Finite groups in which all proper subgroups are cyclic

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

Would you please tell me a reference?

• 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 $q^n$. – Arturo Magidin Nov 2 '16 at 22:37
• 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. – Todd Leason Nov 2 '16 at 22:49

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)

• Your case (3) seems a bit funny. If $r\equiv 1\pmod{p}$ then $r^q\equiv 1\pmod{p}$ automatically? – Neil Strickland Dec 2 '19 at 10:07
• @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. – Yanior Weg Dec 2 '19 at 17:51
• Isn't this what @ArturoMagidin said? – LSpice Dec 2 '19 at 17:58
• @LSp Arturo didn't give the Miller, Moreno reference. – Gerry Myerson Dec 2 '19 at 22:01