1. ## Normal subgroups

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

Would I say:

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

Thanks guys!

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

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

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

4. Originally Posted by sfspitfire23
and because the normal subgroup is abelian i can write the $\displaystyle haka^{-1}h^{-1}$ as $\displaystyle ahkh^{-1}a^{-1}$ 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

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

Heres what I have so far.

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

6. Originally Posted by sfspitfire23
I guess what I dont understandis how the $\displaystyle a$ and $\displaystyle a^{-1}$ get pushed outside.

Heres what I have so far.

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

Take $\displaystyle h(aka^{-1})h^{-1}$ and multiply it both from the left and from the right by $\displaystyle 1=aa^{-1}$, 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