# Can only find 2 of the 4 groups of order 2014?

Let $G$ be a group of order $2014=2\times 19\times 53$. The Sylow Theorems give that the Sylow-19 and 53 subgroups are unique and normal. Let these be $H$ and $K$, respectively. Then since $H$ and $K$ intersect trivially, $HK$=$H \times K$=$C_{19} \times C_{53} \cong C_{1007}$ is a (normal) subgroup of $G$. But this means that $G$ is either the direct product with $C_2$, giving $C_{2014}$, or the semidirect product over the only automorphism of order 2, giving $D_{2014}$.

I guess the 2 groups I'm missing are $C_{19} \times D_{106}$ and $C_{53} \times D_{38}$. Is that correct? In either case, what was the flaw in my argument above?

$$\text{Aut} (C_{1007})\cong C_{18}\times C_{52}$$
so $\;C_2\;$ can act as inversion only on the generator of $\;C_{19}\;$ , only in the generator of $\;C_{53}\;$ or on both. These three non-trivial actions give you three semidirect products...
• Ah, I was assuming the Automorphism group of $C_{1007}$ was cyclic. Thanks! Dec 31, 2013 at 17:33