Let

Step 1 of 3

Let us consider a group of order i.e., .

Suppose has only one subgroup of order and a subgroup of order .

First, let us prove that is cyclic.

Let, be a subgroup of order generated by and be a subgroup of order generated by .

Then, by the definition of cyclic subgroup,

If possible, let us assume that .

Then, for some integer , .

It follows that

Here, is the identity element.

Since and , therefore, .

Now, . Therefore, implies .

By definition of divisibility, there exists an integer such that

Thus,

It shows that