abstract algebra?

Question

letIGI =15 if G has only one subgroup of order 3 and only one of order 5 prove that G is cyclic generize for IGI =pq

Answer 1

There's only one element of order 1.

If there's only one subgroup of order 3, there are no more than 2 elements of order 3, namely the elements of that group. (If there were any others, they would generate their own subgroups of order 3 in the obvious way.)

Similarly, there are only 4 elements of order 5.

Everything else is an element of order 15.

Generalizing this, you have

1 element of order 1
p-1 elements of order p
q-1 elements of order 1
pq > p + q + 1 = 1 + (p-1) + (q-1).

Other than that last inequality, which is obvious once you assume p>q and note that that also implies p>2, we're done!

Q.E.D. (except for checking the inequality)