A common mistake for beginning group theory students is the belief that a quotient of a group $G$ is necessarily isomorphic to a subgroup of $G$. Is there a characterization of the groups in which this property holds?
If this question is too broad, I might ask if such a characterization exists for $p$groups.
History: I originally posed the opposite question, regarding groups for which $\exists N\unlhd G\,:\, \not\exists H \unlhd G\, \text{ s.t. } H \cong G/N$, and crossposted this to MO. I received an answer there to the (now omitted) peripheral question about probability, which shows that most finite groups probably have this property. After this, I changed the question to its current state, as this smaller collection of groups is more likely to be characterizable.
Answer
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: Qdual groups. There are several related definitions, the most relevant of which I will condense as follows.
An Sdual 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 Qdual 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 Sdual and Qdual is selfdual.
Actually, there are two definitions for a selfdual 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 Qdual groups?
To begin with, Qdual groups are rare. Here Derek Holt presents evidence that most groups are not Qdual, 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 Qdual, $g(n)\to 1$ as $n\to \infty$. On the other hand, it’s also important to note that Sdual groups are more rare than Qdual 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 Qdual:

Finite simple groups

Symmetric groups

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

HigmanNeumannNeumann 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 Qdual:
 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 torsionfree hyperbolic group that is residually finite is not Qdual (which may constitute all torsionfree hyperbolic groups; see here)
Relating this to other properties,
 Qdual groups may or may not be solvable and vice versa
 Qdual groups may or may not be nilpotent and vice versa (this relationship discussed in a section below)
 Qdual groups may or may not be $p$groups and vice versa
 the Qdual 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 Qdual. If there exists an element of prime order in the center that is not a commutator, then $G$ has a nontrivial, cyclic direct factor.
This dovetails nicely with the next theorem.
Let $G$ be Qdual. If $G=H\times \langle x \rangle$, $H$ and $\langle x \rangle$ are Qdual.
This lets us reduce away cyclic direct factors.
Building examples of Qdual groups
To build new Qdual groups from old ones, take any Qdual group $H$ that has a unique nontrivial 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 Qdual 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 Qdual 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 Qdual.
It’s possible that this family comprises all nonabelian Qdual $p$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 Qdual 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 Qdual if and only if all of its Sylow subgroups are Qdual.
(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 Qdual $p$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 Qdual $p$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 Qdual 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 Qdual $p$group of class $p$ and exponent $p^2$.
This example shows that the Qdual property is not subgroup closed, via the subgroup $\langle a,c,x,z\rangle$. It also shows that finite Qdual $p$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 $p$.
Attribution
Source : Link , Question Author : Alexander Gruber , Answer Author : Community