Abelian subgroup of a group of order $2002$ Another unsolved question from my studying for quals - 

Show that if $G$ is a group of order $2002=2\cdot 7 \cdot 11 \cdot 13$, then $G$ has an abelian subgroup of index $2$.

I know it has to do with the direct product of the normal subgroups of $G$, but I'm not sure how to relate that to what I need.
 A: 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.
A: We can completely bypass Schur-Zassenhaus theorem to show that $n_{11}=n_7=1$ instead of focusing on $n_{13}$.
Applying Sylow theorems, we have:
$$n_{11}| 2\times 7\times 13\quad  \text{and}\quad n_{11}\equiv 1\pmod {11}$$
 which implies $n_{11}=1$ and $Q_{11}\triangleleft G$, where $Q_{11}$ is Sylow $11$-subgroup.
Now consider $G/Q_{11}$ and since 
$$n_{7}'|2\times 13\quad \text{and}\quad n_7\equiv 1\pmod 7$$
we have $n_7'=1$ where $n_7'$ is the number of Sylow $7$-subgroups of $G/Q_{11}$.
By correspondence theorem, we have only one Sylow $7$-subgroup in $G$, i.e., $n_7=1$. (Details refer to this)
Now $Q_7Q_{11}\triangleleft G$ thus $Q_7Q_{11}Q_{13}\leqslant G $ by Sylow. And the index of $Q_7Q_{11}Q_{13}$ in $G$ is $2$. 
To see that $Q_7Q_{11}Q_{13}$ is cyclic, it is sufficient to notice two facts:


*

*Groups of order $pq$ where $p<q$ are prime with $q\not\equiv 1\pmod p$ are cyclic. (Plug in $p=11$, $q=13$.)

*Any homomorphism $\varphi: Q_{11}Q_{13}\cong\mathbb Z_{143}\to \mathbb Z_{6}\cong \operatorname{Aut}(Q_7)$ is trivial. Thus $Q_7Q_{11}Q_{13}\cong\mathbb Z_{143}\times \mathbb Z_{7}\cong\mathbb Z_{1001}$.
