Best Answer:
Use the Sylow Theorems.

Let m = number of 5-Sylows and n = number of 7- Sylows. Then, (a) m|7 and m = 1 mod 5 ==> m = 1.

(b) n|5 and n = 1 mod 7 ==> n = 1.

Thus, there exist unique and thus normal subgroups of order 5 and order 7 in our group G of order 35.

Thus, G is isomorphic to the direct product

(5-Sylow) X (7-Sylow) = Z_5 X Z_7, since groups of prime order are cyclic.

Finally, by the Chinese Remainder Theorem or otherwise,

Z_5 X Z_7 = Z_35, the cyclic group of order 35.

(One can use the fact that there is only one group up to isomorphism with one normal subgroup of orders 5 and 7), and the fact that Z_35 satisfies these criteria to bypass the last step or two, if you need to do this!)

Good luck!

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