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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 :)

Shaun
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5 Answers5

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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 $F_\infty$.

Alessandro Codenotti
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An example of a first-order property which is inherited by quotients but not by subgroups is $\forall x\exists y[y^2=x]$. It is inherited by quotients because it's a positive sentence; it's not inherited by subgroups because it holds in $(\mathbb Q,+)$ but not in $(\mathbb Z,+)$.

Nobody asked about products but let me also remark that this property, being defined by a Horn sentence, is also preserved by (finite or infinite) direct products, and even reduced direct products.

bof
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  • This is interesting! Thank you! What name does this type of group theory go by? (Please excuse my ignorance.) – Shaun Dec 05 '20 at 00:45
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    I have no idea. (Sorry, I'm not an algebraist, let alone a group theorist!) The much stronger property of being a divisible group would also work here. – bof Dec 05 '20 at 00:47
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Being perfect is such an example. If $G$ is perfect, then $G/N$ is also perfect for all normal subgroups $N$ of $G$ since in general $(G/N)' = G'N/N$. But perfectness is clearly not preserved by subgroups.

the_fox
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The property of having no nontrivial quotient groups (aka being simple or the trivial group).

Ben
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The property of being a $T$- group is an example. That's that every subnormal subgroup (normal subgroup of a normal subgroup) is normal.

The property is inherited by normal subgroups, but not by subgroups in general.

And it's inherited by homomorphic images, hence quotients.