Thm 5.19 (Brauer) in Martin Isaacs' - Character Theory of Finite Groups I would please like some help to understand the proof of thm 5.19 (Brauer) in Isaacs Martin - Character Theory of Finite Groups. It states:
Let $\chi$ be a character of G with $[\chi,1_G]=0$. Let $A,B \subset G$ and suppose:
$[\chi_A,1_A]+[\chi_B,1_B]>[\chi_{A\cap B},1_{A\cap B}] $. Then A and B generate a proper subgroup of G.
Proof: Let U be a $C[G]-$ module which affords $\chi$ and letV,W and Y be the subspaces of fixed points of U under A,B and $A\cap B$ respectively. Then, $V\subset Y$,$W\subset Y$, and $dimV+dimW=[\chi_A,1_A]+[\chi_B,1_B]>[\chi_{A\cap B},1_{A\cap B}]=dimY$. It follows that $V \cap W \neq 0$ and thus $<A,B>$ has non trivial fixed points on U. Since $[\chi,1_G]=0$, we conclude that $<A,B>=G$. 
Plenty of questions here:
1) Is V defined as the subspace of points of U fixed by ALL points of A ?
2)  Why do we have $dimV+dimW=[\chi_A,1_A]+[\chi_B,1_B]>[\chi_{A\cap B},1_{A\cap B}]=dimY$ ? I assume it comes for instance from $dimV=[\chi_A,1_A]$ but why is that?
3) Why does $[\chi,1_G]=0$ allows one to conclude that $<A,B>=G$ ?
 A: *

*Yes, $V$ is the set of elements of $U$ fixed by all points of $A$.

*In general, given a $\Bbb CG$-module $V$ and a point $v\in V$ fixed by all of $G$, $v$ generates a one-dimensional trivial submodule, with character $1_G$. The subspace of all such fixed points is a submodule of $V$ which decomposes as a direct sum of trivial one-dimensional submodules, so the dimension of this subspace is equal to the multiplicity of the trivial submodules of $V$, which is measured by $[\chi,1_G]$, where $\chi$ is the character of $V$. Applying this to the current situation, with $A$, $B$, and $A\cap B$ in place of $G$, it means that $\text{dim }V = [\chi_A, 1_A]$, $\text{dim }W=[\chi_B,1_B]$, and $\text{dim }Y=[\chi_{A\cap B},1_{A\cap B}]$.

*It appears that there is a typo in the last sentence of the proof: It should say $\langle A,B\rangle < G$, not $\langle A,B\rangle=G$. As explained above, $[\chi,1_G]=0$ means that the dimension of the subspace of points
of $U$ fixed by $G$ is 0. This means that there are no non-trivial fixed points of $G$ on $U$, but on the other hand, the non-trivial points in $V \cap W$ are fixed by both $A$ and $B$, and hence by $\langle A,B\rangle$, so if $\langle A,B\rangle=G$, then they would be fixed by all of $G$, a contradiction.

