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I am going through Gallian's book on group theory, and while proving that identity permutation is an Even permutation, author assumes the identity permutation is of $r$ $2$-cycles and in one case, if the $r$ $2$-cycles reduces to $r-2$, then using second principle of mathematical induction, $ r-2$ is even. How is that leap made using second principle of mathematical induction? I understand the assumptions made in first principle induction, but how does it work in second principle?

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Strong mathematical induction allows you to prove a statement $P(n)$ by assuming that $P(k)$ applies to every $k<n$. In this case, he is proving the statement for $n$ $2$-cycles, and when that reduces to $n-2$, it can be assumed to be true because $n-2=k<n$.

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  • $\begingroup$ Ok. But what will the statement be, when k is odd? This is what I am struggling with. And what if my n is odd? n-2 will then be odd, right? please bear with me. $\endgroup$ – jnyan Feb 21 '18 at 1:46

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