$A_4 \oplus Z_3$ has no subgroup of order 18 Here my solution: Suppose there exists and $H \leq A_4 \oplus Z_3$ such that order of H is 18. Now, notice index of H in $A_4 \oplus Z_3$ is 2. therefore, H is normal, and therefore, the $A_4 \oplus Z_3 / H$ exists. Now, for every $\alpha$ in $A_4 \oplus Z_3$, $\alpha^2H = (\alpha H)^2 = H$ since the quotient group has order 2. But notice, $A_4$ has only 3 elements of order 2, therefore, $A_4 \oplus Z_3$ have only 3 elements of order 2, which implies H has 3 elements of order 2 and the rest must have order 1, and this is an absurd.

Is this a correct solution? Do you guys have any other solution? 

thanks,
The image of $H$ under the projection $A_4\oplus \mathbf Z_3\to A_4$ would be a subgroup of $A_4$ isomorphic to $H/(H\cap \mathbf Z_3)$, which is of order $18/\#( H\cap \mathbf Z_3)$. But $A_4$ neither has any subgroup of order $18$ (obviously) nor of order $6$ (requires only slightly more reflection).
$\operatorname{Order}(g1+g2)=\operatorname{lcm}(\operatorname{Order}(g1),\operatorname{Order}(g2))$.

But for any $g_1$ from $A_4$, $\operatorname{Order}(g_1)\leq4$ and for any $g_2$ from $Z_3$, $\operatorname{Order}(g_2)\leq3$.

So, $\operatorname{Order}(g_1+g_2)\leq12$.