Let $G$ be a non-Abelian group such that $|G| \le 15$ with proper subgroups $H$ and $K$ such that $G=H\times K$. Prove that $G$ is isomorphic to $D_6$.
I understand that the two subgroups must be $H=\{e,r_3\}$ and $K=\{e,r_2,r_4,f,r_2f,r_4f\}$ in $D_6$. I also know that $D_6$ has order $12$ but I do not understand why the less than or equal to $15$ is necessary. I'm really not understanding why this must be always isomorphic to $D_6$ no matter the $G$. Any help is greatly appreciated. Thank you.
Why $\le 15$? Because the next example, the product of a cyclic group of order $2$ and either the dihedral or quaternion group of order $8$, has order $16$.
The way to look at this? We have a product $H\times K$, both nontrivial. In order for the group to be non-abelian, one of the factors must itself be non-abelian. WLOG, let that non-abelian factor be $H$. The smallest non-abelian group $S_3$ has order $6$, and the smallest nontrivial group $\mathbb{Z}/2$ has order $2$. Those are the only groups that fit, so that's the only option of that order.
Increasing the order... $13$ is prime. $14=7\cdot 2$, and there's no non-abelian group of order $7$. $15=5\cdot 3$, and there are no non-abelian options there. We have only one option up to order $15$, as desired.
The least trivial part? Showing that the product $S_3\times\mathbb{Z}/2$ is isomorphic to the dihedral group $D_6$ on the hexagon. That comes up because the $180^\circ$ rotation commutes with everything.