My approach: the prime factorization of $2019$ is $3 \times 673$. Let $m$ and $n$ be the number of Sylow-$3$ and $673$ subgroups of $G$ (some group of order $2019$). Then from the Sylow theorems, we have the following relations:
- $m | 673,$
- $m \equiv 1 \pmod 3,$
- $n| 3,$
- $n \equiv 1 \pmod {673}.$ From here, $n = 1$ and $m = 1$ or $m = 672$. For the case $m = 1$, the Sylow $3$ and $673$ subgroups are unique, and let us call them $H$ and $K$ respectively. Since they are unique, $H \unlhd G$ and $K \unlhd G$, and it is not hard to see that in this case, $G \cong H \times K \cong \mathbb{Z}_3 \times \mathbb{Z}_{673}$.
But I am not sure what to do with the other case: $m = 672$. Here are a few observations I have made.
- None of the $672$ subgroups of order $3$ is normal, but the subgroup of order $673$ is.
- From here, we can also infer that $G$ is not abelian.
- Finally, the $673$ subgroups mentioned above are pairwise disjoint.
Can we describe $G$ in this case?
