$\displaystyle K \lhd G$ has index $\displaystyle m$ , show that $\displaystyle g^m \in K$ for all $\displaystyle g \in G$
Thanks guys
Uuuh?? Ok, let us go back to the basics, shall we? First, we must know that $\displaystyle Ka=Kb\Longleftrightarrow ba^{-1}\in K$, so we get that $\displaystyle Kx=K \Longleftrightarrow x\in K$.
Second...what I already wrote in my past messages: as $\displaystyle G/H$ is a finite group of order m, any of its elements raised to the m-th power is the group's unit, and this means that $\displaystyle (Kg)^m=K\,\,\forall\,g\in G$. But $\displaystyle (Kg)^m=Kg^m$ , according to the definition of the operation that makes $\displaystyle G/H$ a group, so this togehter with the first lines in this message gives...
Tonio