Normal subgroups

**sfspitfire23** · October 30th 2009, 07:39 PM

Prove: If K is normal to H and H is normal to G, show that $aKa^{-1}$ is normal to H for all $a\in G$

Would I say:

$hKh^{-1}=k$ and $aHa^{-1}=h$ then try to get H=K and subsitute the K for H? Or is that way off?

Thanks guys!

**proscientia** · October 30th 2009, 07:55 PM

Let $h\in H,\,aka^{-1}\in aKa^{-1}.$ Then

$h(aka^{-1})h^{-1}\ =\ a\underbrace{\overbrace{(a^{-1}ha)}^{\in H}k\overbrace{(a^{-1}h^{-1}a)}^{\in H}}_{\in K}a^{-1}$

**sfspitfire23** · October 31st 2009, 08:48 AM

and because the normal subgroup is abelian i can write the $haka^{-1}h^{-1}$ as $ahkh^{-1}a^{-1}$ as you did correct?

**tonio** · October 31st 2009, 12:02 PM

Originally Posted by sfspitfire23

and because the normal subgroup is abelian i can write the $haka^{-1}h^{-1}$ as $ahkh^{-1}a^{-1}$ as you did correct?

No! Read again, very carefully and writing down the proof ina sheet of paper, the proof: in no place he did write anything from where one can deduce the normal sbgp. is abelian which, of course, is false in general...and he didn't write what you say he did.

Tonio

**sfspitfire23** · October 31st 2009, 01:00 PM

I guess what I dont understandis how the $a$ and $a^{-1}$ get pushed outside.

Heres what I have so far.

We have to show $h(aka^{-1})h^{-1}$ is true. Then, $aha^{-1}=h$ and I can write the thing as $aha^{-1}(aka^{-1})a^{-1}h^{-1}a$ then I get stuck

**tonio** · October 31st 2009, 04:34 PM

Originally Posted by sfspitfire23

I guess what I dont understandis how the $a$ and $a^{-1}$ get pushed outside.

Heres what I have so far.

We have to show $h(aka^{-1})h^{-1}$ is true. Then, $aha^{-1}=h$ and I can write the thing as $aha^{-1}(aka^{-1})a^{-1}h^{-1}a$ then I get stuck

Take $h(aka^{-1})h^{-1}$ and multiply it both from the left and from the right by $1=aa^{-1}$ , and then arrange parentheses accordingly (using associativity) and then use normality of H in G and normality of K in H. That's all the trick.

Tonio

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