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?

1$\begingroup$ I am having trouble understanding your question. Is $S_3$ a valid example? $\endgroup$– S. Carnahan ♦Mar 25 '13 at 9:24
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:
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 semidihedral Sylow 2subgroups, 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 Sylowcyclic, as the group $PSL_2(7)$ demonstrates.)

$\begingroup$ @Nick: Isn't every group with all Sylow subgroups cyclic cyclicbycyclic? I seem to remember this result from the first group theory course. $\endgroup$– user6976Mar 25 '13 at 11:28

$\begingroup$ @Mark, you're right. And, thinking about it, it's easy to prove by considering $F^*(G)$. In light of this, it is perhaps surprising that Holder's result is still referenced. (I read about Holder's result, when I was investigating almost Sylowcyclic groups, and didn't realise that a stronger statement was so easy.) $\endgroup$ Mar 25 '13 at 13:59

$\begingroup$ @Nick,Mark:Is there a classification of Sylowcyclic groups in which cases they are cyclic or noncyclic? $\endgroup$– TomMar 26 '13 at 1:26

$\begingroup$ @Tom: It should be an easy exercise to describe all these groups and to figure out when they are cyclic or Abelian (which I do not have time to solve right now) because the the automorphism groups of cyclic groups are known. $\endgroup$– user6976Mar 26 '13 at 1:40
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.

$\begingroup$ For a complete classification of such extensions (of a cyclic group by a cyclic group), see for example Mark Reeder's notes on Group Theory (www2.bc.edu/~reederma/806.html). They arise naturally as Galois groups of tamely ramified extensions of local fields, for which see Chapter 16 of Hasse's Nnumber Theory. $\endgroup$ Mar 26 '13 at 1:30
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 $G.$

$\begingroup$ @Geoff:If both the Sylow $p$subgroup $P$ and the normal $p$complement are cyclic,here $p$ is the smallest prime divisor of $G$,if we can get $C_G(P)=P$ or what will happen to $G$?And in what cases $G$ can be noncyclic? $\endgroup$– TomMar 26 '13 at 1:17

$\begingroup$ @Tom: Well, for example, you could get a semidirect product with a normal Sylow $31$subgroup with a cyclic group of order $15$ acting faithfully as a group of automorphisms of the group of order $31$. In that case, the smallest prime divisor of the group order is $3,$ there is a normal $3$complement, but that normal complement os not cyclic it is a Frobenius group of order $155$. $\endgroup$ Mar 26 '13 at 7:12