Let G be a group of order pqwhere p and q are prime numbers. Such that p q then, G can havea)atmost one subgroup of order p.b)atleast one subgroup of order q.c)atleast one subgroup of order p.d)atmostone subgroup of order q.Correct answer is option 'C'. Can you explain this answer?

Question

Let G be a group of order pqwhere p and q are prime numbers. Such that p > q then, G can havea)atmost one subgroup of order p.b)atleast one subgroup of order q.c)atleast one subgroup of order p.d)atmostone subgroup of order q.Correct answer is option 'C'. Can you explain this answer?

Answer

By Cauchy's theorem, since p and q are primes dividing the order of the group G, there exist elements a and b in G of order p and q respectively.

Consider the subgroup H = . Since the order of a is p, the order of H is p. Therefore, H is a cyclic subgroup of G of order p.

Similarly, consider the subgroup K = . Since the order of b is q, the order of K is q. Therefore, K is a cyclic subgroup of G of order q.

Now, let's show that H and K are distinct subgroups. Suppose for contradiction that H and K are not distinct, i.e., H = K. Since a is in both H and K, we have a = b^k for some integer k. Taking the order of both sides, we have p = q^k. Since p and q are distinct primes, this is a contradiction. Therefore, H and K are distinct subgroups.

Now, consider the intersection of H and K, denoted by H ∩ K. Since both H and K are cyclic subgroups, their intersection is the identity element e of G. Thus, H ∩ K = {e}.

By Lagrange's theorem, the order of a subgroup divides the order of the group. Therefore, the order of H and K divides the order of G. Since p and q are distinct primes, the only way for both p and q to divide the order of G is if the order of G is pq.

Therefore, G is the direct product of the subgroups H and K, denoted by G = H × K.

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