$A_4 \oplus Z_3$ has no subgroup of order 18

Solution 1:

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).

Solution 2:

The group $G=A4 |+| Z3$ has not a subgroup of 18 order

Proof:

Let $H$ be such a group. Then as the order of $G$ is equal to $12 \cdot 3=36$ index of $H$ is equal to $\frac{36}{18}=2$ so $H$-normal. Therefore $G/H$ of order $2$ exists. If $g$ from $G$ then $(g^2)H=(gH)^2=H$, So the square of any element from $G$ belongs to $H$. But in A4 there are 9 different elements of the form $a^2$ and in $\mathbb{Z}^3$ all three elements are of this form. Then in $G$ in all $9 \cdot 3=27>18$ such elements. CONTRADICTION. The theorem is proved.