Let \(H_1, H_2\) be two distinct subgroups of a finite group G, each of order 2. Let \(H\) be the smallest subgroup containing \(H_1\) and \(H_2\). Then order of \(H\) is

A. always 2

B. always 4

C. always 8

D. none of the above


Let’s check out the simplest example where we suspect things might get wrong.

Take \(S_3\) as the group. It has 3 order 2 subgroups. If we take any two of those then the subgroup generated by (i.e the smallest subgroup containing) should have order greater than 2; so it must have order 3 or 6 and that itself leads us to conclude that the answer is none of the above.

For example take \(<(1 2)>=H_1,<(13)>=H_2\). Then \(H\) contains \( (1 2), (1 3), (1 2)(1 3)=(1 3 2), (1 3)(1 2)=(1 2 3)\). Therefore \(H\) has atleast 4 elements, so it must have 6 elements i.e, \(H=S_3\).