Thread: Factor exponent

$\displaystyle K \lhd G$ has index $\displaystyle m$ , show that $\displaystyle g^m \in K$ for all $\displaystyle g \in G$


Thanks guys

Originally Posted by sfspitfire23

$\displaystyle K \lhd G$ has index $\displaystyle m$ , show that $\displaystyle g^m \in K$ for all $\displaystyle g \in G$


Thanks guys

What is the order of the group $\displaystyle G/K$? Then....

Tonio

order will be $\displaystyle g^m=1$ where m is the copies of g

Originally Posted by sfspitfire23

order will be $\displaystyle g^m=1$ where m is the copies of g

No. The order is m and thus any element in the quotient group raised to the m-th power is the group's unit and thus...

Tonio

So, If Kg was raised to the mth power, it would be come the inverse, bringing Kg back to K

Originally Posted by sfspitfire23

So, If Kg was raised to the mth power, it would be come the inverse, bringing Kg back to K

No. You'd get $\displaystyle (Kg)^m=Kg^m=K$, SO....

Tonio

So, $\displaystyle kg^mk^{-1}=k \in K$ because $\displaystyle K\lhd G$ then $\displaystyle g^m\in K$

Originally Posted by sfspitfire23

So, $\displaystyle kg^mk^{-1}=k \in K$ because $\displaystyle K\lhd G$ then $\displaystyle g^m\in K$

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