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I could not show it, I could not define the h. I need help, please.I'm not very good at math.

Show that $sgn (f) = 1$ if and only if there is h ∈ $S_n$ such that $f = h ◦ h$. Help me.

My proof: -> Suppose that sign(f) = 1 then f is even. Consider $h$ = $h$ o $h^{-1}$

$h$ o $h$ = (($h$ o $h^{-1}$) o ($h$ o $h^{-1}$))(f(x))

= (($h$ o ($h^{-1}$ o $h$) o $h^{-1}$))(f(x)) = (($h$ o $1_h$) o $h^{-1}$))(f(x)) = (($h$ o $h^{-1}$))(f(x)) = (($h$ ($h^{-1}(f(x)$))= f(x) =f .

Therefore there is h in $S_n$ such that f = h o h.

<- there is h in $S_n$ such that $f = $h o $h$, if h is in $S_n$, then h is bijective, but $f = $h o $h$,

therefore f is bijective, then ... $sgn(f)=sgn(ho h)= 1$ :(.

Jean Paul
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  • Also, I believe the result is not true? See https://math.stackexchange.com/questions/266569/how-to-find-the-root-of-permutation – Mike Earnest Aug 09 '18 at 23:29
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    Let $f = (1,2)(3,4,5,6)$. Even though $\text{sign }f=+1$, you can show there is no $h$ for which $f=h\circ h$ by considering all the possible cycle structures of a permutation in $h\in S_6$, and that for each of them, $h\circ h$ does not have the cycle structure of $f$. – Mike Earnest Aug 09 '18 at 23:54
  • Thank you Mike :) – Jean Paul Aug 10 '18 at 00:00

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Hint: Try to show that there is no $h \in S_6$ such that $$h^2= (1,2) (3,4,5,6)=\begin{bmatrix} 1 & 2 &3 &4 &5 &6 \\ 2 & 1 &4 &5 &6 &3 \end{bmatrix}$$

N. S.
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