I see that the exact same question has been asked:$A_4 \oplus Z_3$ has no subgroup of order 18 but I don't really understand the accepted answer. Why is it that the image of $H$ in the projection $A_4 \times Z_3 \to A_4$ would be isomorphic to $H/(H \cap Z_3)$?
Is there also possibly another way to prove this?
This is an application the first isomorphism theorem:
If $\psi:A_4\times Z_3\to A_4$ is the projection, then $\ker(\psi)=Z_3$.
If we restrict this to $H$ with $\psi_H:H\to A_4$ then $\ker(\psi_H)=H\cap\ker(\psi)=H\cap Z_3$. The first isomorphism theorem gives the image of $H$ is isomorphic to $H/(H\cap Z_3)$.
An alternative solution is hinted in one of the other answers:
Write $G=A_4\times Z_3$ and note that $|G|=36$
If $|H|=18$ then $[G:H]=2$ so $[A_4:H\cap A_4]\le 2$.
If $[A_4:H\cap A_4]=1$ then $|H\cap A_4|=12\nmid 18$, so $[A_4:H\cap A_4]=2$.
So, as in the accepted solution, the problem reduces to showing $A_4$ has no subgroup of order $6$.