Abelian subgroup of a group of order $2002$

If $\,n_p=\,$ number of Sylow p-subgroups of $\,G\,$ , then $\,n_{13}=1=n_{11}\,$, thus letting $\,Q_p\,$ be a Sylow p-sbgp., we have that $$Q_{11}\,,\,Q_{13}\triangleleft G\Longrightarrow Q_{11}Q_{13}\triangleleft G\Longrightarrow P:=Q_{11}Q_{13}Q_7\triangleleft G$$ the last one being a subgroup of order $\,7\cdot 11\cdot 13\,=1001$ in $\,G\,$.

Abelianity follows from the fact that the only group of order $\,1001\,$ is the cyclic one, as $\,P\cong C_{11}\times C_{13}\times C_7\,\,,\,\,C_m:=\,$ the cyclic group of order $\,m\,$

Added: Please note the comment by Jack Schmidt below for an alternative approach: by the Schur-Zassenhaus theorem, if we have $\,N\triangleleft G\,$ s.t. $\,\left(|N|,\left|G/N\right|\right)=1\,$ then $\,G\cong\,N\rtimes G/N\,$ .

In our case, take $\,N:=Q_{11}\,$ and apply the above, then take $\,G/N\,\,,\,\left|G/N\right|=2\cdot 7\cdot 13=182\,$ , which has a unique Sylow 7-subgroup, which we know is $\,\,\,\overline Q_7=Q_7N/N\,$ , and again apply the S-Z theorem to obtain $$G/N\cong \overline Q_7\rtimes \left(G/N\right)/\overline Q_7$$ with $\,\,\left(G/N\right)/\overline Q_7\cong G/Q_7Q_{11}\,\,$ , and so on.

Note that we can directly apply Sylow theorems to deduce that $\,\,G/Q_7Q_{11}\,$ has a unique Sylow 13-sbgp. which we cannot do, at least directly or easily, with the original group using only Sylow theorems, which is what Jack writes there.