Haha, how does my proof look?

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


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