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Does $S_3 \times\mathbb Z_6$ have a subgroup of order $9$? where $S_3$ is the symetric group. Clearly $|S_3| \times |\mathbb Z_6|=36$ and $9$ does divide $36$, so its possible. But i can not figure out why or why not there would be a subgroup of order $9$.

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  • $\begingroup$ Does $S_3$ have a subgroup of order $3$? Does $Z_6$ have a subgroup of order $3$? $\endgroup$ – Lord Shark the Unknown Jun 1 at 14:29
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Hint: Both $S_3$ and $\mathbb Z_6$ have subgroups of order $3$.

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If you know about Sylow Subgroups and the existence of these. Then, a 3-Sylow Subgroup of a group of order $36=3^2\times 2^2$ will have order $3^2=9$.

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