a group with all sylow p subgroups cyclic  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?
 A: There is a complete classification of groups with all Sylow-subgroups being cyclic. In fact one can weaken this: we say that a group $G$ is almost Sylow-cyclic if every Sylow subgroup of $G$ has a cyclic subgroup of index at most $2$. Almost Sylow-cyclic groups are fully classified in two papers:

M. Suzuki, On finite groups with cyclic Sylow subgroups for all odd primes, Amer. J. Math. 77 (1955) 657–691.
W.J. Wong, On finite groups with semi-dihedral Sylow 2-subgroups, J. Algebra 4 (1966) 52–63.

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 Sylow-cyclic, as the group $PSL_2(7)$ demonstrates.)
A: Take $p$ and $q$ two prime numbers with $q$ dividing $p-1$. Then there is a nonabelian semi-direct 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 $p-1$ is the order of the automorphism group of $C_p$, when $p$ is an odd prime.
A: 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 group-theoretic analysis in the proof of the Feit-Thompson 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 $|G|.$
