abstract algebra?

Question

suppose that IGI=pq; p and q primes prove that evrey proper subgroup of G is cyclic'

Answer 1

IF S is a proper subgroup of G then s =|S| (assuming this means the order of S) must divide the order of G, pq. THe only numbers that do so are p and q both prime. So the order of S is prime and hence it is Cyclic.

Answer 2

Lagrange stated and proved that if G is a finite group and H is a subgroup of G, Then the order of H is a divisor of the order of G.

Since the order of G = pq, the order of any subgroup must be either p or q (or 1; we know that there exists only one group of order one and it is necessarily cyclic) since p and q are both prime numbers. Any subgroup is itself a group. Groups whose order is a prime number are necessarily unique and are, therefore, cyclic.

Answer 3

Any proper subgroup H of G must have order p or q.
Since p and q are primes, H is a group of prime
order, so it is cyclic.