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Is there any classification of finite group in which all proper subgroups are cyclic?

Would you please tell me a reference?

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    $\begingroup$ In addition to the cyclic groups themselves, you would be looking at the (finite) minimal non-cyclic groups. Geoff Robinson gives an answer at math.stackexchange.com/questions/1934131/… They are: $Q_8$, $C_p\times C_p$, and the unique group of order $pq^n$ with $p\equiv 1\pmod{q}$ with a normal Sylow $p$-subgroup and cyclic groups of order $pq^{n-1}$ and $q^n$. $\endgroup$ – Arturo Magidin Nov 2 '16 at 22:37
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    $\begingroup$ The finite groups those abelian subgroups are cyclic have been classified by Zassenhaus and Suzuki (these are exactly the finite groups with periodic integral cohomology). There are exactly 6 types. A list can be found in Adem, Milgram: Cohomology of finite Groups, Theorem 6.15. Then you can check which of them have all their (proper) subgroups cyclic. $\endgroup$ – Todd Leason Nov 2 '16 at 22:49
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All subgroups of a cyclic group (including the proper ones) are cyclic. And there is the following classification of non-cyclic finite groups, such that all their proper subgroups are cyclic:

A finite group $G$ is a minimal noncyclic group if and only if $G$ is one of the following groups:

1) $C_p × C_p$, where $p$ is a prime

2) $Q_8$

3) $\langle a,b | a^p = b^{q^m} = 1, b^{−1}ab = a^{r}\rangle$, where $p$ and $q$ are distinct primes and $r ≡ 1 \pmod q$, $r^q ≡1 \pmod p$.

This theorem first appeared in "Non-abelian groups in which every subgroup is abelian" by G.A.Miller and H.G.Moreno (1903)

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    $\begingroup$ Your case (3) seems a bit funny. If $r\equiv 1\pmod{p}$ then $r^q\equiv 1\pmod{p}$ automatically? $\endgroup$ – Neil Strickland Dec 2 '19 at 10:07
  • $\begingroup$ @NeilStrickland, thank you for pointing out that sneaky typo. This answer is already 7 month old, but you were the first one who noticed it so far. Now I have corrected it. $\endgroup$ – Yanior Weg Dec 2 '19 at 17:51
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    $\begingroup$ Isn't this what @ArturoMagidin said? $\endgroup$ – LSpice Dec 2 '19 at 17:58
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    $\begingroup$ @LSp Arturo didn't give the Miller, Moreno reference. $\endgroup$ – Gerry Myerson Dec 2 '19 at 22:01

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