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everyone. I am afraid that my question is too trivial. But here it is. The Klein four group is the first counterexample to the the statement: "If all proper subgroups of a group are cyclic, then the group is cyclic." I am looking for other examples, if any. Are there?

Thanks in advance.

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up vote 13 down vote accepted

The example of the Klein group, $\mathbb{Z}_2 \times \mathbb{Z}_2$, generalizes naturally to $\mathbb{Z}_p \times \mathbb{Z}_p$ for any prime $p$. Other finite examples include any nonabelian group of order $pq$, where $p$ and $q$ are primes (such a group exists whenever $p-1$ is divisible by $q$).

But there are even infinite examples. For all sufficiently large primes $p$, there exist infinite groups all of whose proper nontrivial subgroups are cyclic of order $p$. These are the Tarski monsters.

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There is in fact a complete classification available. The harder p-group case is covered in Berkovich's book on groups of prime power order, and the easier non-p-group case follows from general results, and basically gives $C_p\rtimes C_r$, where $r$ is a prime power. – user641 Jan 11 '11 at 7:41
Does Berkovich's book cover the case of infinite p-groups, or just finite ones? – Chris Eagle Jan 11 '11 at 7:54
Berkovich's book is only about finite p-groups (but it covers a huge amount about these) – Tobias Kildetoft Jan 11 '11 at 8:51
The classification says that if $p=2$ then the groups with a cyclic subgroup of index $p$ which are not themselves cyclic are the dihedral, the semidihedral, the quasidihedral and the quaternion groups. For $p$ odd, it says that the only possibility is the group which has he same presentation as the quasidihedral group (with 2 replaced by $p$ in the exponents) – Tobias Kildetoft Jan 12 '11 at 8:45
By considering the Frattini subgroup of a p-group, you can show the only p-group examples are $C_p\times C_p$ and the quaternion group (this follows from Proposition 1.3, pg 25, in Berkovich's Groups of Prime Power Order, vol. I). The non-p-group case can be seen from work by Miller-Moreno on minimal non-abelian groups, which in the non-p-group case are of the form $(C_p)^k\rtimes C_r$, where $r=q^m$, $p$ and $q$ primes. Here we need $k=1$ and the subgroup of index $q$ in $C_r$ to act trivially on $C_p$. – user641 Jan 12 '11 at 9:50

$\Sigma_3$ (the symmetric group on three elements) is the next smallest example.

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That is $S_3$. Than you. – Chulumba Jan 11 '11 at 15:11
This is the $p=3$, $q=2$ case of Chris's second sentence. – Jonas Meyer Jan 11 '11 at 17:27
But isn't $mathbf(Z_6)$ the next smallest? – Chulumba Jan 11 '11 at 17:33
@Chulumba: $\mathbf{Z}_6$ is cyclic. – Jonas Meyer Jan 11 '11 at 17:39
There are many different ways of denoting this group. – Sean Tilson Jan 11 '11 at 18:33

Another collection of examples, not yet mentioned, are the Prüfer groups (also known as quasicyclic groups); it is sometimes denoted $\mathbb{Z}_{p^{\infty}}$. They are infinite, but every proper subgroup is finite (and every nontrivial quotient is isomorphic to the original group). Here are three descriptions:

  1. For a fixed prime $p$, let $\mathbb{Z}_{p^{\infty}}$ be the group of all $p^k$-th complex roots of unity for all $k\geq 0$, with the group operation being multiplication. It is not hard to show that every proper subgroup is generated by a $p^n$th primitive complex root of unity for some $n$, hence is cyclic.

  2. As an alternative description, consider subgroup of the additive group $\mathbb{Q}/\mathbb{Z}$ that consists of all classes represented by a fraction whose denominator is a power of $p$.

  3. A final description: consider the collection of groups $\{\mathbb{Z}/p^n\mathbb{Z}\}_{n\in\mathbb{N}}$, with injections $i_n\colon\mathbb{Z}/p^n\mathbb{Z}\to\mathbb{Z}/p^{n+1}\mathbb{Z}$ given by $i_n(a+p^n\mathbb{Z}) = pa+p^{n+1}\mathbb{Z}$. Then $\mathbb{Z}_{p^{\infty}}$ is the direct limit $\displaystyle\lim_{\longrightarrow}\mathbb{Z}/p^n\mathbb{Z}$ of this direct system.

The additive group $\mathbb{Q}$ is almost such a group, in that every finitely generated subgroup is cyclic, but of course it contains noncyclic proper subgroups (e.g., the pre-image of the Prüfer group for some $p$).

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Thank you. I learned a lot. – Chulumba Jan 20 '11 at 14:09

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