# Groups whose proper quotients are cyclic

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..

• And (vacuously) if $G$ is simple. – hardmath Jan 8 '16 at 6:24
• $S_3$ is a counterexample to your "well known" fact. – user138530 Jan 8 '16 at 6:48
• 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. – Derek Holt Jan 8 '16 at 9:23
• $S_n$ $(n\neq 4)$ is already included in the rest of your list. It is cyclic for $n=2$, dihedral for $n=3$ and almost simple for $n>4$. – verret Jan 8 '16 at 12:17
• @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... – 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.