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Let $G$ be a nonabelian finite simple group all of whose Sylow subgroups of odd order are cyclic.

If we further assume that its Sylow $2$-subgroup is dihedral, then due to Suzuki, we know that $G\cong \operatorname{PSL}(2,p)$ for a prime $p>3$.

Without any further assumption, what is the list of finite nonabelian simple groups all of whose Sylow subgroups of odd order are cyclic?

Secondly, is there any reference for that not appealing to the full classification of finite simple groups?

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Such simple groups are a subset of the finite simple "thin" groups, and the latter have been classified (by Michael Aschbacher in 1976 and 1978). A $2$-local subgroup of a finite group is the normalizer of some non-trivial $2$-subgroup. A finite simple group $G$ is said to be "thin" if all its $2$-local subgroups have cyclic Sylow subgroups for all odd primes.

We may note that Feit and Thompson proved early in their odd order paper (using transfer theorems) that if $G$ is a finite group of odd order which has no elementary Abelian subgroup $p$-subgroup of rank $3$ or more for any prime $p$, then $G$ has a normal Sylow $q$-subgroup where $q$ is the largest prime divisor of $|G|$. Hence (without using the full force of the odd order theorem), it is indeed the case that if a finite non-Abelian simple group $G$ has cyclic Sylow subgroups for all odd primes $p$, then $G$ has even order, and is, in particular, a thin group. However, some thin groups do not satisfy the condition of the question.

According to Aschbacher's classification, the list of thin finite simple groups is ${\rm SL}(2,2^{n})$ and ${\rm PSL}(2,q)$ for $q$ an odd prime power (but we can remove the cases when the odd $q$ is not prime in answer to the present question since the Sylow subgroup of order $q$ is not cyclic), ${\rm PSL}(3,4)$ (but we can remove this in answer to the present question since it has a non-cyclic subgroup of order $9$), ${\rm PSL}(3,p)$ for $p$ of the form $1+2^{a}3^{b}$ (which can be removed in answer to the present question when $p$ is odd, since there are non-cyclic $p$-subgroups in that case), ${\rm PSU}(3,p)$ ($p = 2^{a}3^{b}-1$, $b \in \{0,1\}$ (but these can be omitted in answer to the present question as there are non-cyclic $p$-subgroups), ${\rm PSU}(3,2^{n})$ (but these can be omitted in answer to the present question, since ${\rm PSU}(3,4)$ has an elementary Abelian subgroup of order $25$), ${\rm Sz}(2^{n})$ , the Tits group $^{2}F_{4}(2)^{\prime}$ (which can be omitted in answer to the present question, as it contains an elementary Abelian subgroup of order $9$), the Steinberg triality group $^{3}D_{4}(2)$ (which can be omitted in answer to the present question since it has elementary Abelian subgroups of order $9$), the Mathieu group $M_{11}$ (which can be omitted in answer to the present question as it has elementary Abelian subgroups of order $9$), and the first Janko group $J_{1}.$

Later edit: Notice that Aschbacher's classification of the thin finite simple groups predates the full classification of the finite simple groups. As a matter of historical interest, the classification of finite simple groups in which all $2$-local subgroups have low $p$-rank for all odd primes $p$ is a difficult part of the current proofs (and of ongoing revisions) of the classification of finite simple groups.

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  • $\begingroup$ Thank you very much for your answer. If I understand correctly, the answer is $PSL(2,2^n)$ , $PSL(2,p)$ , $Sz(2,2^n)$ and $J_1$where their Sylow $2$-subgroups are elementer abelian, dihedral, Suzuki 2-group and elementer abelian, respectivly. $\endgroup$ – mesel Apr 4 at 11:18
  • $\begingroup$ Yes, I think that is correct. $\endgroup$ – Geoff Robinson Apr 4 at 11:21
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Partial answer:

Any group whose order is a power of $2$ times a squarefree number must satisfy this condition. This condition holds for some of Suzuki's groups ${\rm Sz}(2^k)$ (with $k \geq 3$ odd), which have order $2^{2k} (2^{2k} + 1) (2^k - 1)$; and it seems from https://en.wikipedia.org/wiki/Suzuki_groups#Conjugacy_classes that even the Suzuki groups whose orders do have odd square factors (such as ${\rm Sz}(2^5)$, of order $2^{10} \, 5^2 \, 31 \!\cdot\! 41$) still have all their odd-order Sylow groups cyclic.

Another example is the first Janko group $J_1$, of order $2^3 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 19$. The appearance of a sporadic group suggests that the full answer will require at least some nontrivial part of the classification theorem.

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    $\begingroup$ NIce! One way to phrase this is that we have $2^{2k}+1 = (2^k + 1 - 2^{ \frac{k+1}{2} } ) (2^k + 1 + 2^{\frac{k-1}{2}})$ and the pairwise gcds of $2^k-1$, $2^k + 1 - 2^{ \frac{k+1}{2} } $, and $2^k + 1 + 2^{ \frac{k+1}{2} } $ are each a power of $2$, so any odd prime power dividing the order divides one of those, and Suzuki found cyclic subgroups of those three orders, so any Sylow subgroup is contained in a cyclic subgroup and hence cyclic. $\endgroup$ – Will Sawin Apr 4 at 1:58
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    $\begingroup$ One family of examples is $\operatorname{PSL}_2(2^n)$, $n > 1$. Also, I think it was shown by Suzuki that every subgroup of odd order in $\operatorname{Sz}(2^{2n+1})$ is cyclic. $\endgroup$ – Mikko Korhonen Apr 4 at 2:28

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