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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.

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    @Serkan: With full details your approach should be shorter (and easier) than the one given by DonAntonio.2012-08-16

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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.

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    why is $n_{13} = 14$ not possible?2012-08-15
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    @Serkan: Hehe, Hall's theorem or Zassenhaus's theorem (both easy). But yes, it is a hassle to show that $n_{13} \neq 14$ just from Sylow counting.2012-08-15
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    DonAntonio: You can fix it by considering $G/Q_{11}$ then $G/Q_7Q_{11}$, then $Q_{13}(Q_7 Q_{11})$ is the needed subgroup.2012-08-15
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    Indeed @JackSchmidt, that seems simpler and I'll add an observation to my answer.2012-08-15
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    Hi, question: how do we know that $n_{13}=1$ and that $P$ is isomorphic to $C_{11}$ X $C_{13}$ X $C_{7}$? Also, each of those subgroups are also cyclic, no? They have prime orders. How exactly does abelianity follow from all of this? (I have not been introduced to those extra theorems yet)2017-11-18
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    PBJ Direct product of abelian groups is abelian, and the rest follows at once from Sylow theorems.2017-11-18
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    Thank you DonAntonio.2017-11-19