How does my elegantly cradfted proof look?

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.