Question

Let G be a group of order 39 such that it has exactly one subgroup of order 3 and exactly one subgroup of order 13. Then, which one of the following statements is TRUE?

  1. G is necessarily cyclic
  2. G is abelian but need not be cyclic
  3. G need not be abelian
  4. G has 13 elements of order 13

Answer: G is necessarily cyclic

Solution :-

Let H and K be the subgroup of G of orders 3 and 13 respectively. So, Given that H and K are exactly one subgroup of G. Hence H and K are unique subgroups of G. And we know that a unique subgroup of any group is always normal in that group. So, subgroups H and K are normal subgroups in group G.

(1). H and K are normal subgroups of G.

Now, H and K are subgroups of G, and we know that the intersection of two subgroups is also a subgroup.
So, H ∩ K is also a subgroup of G. And H ∩ K is subgroup of H and K also because H ∩ K is subset of H and also K.

We know order of any subgroup of a group is divides order of that group.
So, |H ∩ K| divides |H| and |H ∩ K| divides |K| also |H ∩ K| divides 3 and |H ∩ K| divides 13 (because |H| = 3 and |K| = 13)
Hence, |H ∩ K| = 1 because the only positive integer that divides both 3 and 13 is 1.

We know that any group of order 1 contains only identity element. So,
(2). H ∩ K = {e}.

Now, because H and K are subgroups of G, so
|HK| = (|H|·|K|)/|H ∩ K| = (3·13)/1 = 39 = |G|. because 39 is the order of group G. It means HK = G.

(3) HK = G.

Now we say that G is the internal direct product of H and K.
G = H ⊕ K. Also, internal direct product is isomorphic to external direct product. So,
G = H ⊕ K ≅ H × K; here the order of H is 3 and 3 is prime so H is cyclic also the order of K is 13 and 13 is prime so K is also cyclic.

We know that the external direct products of two cyclic groups or subgroups are cyclic if the order of those two groups or subgroups is coprime. Here the order of subgroups H and K are coprime. so, their external direct product is cyclic. Hence, H × K is cyclic. also, a cyclic group is isomorphic to a cyclic group so, H ⊕ K is also cyclic.

Hence, G is cyclic.

So, correct answer is option-1, → G is necessarily cyclic

#Group Theory #IIT JAM PYQ 2024 #Mathematics #IIT JAM PYQs