This question originates from a recent question of Jafri and its answers. Notice that if G is a finite abelian group, then any quotient gp of G is isomorphism to some subgroup of G. The same is true for all finite simple groups. So I was wondering whether this phenomenon also occurs for some other groups. If not precisely for all normal subgroup, is it true for lots of normal subgroups?
I have just noticed that finite symmetric groups have this property. So please avoid these groups. Most probably group with small number of normal subgroups may have this property. But it'll be nice to see more examples.
$S_3$ is an example, and hence any direct product of it with an abelian group.
Very nice! Though there are examples, obvious as well as non-obvious, it'll be interesting (at least to me ) to look into the problem further. I just want to add that a non-abelian group G of order $p^3$ and exponent $p$ has this property. Moreover, if we take an extra-special $p$-group of exponent $p$ and order $p^{2n+1)$, where $n$ any positive integer, then for any normal subgroup N of G having order at least $p^n$, G/N is isomorphic to a subgroup of G.
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Gareth A. Jones
University of Southampton
Let G be the cartesian product of k non-abelian simple groups. This has 2^k normal subgroups, namely the products of subsets of these k factors. Their quotients are cartesian products of the complementary sets, hence isomorphic to subgroups of G.
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1 Recommendation
Saeid Jafari
Shahed University
By Schur-Zassenhaus theorem if N be a normal subgroup of G and (|N|,|G/N|)=1 then G is a semidirect product of N and G/N. i.e. G/N is isomorphic to a subgroup of G.
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Manoj K. Yadav
Harish-Chandra Research Institute
I found the following paper in which self dual groups are classified.
An, Lijian; Ding, Jianfang; Zhang, Qinhai, Finite self dual groups. J. Algebra 341 (2011), 35–44.
A group G is said to be s-self dual if every subgroup of G is isomorphic to some quotient of G. G is said to be q-self dual if every quotient of G is isomorphic to some subgroup of G. G is called self dual if it is both s-self dual as well as q-self dual. The paper contains the following question:
Question. What is the structure of q-self dual groups?
Last year Janko published the following paper:
Janko, Zvonimir, Finite nonabelian p-groups all of whose subgroups are q-self dual.
J. Algebra Appl. 13 (2014), no. 6, 1450008, 7 pp.
Since this is somehow related to the above discussion, I thought it is good idea to mention this here.
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Thong Nguyen Quang Do
Université de Franche-Comté
Dear Manoj,
May I say that your question of Nov.22, 2014, is not precise enough ? You ask about (non abelian, non simple) finite groups G such that for "most" normal subgroups N, the quotient Q = G/N is isomorphic to a subgroup H of G. But if you don't require any condition on the isomorphism in question, this doesn't make much sense. For instance, take G to be a p-group, p being a prime. Then G is solvable, in particular it admits subgroups H of order p, as well as subgroups N of index p, and all these N are automatically normal. Then all the H's above are isomorphic to all the G/N's, because all groups of order p are isomorphic (your G of order p^3 and exponent p belongs to this category of examples). So you must actually put conditions on the isomorphism in your question. The most natural one would be : the subgroup H should be iso. to the quotient Q via the canonical surjection s : G ---> Q. This is easily seen to be equivalent to the existence of a homomorphism j : Q ---> G such that s.j is the identity of Q. Then H = j(Q) is naturally iso. to Q. This is the classical definition of a semi-direct product. A particular case is given by the Schur-Zassenhaus theorem, as quoted by Saied Jafari. ¤
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Manoj K. Yadav
Harish-Chandra Research Institute
Dear Thong nguyen quang,
Thanks. One can very well make the situation more precise by imposing appropriate hypotheses. The situation your have suggested seems interesting. But again this is not precise. If one wants this to happen for all normal subgroups of the group, then it will be a subclass of q-self dual groups. The phrase 'most of normal subgroups' may have different meanings depending on the person. One can make it precise by taking n %, where n may be such that $80 < n < 100$ (say). The situation when $n = 100$ has already been studied in the literature as I have mentioned above.
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