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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. – j.p. Aug 16 '12 at 7:47
up vote 1 down vote accepted

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? – user33321 Aug 15 '12 at 18:56
@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. – Jack Schmidt Aug 15 '12 at 19:16
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. – Jack Schmidt Aug 15 '12 at 19:19
Indeed @JackSchmidt, that seems simpler and I'll add an observation to my answer. – DonAntonio Aug 15 '12 at 23:20

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