This is Exercise 4.6a of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to this search and Approach0, it is new to MSE.
The Details:
From p.116 of Roman's book,
Definition: Let $\mathcal{P}$ be a property of groups. [ . . . ] We write $G\in \mathcal{P}$ if $G$ has property $\mathcal{P}$. [ . . . ] A property $\mathcal{P}$ of groups is inherited by subgroups if $$G\in \mathcal{P}\text{ and }H\le G\implies H\in \mathcal{P},$$ and $\mathcal{P}$ is inherited by quotients if $$G\in \mathcal{P}\text{ and }H\unlhd G\implies G/H\in \mathcal{P}.$$
The Question:
Find a property of groups that is inherited by quotients but not by subgroups.
Thoughts:
My first guess – I don't know why – was to let $\mathcal{P}$ be "nonabelian", which fell quickly in the light that $G/G=\{G\}$ is abelian for all groups $G$.
Having $\mathcal{P}$ be "abelian" won't work, since it is inherited by both quotients and subgroups; out goes "cyclic" along with it.
"Finite" is not an example, obviously.
"Solvable" is inherited by both quotients and subgroups. This can be checked on Wikipedia. (I haven't done the Maths, ironically; but with such a standard property of a group, I reckon it's more reliable here than not).
All nilpotent groups are solvable, so they're out; the same goes for supersolvable groups and, in turn, metacyclic groups. Metabelian groups are also solvable.
"Simple" appears not to work.
It would be fiendish if Roman had not defined a group property as required prior to the exercise, so, if all else fails, I could sieve through the definitions in the first 144 pages; this is cheating though (and quite a daunting task!).
Please help 🙂
Best Answer
Being finitely generated is such a property.
Let $F_n$ denote the free group on $n$ generators. If $F_n\to G$ witnesses that $G$ is f.g. then $F_n\to G\to G/H$ witnesses that the quotient also is, while $F_2$ contains a subgroup isomorphic to $F_\infty$.