral symmetry (breaking) With $3$ flavors of massless quarks, the QCD Lagrangian is invariant under flavor transformations$$SU(3)_V\ \otimes\ SU(3)_A\ \otimes\ U(1)_V\ \otimes\ U(1)_A.$$
Now, this is equivalent to $$SU(3)_R\ \otimes\ SU(3)_L\ \otimes\ U(1)_V\ \otimes\ U(1)_A.$$
Wikipedia states that it is not the $SU(3)_A$ that is spontaneously broken, but the coset space $$\big(SU(3)_R\ \otimes\ SU(3)_L\big)\ \big/\ SU(3)_V,$$which is the non-diagonal part of $SU(3)_R \otimes SU(3)_L$. 

  
*
  
*Why is correct to say that $(SU(3)_R\otimes SU(3)_L)\ /\ SU(3)_V$ is spontaneously broken, instead of $SU(3)_A$?
  
*Why is $(SU(3)_R\otimes SU(3)_L)\ /\ SU(3)_V$ a coset space and not a group?
  
*What’s the meaning of „non diagonal“?
  

 A: You are strictly asking about names used--except I cannot comment on $SU(3)_A$ because there is no such group. I'll skip U(1) because it commutes with all else and is irrelevant for our purposes.
Your QCD text probably describes how the space integrals of  the 3-flavor current algebra 0th components yield 16 conserved charges obeying the Lie algebra of $SU(3)_L\times SU(3)_R$,
$$ [L^a ,L^b ]=if^{ab}_c L^c ,\quad
[L^a ,R^b ]=0,\quad
[R^a ,R^b ]=if^{ab}_c R^c  ~.
$$
So the L and R generators don't know about each other. 
We may make a change of language, by rearranging them,
$$2L^a \equiv V^a -A^a , \qquad       2 R^a \equiv V^a +A^a ,$$
to get the same algebra in a different basis,
$$ [V^a,V^b]=if^{ab}_cV^c \qquad 
[V^a,A^b]=if^{ab}_cA^c ,\qquad
[A^a,A^b]=if^{ab}_cV^c.$$
The V generators close into a subalgebra of the original one, called $SU(3)_V$, generating the diagonal subgroup, a mathematical term mapping all elements g of SU(3) to doublet elements (g,g) of the original $SU(3)\times SU(3)$. Think of it as synchronized swimming, where the angle of each L rotation is identical to the angle of a simultaneous R one.   
Crucially, as evident in the Lie algebra, the 8 generators A do not close into a subalgebra of our original one.  So there is no  group generated by them and calling them collectively "$SU(3)_A$" asks for trouble--your very question. Instead, we call them the "nondiagonal" generators (they are not in the diagonal subgroup above), and they comprise the cosets of the original full algebra/group, denoted by this division symbol employed.  
When you study the action of the exponentials of the Vs and As on generic group elements, you note how truly different they are.
In hadronic physics, $SU(3)_V$ is unbroken by gluonic interactions, but the 8 As, generating the elements of the  coset space $(SU(3)_R\otimes SU(3)_L)\ /\ SU(3)_V$, are SSBroken, and are realized "nonlinearly" in the Nabu-Goldstone mode, a very distinctive and physically significant realization of fields subject to Goldstone's fundamental theorem.
