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Is there a characterization of groups with the property $\exists N \unlhd G : \not \exists H \leq G \;\; \text{s.t.} \;\; H\cong G/N$?


A common mistake among 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 for groups in which this assumption fails?

If this question is too broad, I might ask if such a characterization exists for $p$-groups.

A peripheral question is how likely it is for a finite group to have this property, which I would formulate in the following way: if $g(n)$ denotes the fraction of isomorphism classes of finite groups of order $≤n$ for which the property holds, what happens to $g(n)$ as $n\rightarrow \infty$? (Here is what $g$ looks like for small values.) If we again restrict to $p$-groups, can we say anything about the fraction of isomorphism classes of order $p^k$ with the property, e.g. a bound in terms of the exponent?

EDIT I believe Derek Holt's answer satisfies the peripheral question. The primary question is still open - can we say anything to characterize these groups?

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Isn't the better question the other way round: "Is there a characterization of groups in which all normal subgroups are endomorphism kernels"? –  Stefan Kohl Jan 16 '13 at 9:08
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Suggestion: How about calling a group $G$ boring if all its quotients are isomorphic to some subgroup of $G$? So you are looking for the characterization of non-boring groups. –  j.p. Jan 16 '13 at 12:51
    
Can you say anything about groups that are non-boring for an abelian quotient? (like $Q_8$?) –  j.p. Jan 16 '13 at 12:54
    
A minor comment: All nonelementary hyperbolic groups $G$ have this property. The reason is that such groups are SQ-universal (due to Delzant and Olshansky, who proved this independently). Thus, say, $Z^2$ embeds in a quotient $G/N$, which implies that $G/N$ cannot embed in $G$. Another class consists of infinite finitely generated linear groups over ${\mathbb R}$. The reason is that such groups $G$ are residually finite, so have finite quotient groups of arbitrary order. But Selberg's lemma limits orders of finite subgroups of $G$. –  Misha Jul 9 '13 at 5:29
    
Infinite residually finite groups whose finite subgroups have bounded order have this property (any too big finite quotient can't be a subgroup). This includes infinite f.g. linear groups in characteristic 0 and infinite f.g. hyperbolic groups (as already mentioned by Misha using another argument). But it looks like the question is especially focused on finite groups. –  Yves Cornulier Aug 28 '13 at 8:45
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3 Answers

I would guess that $g(n)$ approaches 1 as $n$ approaches infinity.

The topic of the growth rate of the number of isomorphism classes of groups of order $n$ has been discussed previously, for example in:

How many groups of size at most n are there? What is the asymptotic growth rate? And what of rings, fields, graphs, partial orders, etc.?

Both from looking at these results and from known results for $n \le 2000$, there is very strong evidence that most finite groups are 2-groups. Unfortunately, the imprecision in the results proved is big enough to make it very unlikely that this will be proved any time soon.

The known lower bound on the number of groups of order $2^n$, due to G. Higman, which is about $2^{2n^3/27}$, is proved by considering groups $G$ in which $G'=Z(G)$ and $G/Z(G)'$ and $Z(G)$ are both elementary abelian 2-groups, and where $|Z(G)|^2$ is approximately equal to $|G/Z(G)|$. Sims proved that the total number of 2-groups of a given order is roughly the same as this, but the error term is in the exponent, so it has not been proved that almost all 2-groups are of this form, and that might not be correct.

But looking at groups $G$ of that form, we can see that they will virtually all have quotient groups that are not isomorphic to subgroup - I am sure that claim could be proved. If you quotient out a central subgroup of order 2, then you will usually end up with another $n$-generator 2-group $H$ of order $|G|/2$ with $Z(H) = H'$, and such an $H$ cannot be subgroup of $G$, because any $n$ elements of $G \setminus G'$ that generate $G/G'$ also generate $G$.

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I had read about that result and figured this might be the case. Does this central subgroup technique extend to odd primes (e.g. looking at $g(p,k)=$ the fraction of groups of order $p^k$ with the property)? –  Alexander Gruber Jan 17 '13 at 20:34
    
Yes, the results of Higman and Sims on estimating numbers of groups of order $p^n$ apply to all primes $p$. –  Derek Holt Jan 17 '13 at 21:37
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In short, I do not think there is a reasonable characterization of such groups, beyond restating the definition you gave. Here is why. First, let's give the property of $G$ that every quotient $G/N$ embeds in $G$ "self-universality" (SU).

First, note that there are three classes of groups which have SU:

  1. Simple groups (and groups obtained from them via semidirect products).

  2. Higman-Neumann-Neumann type universal groups: They proved existence of 2-generated groups $G$ which contain every countable subgroup. In particular, such $G$ contains every quotient of $G$.

  3. Note that in (1), the group has too few quotients, while in (2) it has too many subgroups. One can have infinite f.g. groups which have both too few quotients and too few subgroups, like Tarski monsters (every nontrivial proper subgroup is isomorphic to $Z_p$).

Given the diverse reason for these groups to have SU, there does not seem to be any "unifying reason" for the SU property.

Note, however, that, at least among finitely-generated groups, all the examples 1-3 are mostly "artificial"; e.g.: most "naturally occurring" f.g. groups tend not to be simple (or universal in HNN sense, or be "monsters"). Among "natural" f.g. groups, the ones which do not have SU tend to dominate:

a. F.g. (infinite) matrix groups (over the real numbers) do not have SU, by a combination of Malcev's theorem and Selberg's lemma: If $G$ is such a group, it has finite quotients of arbitrarily high orders (Malcev), but $G$ contains only finitely many isomorphism classes of finite subgroups (Selberg).

b. Infinite hyperbolic groups do not have SU: For virtually cyclic groups this follows from (a), while for nonelementary groups it follows from Delzant and Olshansky who proved SQ universality for such groups. In particural, such $G$ always has a quotient containing $Z^2$ and, hence, cannot embed in $G$.

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Search morphic groups. Some papers of Y. Li from Brock University.

EDIT: Thanks to Misha. Let me recall the definition of a morphic group. A group $G$ is called morphic if every endomorphism $\alpha$ of $G$ for which $G\alpha$ is normal in $G$ satisfies $G/G\alpha=\ker(\alpha)$. Let me call the groups in question Quotient-closed groups. It seems that the class of morphic groups and the one of quotients closed groups are different. There is another class of groups introduced in [Journal of Pure and Applied Algebra 214 (2010) 1827--1834] called strongly morphic: For finite abelian groups, it is mentioned in the latter article that (strongly morphic)=(quotient-closed). So a finite abelian group $A$ is non-boring if and only if each Sylow subgroup (primary component) of $A$ is homocyclic (i.e., direct product of cyclic groups of the same order).

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@Alireza: It is a similar but different concept. –  Misha Jan 17 '13 at 6:21
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