Suppose $A,B,H,G$ are groups, with $A \unlhd B, H\unlhd G$ and $A,B,H \subseteq G$.
Is it true that $HA \unlhd HB$?
My attempt: for $a \in A, b\in B, h \in H$ we have $(hb)(ha)(hb)^{-1}=(hb)h(hb)^{-1}(hb)a(hb)^{-1}=(hbhb^{-1}h^{-1})(hbab^{-1}h^{-1})=(*)$. Now, $h'=bhb^{-1} \in H$, $a'=bab^{-1} \in A$ $\implies (*)=hh'h^{-1}ha'h^{-1}=hh'a'h^{-1}=hh'h''a''$ for some $a'' \in A, h'' \in H$ because $HA$ is a subgroup, hence $AH=HA$.
Hence $(hb)(ha)(hb)^{-1} \in HA$, yielding the claim. Is everything ok with this proof? Do you know of any neater argument?