$A_4 \oplus Z_3$ 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,
 A: 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).
A: 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.
A: The normality observation is a good starting point, but I do not follow the reasoning where you take it.
Hint: try to deduce that $H\cap A_4$ would have to be an index-2 subgroup of $A_4$ (of which there are none). You can apply one of the isomorphism theorems to help do this.
A: $\operatorname{Order}(g1+g2)=\operatorname{lcm}(\operatorname{Order}(g1),\operatorname{Order}(g2))$.
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, $\operatorname{Order}(g_1+g_2)\leq12$.
