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Problem: Prove or disprove that $D_3 \times\mathbb Z_4$ has no subgroup of order $6$.

The order of $D_3 \times\mathbb Z_4$ is $24$ and $6$ divides that so it's possible structurally for there to be a subgroup of this order, from Lagrange's theorem.

I know that a subgroup of a direct/Cartesian product ($\times$) of two groups is the direct product of their respective subgroups (so like if $G_1 \leq G$ and $H_1 \leq H$ then $G_1 \times H_1 \leq G \times H$). So I have found many subgroups of $D_3 \times\mathbb Z_4$ this way and none of them have order 6. However, I know that not all subgroups of $D_3 \times\mathbb Z_4$ can be found this way in general, so there may be more I can't find.

How can I find the remaining subgroups and show none have order 6? Or, how can I prove there are no more subgroups other than those formed by the direct product of their respective subgroups? Or alternatively, is there a non-exhaustive way of proving/disproving the existence of a subgroup with a particular order?

Any help would be appreciated!

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    $\begingroup$ Try $G_1=S_3$, $H_1=\{1_H\}$? $\endgroup$ – David Cheng Oct 20 '20 at 1:00
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    $\begingroup$ Isn't $S_3$ itself of order $6$? $\endgroup$ – GEdgar Oct 20 '20 at 1:00
  • $\begingroup$ @GEdgar How silly of me, of course $S_3$ is of order 6! I was so focused on subgroups I didn't look at the whole group itself. Thank you! $\endgroup$ – user102938 Oct 20 '20 at 1:03
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There is an order-$3$ element in $S_3$ and an order-$2$ element in $Z_4$. The element in $S_3×Z_4$ that is a direct combination of these two elements must therefore have order $6$, as the cycle lengths are coprime. Thus a subgroup of order $6$ exists.

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