Is there a characterization of groups in which at least one subgroup is not an endomorphism kernel? - MathOverflow

Is there a characterization of groups in which at least one subgroup is not an endomorphism kernel?

2013-01-16T07:03:13Z

This is a crosspost from MSE:

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?

Answer by Derek Holt for Is there a characterization of groups in which at least one subgroup is not an endomorphism kernel?

2013-01-16T10:06:22Z

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:

http://mathoverflow.net/questions/21265/

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 $G$.

Answer by Alireza Abdollahi for Is there a characterization of groups in which at least one subgroup is not an endomorphism kernel?

2013-01-17T05:01:12Z

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).