Factor exponent

**sfspitfire23** · November 7th 2009, 01:01 PM

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

Thanks guys

**tonio** · November 7th 2009, 01:05 PM

Originally Posted by sfspitfire23

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

Thanks guys

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

Tonio

**sfspitfire23** · November 7th 2009, 01:23 PM

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

**tonio** · November 7th 2009, 01:39 PM

Originally Posted by sfspitfire23

order will be $g^m=1$ 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

**sfspitfire23** · November 7th 2009, 01:48 PM

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

**tonio** · November 7th 2009, 02:07 PM

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 $(Kg)^m=Kg^m=K$ , SO....

Tonio

**sfspitfire23** · November 7th 2009, 02:23 PM

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

**tonio** · November 7th 2009, 03:11 PM

Originally Posted by sfspitfire23

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

Uuuh?? Ok, let us go back to the basics, shall we? First, we must know that $Ka=Kb\Longleftrightarrow ba^{-1}\in K$ , so we get that $Kx=K \Longleftrightarrow x\in K$ .
Second...what I already wrote in my past messages: as $G/H$ 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 $(Kg)^m=K\,\,\forall\,g\in G$ . But $(Kg)^m=Kg^m$ , according to the definition of the operation that makes $G/H$ a group, so this togehter with the first lines in this message gives...

Tonio

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