I have seen the proof for why the identity permutation $\epsilon$ is even...but from a superficial level, I find the methods used in the proof to be sort of contradictory. Consider the following 3 claims, which I believe are all true:
Permutations can be defined as even or odd on the basis of the number of transpositions required to build them.
A transposition is a type of cycle.
The identity permutation $\epsilon$ is not a cycle.
Using these three points, it strikes me as contradictory to associate a parity with the identity permutation $\epsilon$.
The claims that trip me up are the combination of 2. and 3. Specifically, if we define a transposition as a type of cycle, but then declare that $\epsilon$ is not a cycle, then I would think that it is illogical to construct $\epsilon$ using the composition of some arbitrary number of transpositions given that we know $\epsilon$, itself, is not a cycle.
Stated differently, *I would think that because the identity permutation is not a cycle that it would be illogical to construct it from cycles.*
The only thing I can think of that is sort of analogous to this idea is the following:
The number $1$ is not negative, but I can construct it from negative numbers (e.g. $-1 * -1$)
Could someone please help restate these three claims (or, perhaps, redirect my logic-assessment) so that this contradiction disappear?
The map $\epsilon : S_n \rightarrow \:${$\pm1$} is a homomorphism. Say $\sigma \in S_n$, so $\epsilon(1 \: \sigma) = \epsilon(1) \epsilon(\sigma) = \epsilon(\sigma)$ implies identity permutation must have even parity = 1.
As for whether identity is a transposition or not doesn't matter anymore because for the homomorphism above to work, it must have even parity.