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..
 A: 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.
