Show that if G is a group of order 2002 2 7 11 13 then G

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

Solution

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If np=np= number of Sylow p-subgroups of GG , then n13=1=n11n13=1=n11, thus letting QpQp be a Sylow p-sbgp., we have that

Q11,Q13GQ11Q13GP:=Q11Q13Q7GQ11,Q13GQ11Q13GP:=Q11Q13Q7G

Abelianity follows from the fact that the only group of order 10011001 is the cyclic one, as PC11×C13×C7,Cm:=PC11×C13×C7,Cm:= the cyclic group of order m