Background
Two years ago, I thought about a group of 689 elements.^{1} I only managed to show the existence of such a group.
Problem
Inspired by the use of Sylow III to show that a group of order 15 has only one structure: $\Z_{15} \cong \Z_3 \times \Z_5$, I wondered if $\Z_{689}$ is the only possible structure for a group of order 689.
Solution
Denote $G$ as a group of order 689. First, apply Sylow III to $\abs{G}$, whose largest prime divisor 53. Since $\lvert G \rvert$ is a product of two primes 13 and 53, we conclude that $n_{53} = 1$. Second, we have
Since $53 \equiv 1 \pmod{13}$, we have to consider two cases:
 $n_{13} = 1$: In this case, $G$ has a unique 13Sylow $H$ and a
unique 53Sylow $K$. Both $H$ and $K$ are normal in $G$. Their
intersection $H \cap K$ is, by Lagrange’s Theorem, the identity
element $e_G$. For each $h \in H$ and $k \in K$, the commutator
\begin{equation*} [h,k] = hk h^{1} k^{1} = \underbrace{hk h^{1}}_{\in K} k = h \underbrace{k h^{1} k^{1}}_{\in H} \in H \cap K \end{equation*}
is reduced to $e_G$, implying that $H$ commute with $K$. Therefore, an internal direct product of $H$ and $K$ can be set up, and $\abs{H} \times \abs{K} = 13 \times 53 = \abs{G}$, so $G = H \times K \cong \Z_{13} \times \Z_{53}$.

$n_{13} = 53$: In this case, $G$ has 53 distinct 13Sylows $H_1, H_2, \cdots, H_{53}$. Each of these $H_i$’s is isomorphic to the cyclic group $C_{13}$. Thus, $G$ possess $53 \times 12$ elements of order 13, 52 elements of order 53, and the identity—they add up to $\abs{G}$.
Not knowing how to continue, I finally searched for “nonabelian group of order $pq$” and found this answer on Mathematics Stack Exchange, which showed that my guess is wrong. In fact, $\Aut H \cong \Z_q^*$ and its elements are $\gamma_\lambda: h \mapsto \lambda h$. $\phi(K) = \psi(K)$ because they are both subgroup of order $p$, but it’s unique in $\Aut H$. Since homomorphisms $\phi,\psi: K \to \Aut H$ are supposed to be nontrivial, $\ker \phi, \ker \psi \ne \abs{K} = p$, so they are injective. This proves the existence of a nonabelian group $G \cong H \rtimes_\psi K$.
To conclude, even though the arguments are “too simple” for a mathematicien, a group of order 689 is never simple.