Haha, how does my proof look?

Q- If K is normal to H and H is normal to G show that aKa^{-1}\lhd H for all a in G.


Pf:

Let h\in H and aka^{-1}\in aKa^{-1}. Now, if aka^{-1}\lhd H then haka^{-1}h^{-1}\in aKa^{-1}. Now, pull out the inverse hak(ah)^{-1}. This gives ah_1k(h_1a)^{-1}. Since h_1kh_1^{-1}=k_1 we have ak_1a^{-1}\in  aKa^{-1}. and thus aKa^{-1} \lhd H as desired.