What does U(N)xU(N)/U(1) mean? I am studying a model in which SSB occurs and the original symmetry group is U(N)xU(N)/U(1) it acts as:
 $M'=AMB^{-1}$ (M is an NxN matrix containing the fields)
The groud state i have found is invariant under this transformation only if  $A=B$ but i am not sure what subgroup this is. My aim here is to find out the dimensions of the original group and of the invariant subgroup (and also what group this is) so i can find out how many goldstone bosons there are.
My questions are:


*

*What does factorizing by U(1) mean?

*Are the dimensions of the original group $2N^2-1$?

*Is the subgroup with $A=B$ simply U(N) acting as $M'=AMA^{-1}$, and if this is correct what do i do with the factor $U(1)$? Is the final invariant subgroup $U(N)/U(1)$ or simpy $U(N)$?


My attempt at a solution is:  $dim(U(N)xU(N)/U(1))=2N^{2}-1$ and $dim(U(N)/U(1))=N^{2}-1$ so i get $N_{goldstone}=2N^{2}-1-(N^{2}-1)=N^{2}$ but i have a terrible feeling this is wrong
How SSB works is quite clear to me but i have a hard time with the group theory machinery since i am not familiar with this kind of group and what i normally do is let the generators act on the ground states and see which ones are broken, this i cannot do since i haven't even understood what kind of a group i am working with (I also already know that U(N) = SU(N) × U(1)/Z but i still don't know what to do with the U(1) factor!)
 A: The group $U(N)\times U(N)$ acts on your original space, but some group elements act in the same way: the action is not faithful, in other words, this groups maps onto the symmetry group (this is implicitly assumed by what you wrote), but not injectively. 
Specifically, the diagonal subgroup of scalar matrices (i.e. matrices that are scalar multiples of the identity, these are the only matrices that commute with all other matrices) acts trivially. Its elements are pairs $(A,A)\in U(N)\times U(N)$ where $A$ is a scalar matrix, hence an element of $U(1)$, and assuming that $M$ lives in a sufficiently large subspace of all $N\times N$ matrices, no other subgroup acts trivially. This is a subgroup that is trivially isomorphic to $U(1)$. 
More generally (though in fact equivalently), two elements $(A,B), (A',B')\in U(N)\times U(N)$ act in the same way (i.e. define the same symmetry) exactly when they differ by an element of this diagonal subgroup isomorphic to $U(1)$. This means that the full symmetry group is a quotient set of $U(N)\times U(N)$, whose elements are subsets consisting of all those elements of the original group that induce the same symmetry, i.e. that have the same effect on all possible $M$. In our case the elements are the sets of the form $(A,B)H$, called cosets where $H$ is this diagonal subgroup. 
When this subgroup is normal, as it is in this example, the group structure of the original group naturally descends to the quotient set, making it into a group again. 
When the original group is a Lie group and the subgroup is closed, as it is in this example, the differentiable structure of the original group naturally descends to the quotient set, making it into a Lie group again. 
Note that our group is not the only subgroup that is isomorphic to $U(1)$, so without context the notation $U(N)\times U(N)/U(1)$ strictly speaking is ambiguous. 
You assumption about the dimension of the full symmetry group is correct: when you factor out an $m$-dimensional subgroup of an $n$-dimensional group, the quotient is $(n - m)$-dimensional.
The diagonal subgroup $A = B$ is a subgroup of this quotient group. If you want to forget about all of the preceding, you could also reason again that the elements of this other group $H$, which is a subgroup of the full diagonal subgroup, acts trivially. Your dimension counts look correct to me.
A: The quotient $G/H$ is a system of cosets so that for $h~\in~H$ and $g$ a set of group elements in the $gh~\in~H$ for left cosets and $hg~\in~H$.  We then have $gH~=~Hg$ or that the subgroup $H$ is a normal subgroup.  So for $SU(n)/U(1)$ we may look at $SU(2)$ and $SU(3)$. The Pauli matrices  $\sigma_i,~i~=~1,2,3$ generate transformations of the form 
$$
g_i~=~exp\left(-\frac{i}{2}\sigma_i\theta_i\right).
$$
and we may now consider the element $h~=~exp(i\tau\phi)$ that commutes with this. Consider small angles so that we have $g_i~=~1~+~\frac{i}{2}\sigma_i\theta_i$ and $h~=~1~+~i\tau\phi$ and the commutator leads to a commutator of the structure elements $[\sigma_i,~\tau]~=~0$ and we see that the generator of the $U(1)$ is one of the Pauli matrices, where the choice of $\sigma_3$ is the Cartan center that is the abelian center. The $\sigma_3$ elements define the eigenstates of a system, such as spin state of an electron isospin of a particle.
For $SU(3)$ we have the Cartan center matrices
$$
\lambda_3~=~\left(\begin{array}{ccc}1 & 0 &0 \\ 0 & -1 & 0\\ 0 & 0 & 0\end{array}\right),~
\lambda_8~=~\frac{1}{\sqrt{3}}\left(\begin{array}{ccc}1 & 0 &0 \\ 0 & 1 & 0\\ 0 & 0 & 2\end{array}\right)
$$
We similarly have for $SU(3)$ group elements $g_i~=~exp\left(\frac{i}{2}\lambda_i\theta\right)$ that for $i~=~3,~8$ there Cartan matrices are $U(1)$. The quotient is then a way of “commuting away” a set of eigenvalues. We might in some ways think of this as a sort of degeneracy that is imposed on the system.
For $G\times G'/H$ we have two group elements, even though $G~\simeq~G'$, that act on $H$ so that for $g_1~\in~G$ and $g_2~\in~G'$ then $g_1g_2H~=~Hg_1g_2$. We may think of this as
$$
g_1g_2Hg^\dagger_2g^\dagger_1~=~H,
$$
It is clear that in the middle that $g_1(g_2Hg^\dagger_2)g^\dagger_1~=$ $ g_1Hg^\dagger_1~=~h$, and so this means we have a coset situation with one abelian element of the Cartan center of $G\times G'$. this also means that $U(n)\times U(n)/U(1)$ there is one dimension removed as is the same with $U(n)/U(1)$
