# $A_4 \oplus Z_3$ 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 u guys have any other solution?

thanks,

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CORRECTION: A4⊕Z3 have only 3 elements of order 2, which implies H has 3 elements of order 2 –  Marvin Gaye Oct 6 '12 at 16:55
Why does that imply H has 3 elements of order 2? –  anon Oct 6 '12 at 17:03
since every $\alpha$ is in H –  Marvin Gaye Oct 6 '12 at 17:04
Why is every element of order $2$ contained in $H$? –  anon Oct 6 '12 at 17:10
for every α in A4⊕Z3, $α^2H=(αH)^2=H$ –  Marvin Gaye Oct 6 '12 at 17:11

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).
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.
One of them says that in a parent group $G$, take $A$ and $B$ to be subgroups, with $B$ normal. Then $(AB)/B\cong A/(A\cap B)$. Apply this with $A=A_4$ and $B=H$. You will have to think about the possibilities for what $AB$ might be. There are two cases: one of which is ruled an impossibility by my hint, and the other is ruled an impossibility by $H$'s index and $A_4$'s index. –  alex.jordan Oct 6 '12 at 17:51