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The book that I'm reading states that the identity permutation is an even permutation. But it gives no example, and at this point, I'm confused. So, for example, if we have the identity permutation $\varepsilon=(1)(2)(3)(4)(5)$, how do we write its product of transpositions? I tried $(12)(34)(52)$, and so on, which is obviously incorrect. Any insight and/or example would great!

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    $\begingroup$ a) empty product. b) $(12)(12)$. $\endgroup$ – Daniel Fischer Oct 10 '15 at 19:48
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A permutation is even iff it is a product of an even number of transpositions. The identity permutation can be represented as a product of zero transpositions - and zero is certainly even.

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    $\begingroup$ My book never made this clear, but is the zero transpositions the only way to write the identity permutation or is there any other way? $\endgroup$ – user65422 Oct 12 '15 at 4:03
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    $\begingroup$ There are lots of ways. For example, any transposition followed by itself gives the identity permutation. $\endgroup$ – Tad Oct 12 '15 at 4:05
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I suggest you read some book on group theory to understand permutations and their different representations. Pages 2-9 of Rotman's Intro to the theory of groups is a great place to look, specially the sections on cycles and factorization into Disjoint Cycles.

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Suppose an identity permutation as $(1)(2)(3)$ , then it can be written as $(23)\times(32)$ by remaining the position of $1$ unaltered. Or in similar ways it also can be written as $(12)\times(21)$ or as $(13)\times(31)$.

But this contradicts the fact that "every permutation on a finite set is either a cycle or it can be expressed as the product of disjoint cycles".

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