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

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!

• Try $G_1=S_3$, $H_1=\{1_H\}$? – David Cheng Oct 20 '20 at 1:00
• Isn't $S_3$ itself of order $6$? – GEdgar Oct 20 '20 at 1:00
• @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! – user102938 Oct 20 '20 at 1:03

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.