Mi se da o permutare alfa α ce are pe linia I elementele : 1 2 3 4 5 si pe linia a II a elem. : 2 3 4 5 1
La unul din puncte mi se cere sa arat ca T*α =α*T stiind ca T apartine S5 si
T*α^2 =α^2*T. Cum fac asta? Am calculat cateva puteri ale lui α si am observat faptul ca α^5=e (permutare identica)
Plecand de la constatarea ta ca σ^5=e atunci σ^6=σ considerand pe T=σ^3 putem avea;
T. σ=σ^3.σ^6=σ.T=σ^6.σ^3 si la fel; T. σ^2=σ^3.σ^7=σ^2.T=σ^7.σ^3 DeciT=σ^3=(■(1 2 3 4 5@4 5 1 2 3))
OBS; Solutii pot fi ; T=σ^k, pentru k={1,2,3,4,5}
Deoarece
egalitatea cerută devine ![Rendered by QuickLaTeX.com T\circ \alpha^6=\alpha^6\circ T.](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-78e508e7714779abacbbb6a1e310ad0d_l3.png)
Vom arăta asta folosind ipoteza![Rendered by QuickLaTeX.com T \circ \alpha^2=\alpha^2 \circ T.](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-e10911c5865a55f0b992295eb1793092_l3.png)
Astfel, avem![Rendered by QuickLaTeX.com T\circ \alpha^6=T \circ\alpha^2\circ\alpha^2\circ\alpha^2=\alpha^2\circ T\circ \alpha^2\circ \alpha^2=\alpha^2 \circ \alpha^2 \circ T \circ \alpha^2=\alpha^2 \circ \alpha^2 \circ \alpha^2 \circ T=\alpha^6 \circ T.](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-81eff218801f70b1a025f4e4da3d1155_l3.png)