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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..

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    $\begingroup$ And (vacuously) if $G$ is simple. $\endgroup$
    – hardmath
    Jan 8, 2016 at 6:24
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    $\begingroup$ $S_3$ is a counterexample to your "well known" fact. $\endgroup$
    – user138530
    Jan 8, 2016 at 6:48
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    $\begingroup$ 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, 2016 at 9:23
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    $\begingroup$ $S_n$ $(n\neq 4)$ 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, 2016 at 12:17
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    $\begingroup$ @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, 2016 at 12:43

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Levi
    Oct 26, 2019 at 11:48
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    $\begingroup$ 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, 2019 at 17:26

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