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Let $S_n$ be a symmetric group. Define the sign of a permutation $\sigma$ is denoted $sgn(\sigma)$ and defined as $+1$ if $\sigma$ is even and $−1$ if $\sigma$ is odd. I try to show that any the sign of $k$-cycle is $(-1)^{k-1}$.

My proof is that let $(i_1 i_2\cdots i_k)$ be a $k$-cycle.

Note that $$ (i_1 i_2\cdots i_k)=(i_1 i_2)(i_2 i_3)\cdots (i_{k-1} i_k). $$

By Theorem 5.4, (see Alternative proof that the parity of permutation is well defined?)

If a permutation $\alpha$ can be expressed as a product of an even number of $2$-cycles, then every decomposition of $\alpha$ into a product of $2$-cycles must have an even number of $2$-cycles.

this decomposition of $(i_1\cdots i_k)$ is always even or odd.

Hence $$ sgn((i_1 i_2\cdots i_k))=(-1)^{k-1}. $$

Question: Does this proof work?

Hermi
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    It depends on your audience, but they can probably see why the cycle can be factored in that way. But how do you know the sign of a product of permutations equals the product of the signs? (And what is your definition of "even" and "odd" permutations?) – Karl Oct 05 '23 at 06:33
  • @Karl So how to modify the proof? The definition of even permutation is if it can be expressed as a product of even number of transpositions. – Hermi Oct 05 '23 at 14:58
  • @Karl There is a Proposition that the sign of a transposition is -1 – Hermi Oct 05 '23 at 14:59
  • Several issues: (i) the $a_i$ were never defined. Were they the $i_k$? (ii) Have you proven that any two decompositions of a cycle into a product of transpositions have number of factors with the same parity (if not, then your proof is incomplete)? (iii) The $k$-cycle is not equal to the product of the signs of the transpositions; perhaps you meant that it is equal to the product of the transpositions themselves. – Arturo Magidin Oct 05 '23 at 19:23
  • @ArturoMagidin (1) Yes. They are i not a_i. Sorry for typo. (2) No...I didn't prove that. How to prove that? (3) What do mean that? Here because the k-cycle can be decomposed into (k-1) transposition. So the sign of this k-cycle is the product of these transposition. – Hermi Oct 05 '23 at 20:57
  • Then you need to prove (2). Otherwise, you don't know if a permutation can be both odd and even, in which case the "sign" is not a well-defined function. As to (3), you wrote "this $k$-cycle can be represented $\prod_{i=1}^{k-1}\mathrm{sgn}([a_ia_{i+1}])$." The cycle cannot be represented as the product of the signs of the tranpositions (what you wrote); you probably meant "$\prod_{i=1}^{k-1}[a_ia_{i+1}]$". – Arturo Magidin Oct 05 '23 at 21:04
  • If the question "really" is whether the parity of a permutation is well-defined, then it's a duplicate of this. – Arturo Magidin Oct 05 '23 at 21:05
  • (And if it was a typo, then don't just apologize in comments: edit the post and fix it!) – Arturo Magidin Oct 05 '23 at 21:05
  • @ArturoMagidin Oh. I see. After the proof of the permutation can be only odd or even. How do we know that the sign of $k$-cycle is $(-1)^{k-1}$? Is the reason that the sign of each transposition is $(-1)$, so the sign of $k$-cycle is same as the sign of $\prod_{I=1}^k[a_ia_{I+1}]=(-1)^{k-1}$? – Hermi Oct 05 '23 at 21:34
  • It's because you defined the sign as $1$ if it can be expressed as a product of an even number of transpositions, and $-1$ if it can be expressed as a product of an odd number of transpositions. – Arturo Magidin Oct 05 '23 at 21:35
  • @ArturoMagidin I am still confused. I mean the sign of cycle is the product of the signs of the transpositions but not the cycle. I believe we have $sign(\sigma \sigma')=sgn(\sigma)sgn(\sigma')$. So why do not we have $sgn([a_1a_2\cdots a_k])=sgn([a_1a_2])sign([a_2a_3])\cdots sign([a_{k-1}a_k])=(-1)^{k-1}$? – Hermi Oct 05 '23 at 21:39
  • @ArturoMagidin If so, why do we have sign of the $k$-cycle is $(-1)^{k-1}$ but not $(-1)^{k+1}$? – Hermi Oct 05 '23 at 21:42
  • You realize that $(-1)^{k-1}=(-1)^{k+1}$? You define the sign as $1$ when the permutation is even, $-1$ when the permuation is odd, so given the definition of "even" and "odd", it is even when you can express it as a product of an even number of transpositions, and odd when you can express it as an odd number of transpositions. That the sign of the product is the product of the signs follows from that. – Arturo Magidin Oct 05 '23 at 21:45
  • @ArturoMagidin Can you please check my update proof now? Does it make sense? – Hermi Oct 05 '23 at 22:09
  • (i) It's not a "partition", it's a "decomposition". (ii) The final line is left over from your previous version, still contains the error I pointed out from the very beginning, and is a non-sequitur. – Arturo Magidin Oct 05 '23 at 22:35
  • @ArturoMagidin I have fixed that. Can you please check that? Thank you! – Hermi Oct 05 '23 at 22:54
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    Now all you are doing is saying "I can express it as a product of $k-1$ transpositions, therefore its sign is $(-1)^{k-1}$." It's a trivial observation. What are you unsure about. – Arturo Magidin Oct 05 '23 at 22:55

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You can remember that the sign of a cycle $\sigma$ of odd length $k$ is $1$ by noticing that $\sigma^k = \textrm{Id}$, of sign $1$.

orangeskid
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