Please solve this question part a b and c Thanks Let G be ab

Please solve this question, part a, b and c. Thanks!

Let G be abelian with G 441 72.3

(i) |G| = 441 meaning p can only be 49 and 9

number of possible Sylow 3-subgroups of G = 1 (since 4 does not divide 441). therefore G has a normal subgroup of order 3, and in particular cannot be simple.

number of possible Sylow 7-subgroups of G = 1 (since 8 does not divide 441). again this means that this subgroup is normal, and G is not simple.