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This question pertains to the proof mentioned in this question. What is the inductive hypothesis here? As I can decipher it, it seems like:

If $i$ is the number of terms in 2-cycle identity permutations and $i<r$ then $i$ is even.

Can someone verify this? Or if wrong mention the correct hypothesis.

Shyam Tripathi
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    It says if the identity permutation is written as a product of $r$ transpositions, then $r$ is even. There is no such thing as a two cycle identity permutation because the identity leaves everything fixed and a transposition does not. – John Douma Oct 08 '23 at 17:00
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    $\Large\color{\green}{\checkmark}$ Your guess is correct. – Anne Bauval Oct 08 '23 at 17:06
  • @JohnDouma So you are implying that (12)(12) is not an identity permutation? The question never mentioned the cycles to be disjoint. – Shyam Tripathi Oct 08 '23 at 22:45
  • @AnneBauval Thanks for confirming! – Shyam Tripathi Oct 08 '23 at 22:46
  • @ShyamTripathi No. That would be two transpositions which is an even number of them. I am saying $(12)$ or any other single transposition is not the identity. What you wrote are two $2$-cycles. – John Douma Oct 08 '23 at 23:00

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This is an example of what is sometimes called strong induction. The hypothesis of strong induction is that the theorem is true for all values less than $r$. One place you see the strong induction hypothesis used is where it says "By induction, $r-2$ is even; hence, $r$ must be even." Your statement does, indeed, appear to be identical to the strong induction hypothesis.

Lee Mosher
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