TITLE: Abelian subgroup of a group of order $2002$
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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 $Q_7Q_{11}Q_{13}\cong\mathbb Z_{143}\times \mathbb Z_{7}\cong\mathbb Z_{1001}$.