I'm reading Group Theory in Physics by Wu-Ki Tung and on page 69 in the proof of Theorem 5.3 he makes a group theory statement that I don't get. Let me try give some notation and explanation (coincidentally, I found the precise page online: http://www.physics.utah.edu/~detar/phys7550/scanned_notes/wkt_symm.pdf).
In discussing (normal) Young tableaux, we consider the group of all horizontal permutations (that is, products of permutations of elements only in the same row), and vertical permutations. Defining the Young symmetriser as $c := \sum_{h,v}hv(-1)^v$ where $h$ is any horizontal permutation, and $v$ any vertical one.
The author then proves that the symmetriser is essentially idempotent, that is, $c^2=\mu c$ for some complex $\mu$. My question is w.r.t. the next part that claims $\mu\neq0$. He shows that indeed $c^2=\mu c$ by an unrelated argument, and then explains that determining the coefficient $\mu$ can be done by considering the proof of the unrelated argument above, which shows that $\mu$ is the coefficient of the identity permutation, $e$, in the expansion of $c^2 = \displaystyle\sum_{h,v,h',v'}hvh'v'(-1)^v(-1)^{v'}$ as an element of the group algebra.
Since the horizontal/vertical permutations form subgroups they can be the identity permutations, so that contributes 1 to the coefficient in the expansion. Then he says (and this is the argument I don't see), that if $e$ appears again in the sum, it must have a positive coefficient, therefore we can only be adding to $\mu$ and so it must be some positive integer.
Why must the coefficient be positive? It seems he's saying that $hvh'v'=e \implies (-1)^v(-1)^{v'}=1$, but I don't see what guarantees that $v$ and $v'$ be of the same parity. If $v=v'$ the contribution is positive, or if one of the vertical permutations commutes with one of the horizontal ones then I see, but in general I don't see why this is true.
P.S. I know there are other proofs. I am familiar with the trace argument that is stronger because it gives the actual value of $\mu$ but I'm interested in his claim. Thanks.
If $v_\lambda$ and $v_\lambda'$ have different signs, how are you going to have $h_\lambda v_\lambda h_\lambda' v_\lambda' = e$? Something's going to end up in the wrong row, and you can't fix that by playing with $h_\lambda$ and $h_\lambda'$ since they don't change what row stuff is in.