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We have |G|=60=22⋅3⋅5|G|=60=22⋅3⋅5 Let npnp be the number of Sylow pp-subgroups of GG. By Sylow's third theorem, we have n3∈{1,4,10}n3∈{1,4,10}. But GG contains 2020 elements of order 33, each of which generates a Sylow 33-subgroup, so n3=10n3=10 Similarly n5∈{1,6}n5∈{1,6} and contains 3030 elements each generating a Sylow 55-subgroup so n5=6n5=6 Consider n2n2. By Sylow's third theorem, we have n2={1,3,5,15}