# group permutations 3

• Apr 24th 2009, 08:58 AM
mpryal
group permutations 3
Prove that f and g^-1fg, for any f,g in Sn, are of the same parity.

Notice $g^{-1}(1,2)g = (g^{-1}(1),g^{-1}(2))$.
Write $f=t_1...t_m$ for transpositions $t$. Then $g^{-1}fg = (g^{-1}t_1g)(g^{-1}t_2g)...(g^{-1}t_mg)$. Each $g^{-1}t_jg$ is a transposition by the above sentence. Thus, $f,g^{-1}fg$ both have $m$ components in its transpotitions so they have the same parity.