Question

In a group of order 15 the number of subgroups of order 3 is

Answer 1

Using Sylow's theorem we show that there are subgroups H and K of order 5 and 3 respectively (and the are cyclic of course). Due to 15=3*5 there is only one subgroup of order 5 and only one subgroup of order 3 (Sylow's theorem says that Np≡1 mod p, where Np is the number of subgroups with order p, p∈{3,5}, and Np divides |G| so Np=1). Moreover, these subgroups are normal. That is why G=K∗H and cyclic.