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!