Classifying all groups of order $2013$ (up to isomorphism) 
Let $G$ be a group such that $|G|=2013$, how would you classify, up to isomorphism, all groups $G$? 

 A: Since $2013= 3 \times 11 \times 61$, using Sylow theorems you find that $G$ has only one $11$-Sylow subgroup $S_1$ and only one $61$-Sylow subgroup $S_2$. Let $H$ be the subgroup generated by $S_1$ and $S_2$, and $K$ be a $3$-Sylow subgroup.
Then $H$ is a normal subgroup, $H \cap K = \{1\}$ and $HK= G$. Therefore, $$G = H \rtimes K \simeq (\mathbb{Z}_{61} \times \mathbb{Z}_{11}) \rtimes \mathbb{Z}_3.$$
Because $$|\text{Hom}(\mathbb{Z}_3, \text{Aut}(\mathbb{Z}_{61} \times \mathbb{Z}_{11}))| = |\text{Hom}(\mathbb{Z}_3, \mathbb{Z}_{61}^{\times} \times \mathbb{Z}_{11}^{\times})| = |\text{Hom}(\mathbb{Z}_3,\mathbb{Z}_{61}^{\times})| = 3,$$
You find only three semi-direct products: $$\mathbb{Z}_{2013}, \ (\mathbb{Z}_{61}\times \mathbb{Z}_{11}) \rtimes_{\varphi} \mathbb{Z}_3 \ \text{and} \ (\mathbb{Z}_{61}\times \mathbb{Z}_{11}) \rtimes_{\phi} \mathbb{Z}_3$$ where $\varphi,\phi : \mathbb{Z}_3 \to \text{Aut}(\mathbb{Z}_{61} \times \mathbb{Z}_{11})$ are defined by $$\varphi(1) : (h,k) \mapsto (47 \cdot h,k) \ \text{and} \ \phi(1) : (h,k) \mapsto (13 \cdot h,k).$$ By the way, notice that $13=47^2$.
However, we have the following isomorphism:  $$\left\{ \begin{array}{ccc} (\mathbb{Z}_{61} \times \mathbb{Z}_{11}) \rtimes_{\varphi} \mathbb{Z}_3 & \to & (\mathbb{Z}_{61} \times \mathbb{Z}_{11}) \rtimes_{\phi} \mathbb{Z}_3 \\ ((x,y),z) & \mapsto & ((13x,y),z) \end{array} \right.$$
Consequently, there exist only two groups of order $2013$ (up to isomorphism): $$\mathbb{Z}_{2013} \ \text{and} \ (\mathbb{Z}_{61} \times \mathbb{Z}_{11}) \rtimes_{\varphi} \mathbb{Z}_3.$$
A: $$2013=3\times 11 \times 61$$
By the third Sylow theorem, the Sylow $11$-subgroup $P_{11}$ of $G$ is normal in $G$.  Since $P_{11}$ is cyclic and $G/P_{11}$ is solvable (by Burnside's theorem, or more simply by Sylow theory), we know that $G$ is solvable.  This allows us to use Hall subgroups.  If $G$ isn't cyclic, then because $|G|$ is squarefree, a subset $\{p,q\}\subset \{3,11,61\}$ exists for which $G$ contains no element of order $pq$, and thus a Hall $\{p,q\}$-subgroup of $G$ is a Frobenius group.  In this case, we must have that $q$ divides $p-1$, so the only possibility is that $q=3$ and $p=61$.  Thus there are only two groups of order $2013$: $$\mathbb{Z}_{2013}\hspace{6pt}\text{ and }\hspace{6pt}\mathbb{Z}_{11}\times \left(\mathbb{Z}_{61}\rtimes \mathbb{Z}_3\right).$$
A: 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.
