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Say you have 2 permutations, f and g, how would one calculate $g^f$? Also can you multiply these permutations and how?

Jean Marie
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Usually in group theory $g^f$ is used notation for conjugate of $g$ by $f$, i.e. $g^f := f^{-1}gf$. You can calculate this product in the following manner. If $g = (a_1a_2\cdots a_n)$, then $g^f = f^{-1}gf = (f^{-1}(a_1)f^{-1}(a_2)\cdots f^{-1}(a_n)$).

On the otherside multiplication of permutations is composition of them. To find the actual product, you would need to do term by term caclulations, i.e. $g \cdot f (a) = g(f(a))$

Stefan4024
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    Isn't it more common to have $g^f = f^{-1}gf$ and ${}^fg = fgf^{-1}$ so that $(g^f)^h = g^{fh}$ and ${}^{fh}g = {}^f({}^hg)$? – Daniel Fischer Feb 26 '17 at 13:28
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    @DanielFischer To be fair I have never paid much attention to it and I tend to use $g^f = fgf^{-1}$. I'm sure some authors use this notation too, which might have influenced my thoughts.

    Anyway now that you've mentioned it, it seems that your definition might be more useful and convinient.

    – Stefan4024 Feb 26 '17 at 13:43