 Solution

Let $L$ be the subgroup of $H\mathrm{wr}{A}_m$ generated by all its elements of order $3$. The top group $A_m$ is generated by 3-cycles, also when $m=3$, so $A_m\le L$.

Take $\sigma =(123)$. For any $h\in H$, the conjugate $h_1\sigma h_1^{-1}$ has order $3$, as does ${\sigma }^{-1}$. Thus their product belongs to $L$ and equalsUsing ${\sigma }^{-1}$ instead supplies $d_{13}(h)=h_1{h}_3^{-1}$. Since distinct coordinate groups commute,The hypothesis that $H$ is perfect says that products of its commutators generate $H$, so $H_1\le L$. Transitivity of $A_m$ now gives $H_i\le L$ for every coordinate. Hence $H^m\le L$ and $A_m\le L$, provingThe proof does not require $H$ itself to be finitely generated, nor does it require $A_m$ to be perfect for $m=3,4$.