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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?

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I am having trouble understanding your question. Is $S_3$ a valid example? –  S. Carnahan Mar 25 '13 at 9:24
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3 Answers

up vote 3 down vote accepted

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.)

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@Nick: Isn't every group with all Sylow subgroups cyclic cyclic-by-cyclic? I seem to remember this result from the first group theory course. –  Mark Sapir Mar 25 '13 at 11:28
    
@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 Sylow-cyclic groups, and didn't realise that a stronger statement was so easy.) –  Nick Gill Mar 25 '13 at 13:59
    
@Nick,Mark:Is there a classification of Sylow-cyclic groups in which cases they are cyclic or noncyclic? –  Tom Mar 26 '13 at 1:26
    
@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. –  Mark Sapir Mar 26 '13 at 1:40
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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 $|G|.$

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@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? –  Tom Mar 26 '13 at 1:17
    
@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 $155$. –  Geoff Robinson Mar 26 '13 at 7:12
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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.

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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. –  Chandan Singh Dalawat Mar 26 '13 at 1:30
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