If there exist a non cyclic group $G$ with all sylow $p$subgroups cyclic,and the normal $p_1$complement $M$ for $G$ is cyclic,here $p_1$ is the smallest factor of $G$?And when does it always exist?

There is a complete classification of groups with all Sylowsubgroups being cyclic. In fact one can weaken this: we say that a group $G$ is almost Sylowcyclic if every Sylow subgroup of $G$ has a cyclic subgroup of index at most $2$. Almost Sylowcyclic groups are fully classified in two papers:
You may also be interested in an old paper by Holder from 1895 who proved that every group with all Sylow subgroups cyclic is solvable. (This is not true under the weaker supposition that a group is almost Sylowcyclic, as the group $PSL_2(7)$ demonstrates.) 


Take $p$ and $q$ two prime numbers with $q$ dividing $p1$. Then there is a nonabelian semidirect product $C_p \rtimes C_q$ which seems to be what you want, if i understand the question well. Here $C_n$ is the cyclic group of order $n$, and note that $p1$ is the order of the automorphism group of $C_p$, when $p$ is an odd prime. 


If the finite group $G$ has a cyclic Sylow $p$subgroup $P,$ where $p$ is the smallest prime divisor of $G,$ then $G$ always has a normal $p$complement by (for example) Burnside's transfer theorem, though that normal $p$complement need not be cyclic. However,if the remaining Sylow subgroups of $G$ are also cyclic and $C_{G}(P) = P,$ the normal $p$complement will also be cyclic. In the early grouptheoretic analysis in the proof of the FeitThompson odd order theorem, it is proved that if $G$ is a finite group of odd order and $G$ contains no elementary Abelian subgroup of rank $3$ for any prime, then $G$ has a normal Sylow $q$group where $q$ is the largest prime divisor of $G.$ 

