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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?


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CORRECTION: A4⊕Z3 have only 3 elements of order 2, which implies H has 3 elements of order 2 – ILoveMath 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 – ILoveMath 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, $α^2H=(αH)^2=H$ – ILoveMath Oct 6 '12 at 17:11
up vote 3 down vote accepted

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).

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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.

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I havent yet learned the isomorphism thms :( – ILoveMath Oct 6 '12 at 17:43
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

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


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.

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