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2:27 AM
in CRUDE, 54 secs ago, by Shaun
My accepted answer is wrong . . .
2
A: Find $f,g$ s.t. $f\circ g=\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\ 10 & 4 & 5 & 7 & 8 & 9 & 2 & 6 & 3 & 1\end{pmatrix}.$

ShaunThere is a well-known algorithm for decomposing any given permutation as a product of (not necessarily disjoint) $2$-cycles/transpositions. Such a decomposition of a given $f\circ g$ would, in general, give strong hints about (if not completely determine) the nature of $f$ and $g$. Why? Because...

Please would the moderators delete it for me? (I've flagged it already but still . . . )
 
3:00 AM
I think it's right after all . . .
2
A: Find $f,g$ s.t. $f\circ g=\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\ 10 & 4 & 5 & 7 & 8 & 9 & 2 & 6 & 3 & 1\end{pmatrix}.$

bofA permutation $f$ is an involution if $f\circ f=id$. As you know, any permutation can be written as a product of disjoint cycles; your permutation is $(1\ 10)(2\ 4\ 7)(3\ 5\ 8\ 6\ 9)$. In order to write an arbitrary permutation as a product of two involutions, it suffices (since disjoint permuta...

 

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