Räkna ut fog i abstrakt algebra (permutationer.)

Följ varje tal i definitionsmängden:

$$\begin{gathered} g\left(1\right): 1\to 3 \\ f\left(g\right(1\left)\right)=f\left(3\right): 3\to 1 \\ g\left(2\right): 2\to 1 \\ f\left(g\right(2\left)\right)=f\left(1\right): 1\to 2 \\ g\left(3\right): 3\to 2 \\ f\left(g\right(3\left)\right)=f\left(2\right): 2\to 3 \end{gathered}$$

Totalt ger det 

$$\begin{gathered} f\circ g\left(1\right)=1 \\ f\circ g\left(2\right)=2 \\ f\circ g\left(3\right)=3 \end{gathered}$$

:)

Smutstvätt skrev:

Följ varje tal i definitionsmängden:

$$\begin{gathered} g\left(1\right): 1\to 3 \\ f\left(g\right(1\left)\right)=f\left(3\right): 3\to 1 \\ g\left(2\right): 2\to 1 \\ f\left(g\right(2\left)\right)=f\left(1\right): 1\to 2 \\ g\left(3\right): 3\to 2 \\ f\left(g\right(3\left)\right)=f\left(2\right): 2\to 3 \end{gathered}$$

Totalt ger det 

$$\begin{gathered} f\circ g\left(1\right)=1 \\ f\circ g\left(2\right)=2 \\ f\circ g\left(3\right)=3 \end{gathered}$$

:)

oooh! snyggt :) tack