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.

Alexander Gruber
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  • how do you define the probability space? – Koushik Dec 30 '12 at 09:20
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    Note that all quasisimple groups have the property. – Geoff Robinson Dec 30 '12 at 09:58
  • All finite abelian groups *don't* have the property, and non-abelian groups of order $\,p^3\,\,,\,\,p$ a prime have it. – DonAntonio Jan 15 '13 at 23:31
  • @DonAntonio: Are you sure about your statement for non-abelian groups of order $p^3$? $Q_8$ has the property, but a non-abelian group of order $p^3$ and exponent $p$ (implying $p>2$) doesn't have it. – j.p. Jan 16 '13 at 12:47
  • I made a mess with "the property", which by the OP must be that there exists a subgroup such that non-homomorphic image of the grous is isomorphic to it, so you are right: $\,Q_8\,$ does have it taking $\,\langle i\rangle\cong C_4\leq Q_8\,$. I confused things this since "the property" in this case is defined by negation: NOT having a subgroup...etc. Thanks. – DonAntonio Jan 16 '13 at 13:14
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    There is a discussion of this here: http://groupprops.subwiki.org/wiki/Quotient_group_need_not_be_isomorphic_to_any_subgroup – Douglas B. Staple Mar 11 '13 at 18:02
  • A rather trivial restatement of the property, but with obvious routes to generalization: consider all subnormal series $1 =H_0 \triangleleft H_1 \triangleleft \cdots \triangleleft H_n = G$ of $G$. The property now says that the set of top factors $G / H_{n-1}$ has an element that is not isomorphic to any in the set of bottom factors $H_1 / 1$. Are there any results on when composition factors are freely permutable? – yatima2975 Mar 13 '13 at 13:54
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    A genuinely quasisimple group $G$ , that is one with a non-trivial center, has a homomorphic image which is not a subgroup, namely $G/Z(G).$ – Geoff Robinson Apr 13 '13 at 17:10
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    If $G$ is a free group of rank $\ge 2$ then $G'$ has the required property. – Boris Novikov May 18 '13 at 12:07
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    @yatima2975 I don't think that is in fact a restatement unless every subgroup $H$ can be put into a subnormal series. This won't be the case in general, for instance in any finite non-nilpotent group. – Avi Steiner May 22 '13 at 06:11
  • Every torsion-free hyperbolic group $G$ has such a subgroup $N$ *if* all hyperbolic groups are residually finite. This is because all hyperbolic groups being residually finite is equivalent to all hyperbolic groups having a subgroup of finite index. If $G$ is torsion-free hyperbolic and $N$ is a finite-index subgroup then $G/N$ is torsion and so cannot embed into $G$. – user1729 Jul 02 '13 at 19:42
  • Take any vector space $V$, and $W \subset V$ a subspace of $V$, and let $W'$ be the orthogonal complement of $W$, $W' \cong V/W$. – user614287 Nov 27 '18 at 22:09
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    I found this paper https://link.springer.com/article/10.1007%2FBF01228254 – Charles Hudgins Apr 25 '19 at 18:48

1 Answers1


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.


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 $p$.

Alexander Gruber
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