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In one of your hang you are already assuming that $P_{ij}|\psi\rangle = |\psi\rangle$, which not true. One way to see that the permutation operator has possible eigenvalues $\pm 1$ is by using that exchanging two identical particles and then exchanging them back should give back the original state, $$P_{ij}P_{ij}|\psi\rangle = P_{ij}\lambda|\psi\rangle = \...

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