TITLE: Classifying all groups of order $2013$ (up to isomorphism)
QUESTION [5 upvotes]: Let $G$ be a group such that $|G|=2013$, how would you classify, up to isomorphism, all groups $G$?

REPLY [2 votes]: Here is another solution building up $G$ as (semi)direct products of the Sylow subgroups.
The building order is different from the one in the solution of Seirios, which allows us to use the classification of $pq$-groups to avoid the isomorphism test of semidirect products.
We have $2013 = 3\cdot 11\cdot 61$. For $p\in\{3,11,61\}$, let $P_p$ be a $p$-Sylow subgroup of $G$.
By Sylow, $P_{61}$ is normal in $G$.
Thus, $G$ has a subgroup $H = P_3 P_{61}$ of order $3\cdot 61 = 183$.
By the classification of $pq$-groups, $H \cong \mathbb{Z}_{183}$ or $H \cong \mathbb{Z}_{61}\rtimes \mathbb{Z}_3$, where the second group is the unique non-abelian group of order $183$.
By Sylow, also $P_{11}$ is normal in $G$.
Since $\gcd(183,11) = 1$, $H$ is a complement of $P_{11}$ in $G$, so $G\cong P_{11}\rtimes_\varphi H$, where $\varphi : H\to\operatorname{Aut}(P_{11})$ is a homomorphism.
Since $11$ is prime, $P_{11}\cong\mathbb{Z}_{11}$ and thus $\operatorname{Aut}(P_{11}) \cong \mathbb{Z}_{10}$.
Because of $\gcd(\left|\operatorname{Aut}(P_{11})\right|,\left|H\right|) = 1$, $\varphi$ must be the trivial homomorphism.
So the semidirect product $P_{11}\rtimes_\varphi H$ is in fact a direct product, showing $G\cong \mathbb{Z}_{11}\times H$.
Now plugging in the two possibilities for $H$ gives the two isomorphism types
$$\mathbb Z_{2013}\quad\text{and}\quad\mathbb Z_{11}\times(\mathbb Z_{61}\rtimes \mathbb Z_3)$$
 of groups of order $2013$.
Since only the first group is abelian, they are not isomorphic.