It is well known that a finite group whose all proper subgroups are cyclic is either cyclic or direct product of quaternion group with cyclic group of odd order (am I correct?)

Question: What are the finite groups, whose proper quotients are cyclic?

[Here proper quotient of $G$ means the quotient of $G$ by non-trivial subgroup]

My (incomplete) answer is: if $G$ is a $p$-group then it should be cyclic or $C_p\times C_p$. In non-nilpotent group, some examples are $D_{2p}$ (dihedral groups of order $2p$, $p$ being prime).

Here is a list of such groups including those appearing in comments; I do not know whether they are classified.

  • Cyclic, dihedral of order $2p$ where $p$ is prime,

  • $S_n$ ($n\neq 4$),

  • simple groups,

  • $N\rtimes C_p$ where $N$ is non-abelian simple group and semi-direct product is not direct product..

  • 1
    $\begingroup$ And (vacuously) if $G$ is simple. $\endgroup$ – hardmath Jan 8 '16 at 6:24
  • 1
    $\begingroup$ $S_3$ is a counterexample to your "well known" fact. $\endgroup$ – user138530 Jan 8 '16 at 6:48
  • 5
    $\begingroup$ More generally, groups $A$ with $S \le A \le {\rm Aut}(S)$, where $S$ is nonabelian simple and $A/S$ is cyclic. There are examples of these that are not semidirect products $S \rtimes C_m$. All non-solvable examples are of this type. $\endgroup$ – Derek Holt Jan 8 '16 at 9:23
  • 1
    $\begingroup$ $S_n$ $(n\neq 4)$ is already included in the rest of your list. It is cyclic for $n=2$, dihedral for $n=3$ and almost simple for $n>4$. $\endgroup$ – verret Jan 8 '16 at 12:17
  • 3
    $\begingroup$ @Derek, these aren't the only nonsoluble examples, take the wreath product of a simple group with a cyclic group, for example. These are more or less the only examples though, up to some mild "twisting". Clearly, a non-soluble example must have trivial soluble radical and a unique minimal normal subgroup... $\endgroup$ – verret Jan 8 '16 at 12:43

I claim that $G$ is one of the following:

  • Cyclic

  • $C_p^2$

  • A group of the form $(C_p^n).C$, where $C$ is cyclic, and acts faithfully and irreducibly on $C_p^n$.

  • A group of the form $T^n.C$, where $C$ is cyclic, $T$ is non-abelian simple, $G$ acts transitively on the $n$ copies of $T$, and $T^n$ has trivial centraliser in $G$.

Note that this is a characterisation: every group above has the required property.

Here's a sketch of a proof. If $G$ is non-soluble, then it has a unique minimal normal subgroup, which must be of the form $T^n$, have trivial centraliser, and the quotient is cyclic, and we get the last case.

We now assume $G$ is soluble. Let $P=O_p(G)$ for some prime $p$. Note that this is characteristic in $G$, and thus so is its Frattini subgroup $\phi(P)$. It follows that either $\phi(P)=1$ or $P/\phi(P)$ is cyclic. In the first case, $P$ is elementary abelian, and in the second $P$ is cyclic.

This is true for each prime, and it follows that the Fitting subgroup $F$ of $G$ (which is the direct product of thte $O_p$'s as we run over the primes $p$) is either cyclic or elementary abelian. (Otherwise, we get a non-cyclic quotient.) In particular, $F$ is abelian and $G/F$ acts faithfully on $F$, since $F$ is the Fitting subgroup of a soluble group.

Suppose first that $F$ is cyclic. If $F=G$, then $G$ is cyclic. Otherwise, $G/F$ acts non-trivially on $F$ so, unless $F=C_p$, we get a non-trivial non-cyclic quotient. So $G\leq \mathrm{AGL}(1,p)$, which is included in the third case.

Finally, suppose that $F$ is elementary abelian, say $F=C_p^n$. If $F=G$, then $n\leq 2$. Otherwise, we fall in the third case.

| cite | improve this answer | |
  • $\begingroup$ I struggle to understand the notation in your answer. With the dot symbol, do you mean a subdirect product? What is $O_p(G)$? Do you maybe have a reference for this result? $\endgroup$ – rawbacon Oct 26 '19 at 11:48
  • 1
    $\begingroup$ A dot just denotes an arbitrary extension. (This is sometimes called Atlas notation.) $O_p(G)$ is the $p$-core, that is the largest normal $p$-subgroup. (Equivalently, the intersection of the Sylow $p$-subgroups.) $\endgroup$ – verret Oct 26 '19 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.