In the comments, Charles Hudgins was kind enough to link me to Ying, J.H. Arch. Math (1973) 24: 561. This was a paper leading up to John Hsiao Ying's 1973 PhD thesis, *Relations between subgroups and quotient groups of finite groups*. As far as I can tell, the thesis is not available online anywhere; however, I was recently able to track it down from the library at State University of New York at Binghamton. I will summarize here what I learned from reading his research, and also aggregate some of the information from the help I've gotten from you lovely people. Shout out to Dr. Ying, if you're out there somewhere.

There are still open questions littered throughout this answer and around this topic in general, which will hopefully draw some interest from all of you. Please feel free to ask new questions based on this one, or to edit this answer with updated information if you know something I don't.

# Definitions

First of all, it turns out there is some existing terminology for studying the relationship between subgroups and quotients. As we have seen in the history of this question on MSE and MO, it is easy to confuse the condition defining these groups with a number of subtley different conditions, each which have different consequences. The paper introduces the groups that are the topic of this question as those "satisfying condition (B)," but the thesis goes on to give them a name: **Q-dual groups.** There are several related definitions, the most relevant of which I will condense as follows.

An **S-dual group** satisfies $\forall H\leq G,\:\exists N\unlhd G\text{ s.t. }H\cong G/N$ -- that is, each subgroup of $G$ is isomorphic to a quotient group of $G$.

A **Q-dual group** satisfies $\forall N\unlhd G,\:\exists H\leq G\text{ s.t. }H\cong G/N$ -- that is, each quotient group of $G$ is isomorphic to a subgroup of $G$.

A group which is S-dual and Q-dual is **self-dual**.

Actually, there are two definitions for a self-dual group. The one I've picked comes from Fuchs, Kertész, and Szele (1953). In the context of Abelian groups, there is an alternative definition, which Baer will roar at you about in these papers if you care.

# Examples and nonexamples

So, what do we know about Q-dual groups?

To begin with, Q-dual groups are rare. Here Derek Holt presents evidence that most groups are not Q-dual, where "most" is defined in the sense that if $g(n)$ denotes the fraction of isomorphism classes of finite groups of order $\leq n$ that are not Q-dual, $g(n)\to 1$ as $n\to \infty$. On the other hand, it's also important to note that S-dual groups are more rare than Q-dual groups, which is something that Ying points out.

Let's look at some examples, some from the comments, some from papers. Here are some groups which are Q-dual:

Finite simple groups

Symmetric groups

Hall-complemented groups (for each $H\leq G$, there is a $K\leq G$ with $H \cap K = \mathbf{1}$ and $G=HK$)

Higman-Neumann-Neumann type universal groups (see Higman, Neumann, and Neumann (1949))

The semidirect product $A\rtimes \langle z \rangle$ of an abelian $p$-group $A$ with a power automorphism $z$ of prime power order.

Some groups that are not Q-dual:

- Some simple examples: $C_3\rtimes C_4$, $C_4\rtimes C_4$, $C_5\rtimes C_4$ with a nontrivial center
- Any quasisimple group (that isn't simple)
- The commutator subgroup of a free group of rank $2$
- Finitely generated linear groups over a field characteristic $0$. More generally, by Selberg's lemma and Malcev's theorem, any torsion-free hyperbolic group that is residually finite is not Q-dual (which may constitute all torsion-free hyperbolic groups; see here)

Relating this to other properties,

- Q-dual groups may or may not be solvable and vice versa
- Q-dual groups may or may not be nilpotent and vice versa (this relationship discussed in a section below)
- Q-dual groups may or may not be $p$-groups and vice versa
- the Q-dual property is not subgroup closed (see example in last section)

# Reduction to smaller order

It's often desirable to throw away irrelevant parts of a group when studying a property.

Let $G$ be Q-dual. If there exists an element of prime order in the center that is not a commutator, then $G$ has a non-trivial, cyclic direct factor.

This dovetails nicely with the next theorem.

Let $G$ be Q-dual. If $G=H\times \langle x \rangle$, $H$ and $\langle x \rangle$ are Q-dual.

This lets us reduce away cyclic direct factors.

# Building examples of Q-dual groups

To build new Q-dual groups from old ones, take any Q-dual group $H$ that has a unique non-trivial minimal normal subgroup. Given a prime $p\not\mid |H|$, $H$ has a faithful irreducible representation on an elementary abelian $p$-group $C_p^{\; n}$, so we can build $G=C_p^{\; n}\rtimes H$, which is Q-dual by the theorem. You can keep going with that, tacking on as many primes as you like.

Next we have a large class of easily constructed Q-dual groups.

Let $A$ be an elementary abelian $p$-group and $\varphi\in\operatorname{Aut}(A)$ have prime order. Then $A\rtimes \langle \varphi \rangle$ is Q-dual.

It's possible that this family comprises all nonabelian Q-dual $p$-groups of class $2$ with exponent $p$, but this is open.

# Relationship to nilpotency

Ying's work focuses largely on nilpotent groups, based on the observation that the Q-dual condition is most pronounced in groups with many normal subgroups. Together with the (almost surely true) conjecture that most finite groups are nilpotent, this seems like a pretty good place to start.

A nilpotent group $G$ is Q-dual if and only if all of its Sylow subgroups are Q-dual.

(Actually this is proven in a paper by A.E. Spencer, which I have yet to get ahold of. I'll post a link when I do.)

This is fair enough, and lets us reduce to studying $p$-groups. From here, he delves into nilpotency class.

Let $G$ be an odd order Q-dual $p$-group of class $2$. Then $G'$ is elementary abelian.

The additional hypotheses that $p$ be odd and the nilpotency class be $2$ are important. The counterexample given is the dihedral group of order $16$, whose commutator subgroup is cyclic of order $4$, and which has nilpotency class $3$. It's important to note that this is a $2$-group, however, and it may be that this is what causes the generalization to fail, not the higher class. In particular, it is still open (as of 1973!) whether odd $p$-groups of class greater than $2$, *or* $2$-groups of class $2$, have elementary abelian commutator subgroups.

Let $G$ be an odd order Q-dual $p$-group of class $p$. Furthermore, suppose $\Omega_1(G)$ is abelian. Then $G=A\rtimes \langle z \rangle$ where $A$ is abelian, $z$ has order $p$, and $[a,z]=a^{\operatorname{exp}(A)/p}$ for all $a\in A$.

When $\operatorname{exp}(G)>p^2$, this becomes an if and only if.

Let $G$ be a $p$-group of class $2$ with $\operatorname{exp}(G)>p^2>4$. Then $G$ is Q-dual if and only if $G=A\rtimes \langle z \rangle$ where $A$ is abelian, $z$ has order $p$, and $[a,z]=a^{\operatorname{exp}(A)/p}$ for all $a\in A$.

That is a pretty thorough characterization of this special case. When $\operatorname{exp}(G)\leq p^2$, things get more complicated.

**Example.** Let $p$ be an odd prime, $|a|=p^2$, $|b|=|c|=p$, $[a,x]=a^p$, $[a,y]=b$, $[c,z]=a^p$, and all other commutators between $a,b,c,x,y$ and $z$ equal to $1$. Then $\left(\langle a\rangle\times \langle b\rangle\times \langle c\rangle\right)\rtimes \langle x,y,z\rangle$ is a finite Q-dual $p$-group of class $p$ and exponent $p^2$.

This example shows that the Q-dual property is not subgroup closed, via the subgroup $\langle a,c,x,z\rangle$. It also shows that finite Q-dual $p$-groups of class $2$ need not contain an abelian maximal subgroup. It is still open whether there are counterexamples of this nature for odd primes $p$.