Is there a characterization of groups in which at least one subgroup is not an endomorphism kernel? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T05:07:32Z http://mathoverflow.net/feeds/question/119045 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119045/is-there-a-characterization-of-groups-in-which-at-least-one-subgroup-is-not-an-en Is there a characterization of groups in which at least one subgroup is not an endomorphism kernel? Alexander Gruber 2013-01-16T07:03:13Z 2013-01-25T20:10:16Z <p>This is a crosspost from MSE: </p> <p><a href="http://math.stackexchange.com/q/267521/12952" rel="nofollow">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></p> <hr> <p>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?</p> <p>If this question is too broad, I might ask if such a characterization exists for $p$-groups.</p> <p>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 $n\rightarrow \infty$? (<a href="http://i.stack.imgur.com/Lpc71.png" rel="nofollow">Here</a> 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?</p> <p><em>EDIT</em> I believe Derek Holt's answer satisfies the peripheral question. The primary question is still open - can we say anything to characterize these groups?</p> http://mathoverflow.net/questions/119045/is-there-a-characterization-of-groups-in-which-at-least-one-subgroup-is-not-an-en/119048#119048 Answer by Derek Holt for Is there a characterization of groups in which at least one subgroup is not an endomorphism kernel? Derek Holt 2013-01-16T10:06:22Z 2013-01-17T08:56:15Z <p>I would guess that $g(n)$ approaches 1 as $n$ approaches infinity. </p> <p>The topic of the growth rate of the number of isomorphism classes of groups of order $n$ has been discussed previously, for example in:</p> <p><a href="http://mathoverflow.net/questions/21265/" rel="nofollow">http://mathoverflow.net/questions/21265/</a></p> <p>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.</p> <p>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.</p> <p>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 $G$.</p> http://mathoverflow.net/questions/119045/is-there-a-characterization-of-groups-in-which-at-least-one-subgroup-is-not-an-en/119137#119137 Answer by Alireza Abdollahi for Is there a characterization of groups in which at least one subgroup is not an endomorphism kernel? Alireza Abdollahi 2013-01-17T05:01:12Z 2013-01-25T20:10:16Z <p>Search morphic groups. Some papers of Y. Li from Brock University.</p> <p>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).</p>