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[(1,2,3,4) ○ (1,2)(3,5)]^-1

The solution has the following steps:
f=(1,2,3,4) g=(1,2)(3,5)
Compute f○g:
enter image description here

Where can I find information regarding what the steps are doing in this problem? I understand how to compute the composition of two functions but I can't follow what's done for sets.

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dukevin
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1 Answers1

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Except for the odd style of writing things backward, this is the same as usual composition. $f \circ g$ means as usual first do $g$ then do $f$. For example $g=(12)(35)$ takes $1$ into $2$, then one sees what $f=(1234)$ takes that intermediate result $2$ into, and sees it is $3$. This shows $f \circ g$ carries $1$ to $3.$ If the cycle decomposition is desired the next one to do would be to see what $f \circ g$ takes $3$ into, in the above way, and so on to get the cycles for $f \circ g.$ After that is all done, the usual trick of reversing all the cycles in $f\circ g$ will give the cycle decomposition for $(f \circ g)^{-1},$ and if one likes one can shuffle the representations of the cycles now to place least elements first, etc.

coffeemath
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  • @MauroALLEGRANZA, I think $f$ implicitly behaves like the identity on $5$ – goblin GONE Jun 12 '14 at 08:29
  • The usual definitions of cycle form are that a number not mentioned in a cycle decomposition are assumed to be carried to themselves. – coffeemath Jun 12 '14 at 08:30
  • @coffeemath Thanks for the explanation! Could you elaborate a bit on what you mean by: "... takes 1 into 2 then 1 sees what f has and takes that inermediate result 2 into, and sees it as 3" (how did 1 become 3?) – dukevin Jun 12 '14 at 09:10
  • The term "takes" is informal for "maps". Since if $g=(12)(35)$ then $g$ maps (takes) 1 to 2, and maps 2 to 1, and maps 3 to 5, and maps 5 to 3, by definition of what a cycle is. When applying $f \circ g$ to an element x, by definition of composition one first finds what x maps to under $g$, and I informally call that the "intermediate result", say it is called z. Then one applies $f$ to this intermediate result z, to get the result y of the composition, so that $f \circ g$ maps (or takes) x to y (eliminating the intermediate result z). – coffeemath Jun 12 '14 at 22:05