Let $|G|=30$. I have prove that there is the only subgroup of order $15$, which I'll denote $H$. Now I do know how to classify the group. After thinking, I made the following steps.
1) Possible order of subgroup $K$ of $G$ of order 2 are 1, 3, 5, 15.
Case 1. if $G$ contain only one element of order 2, then $G \cong Z_{30}$.
Now I cannot solve for the next steps. Please give me any hints or any other method.
Hints:
1) There is only one possible abelian group of order $\,30\,$
2) Any group $\,G\,$ of order 30 has a subgroup $\,H\,$ of order $\,15\,$, which is normal and abelian -- in fact, cyclic -- (why and why?), and thus $\,G\cong H\rtimes Q\,$ , for some subgroup $\,Q:=\langle\,q\,\rangle\,$ of order two.
Since $\,\operatorname{Aut}(H)\cong C_2\times C_4\,$ (why?) , there are at least four possible homomorphisms $\,Q\to\operatorname{Aut}(H)\,$ , all of them convolutions: (i) mapping $\,q\,$ to the generator of the factor $\,C_2\,$ , (ii) to $\,p^2\;,\;\;p=$ the generator of $\,C_4\,$ , and (iii) to the element $\,(q,p)\in C_2\times C_4\,$ (the trivial homomorphism gives the abelian group we already had before).
Check the above three non-trivial homomorphisms give three non-isomorphic groups of order $\;30\;$ .
There is a general description of groups with cyclic Sylow subgroups:
Marshall Hall, The Theory of Groups - Theorem 9.4.3.
