I am reading Dummit & Foote, Abstract Algebra, 3e, p.103ff. We know that the first part of Jordan-Hölder program, the classification of finite simple groups, are finished. But it is not written whether the second part, roughly how to build any group from simple groups [edited] and cyclic groups, are finished.

I say that a finite group $G$ is tangible if: [edited] $G$ is a cyclic group, a finite simple group, or $G$ can be written as a semidirect (possibly direct) product $$ G \cong H_1 \rtimes \dotsc \rtimes H_M $$ with $H_m$ tangible, $m =1,\dotsc,M$. (Tell me if there is an existent terminology.) Note this has included all abelian groups.

They seem to say not all groups are tangible, because some do not have complementary subgroups (D&F p.180). Indeed, the quaternion group $Q_8$ is not tangible (D&F p.181). However, every finite group is the image of a homomorphism from a free group (D&F p.217), and can be presented as such quotient of the free group and some words (D&F p.218)

What are some other groups that are proved not tangible? If not, how can we say that Jordan-Hölder program is finished? Isn't it true that the existence of a composition series still does not concretely describe a group (D&F p.103)?

Edit: System recommends Smallest non-p-group that isnt a semidirect product, which answers my question.

  • $\begingroup$ Any cyclic group of order $p^n$, $p$ prime and $n>1$, is not tangible. $\endgroup$ – user1729 Feb 5 '20 at 13:11
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    $\begingroup$ Semidirect product is not associative, so it's dangerous to write a non-parenthesized expression $G \cong H_1 \rtimes \dotsc \rtimes H_M$. For example, nilpotent groups are groups of the form $(((H_1 \rtimes H_2) \rtimes H_3) \rtimes \dotsc \rtimes H_M)$ where each successive quotient $H_k \, / \, ((H_1 \rtimes H_2) \rtimes \dotsc \rtimes H_{k-1})$ is abelian, whereas solvable groups are groups of the form $H_1 \rtimes (H_2 \rtimes (H_3 \rtimes \dotsc \rtimes H_M)))$ where each successive quotient $(H_k \rtimes (H_{i+1} \dotsc \rtimes H_M)) \, / \, H_{k-1}$ is abelian. $\endgroup$ – Lee Mosher Feb 5 '20 at 13:31
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    $\begingroup$ Also, every group is the image of a homomorphism from a free group. $\endgroup$ – Lee Mosher Feb 5 '20 at 13:38
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    $\begingroup$ @Violapterin well, I understand that Dummit & Foote call it that way but it looks like an exaggeration to me. There were people working on classification of simple groups. There are people working on extension problem. There's no "program" as in: people actually grouping and working together on a problem under a leader(s). Or at least I've never heard of it. Or maybe I misunderstand the word "program". Either way, it doesn't really matter. $\endgroup$ – freakish Feb 5 '20 at 13:57
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    $\begingroup$ Possible duplicate : math.stackexchange.com/questions/451249/… $\endgroup$ – Arnaud D. Feb 5 '20 at 14:09

I think the fundamental issue you are facing is that you are confusing group extensions with semi-direct products. The Jordan-Holder theorem can be understood as the fact that every finite group can be built by successive extensions by simple groups. But certainly not semi-direct products.

If $G$ is a group, and $K$ is a normal subgroup, then we say that $G$ is an extension of $K$ by $Q=G/K$. In other words, an extension consists of a short exact sequence $$1\to K\to G\to Q\to 1.$$

A semi-direct product is a very special kind of extension where the short exact sequence is split: there is a section $s:Q\to G$ of the surjection $G\to Q$, where $s$ is a group morphism.

For instance, you had to edit your definition so $\mathbb{Z}/p^2\mathbb{Z}$ would be a "tangible" group. Of course $\mathbb{Z}/p^2\mathbb{Z}$ is an extension of $\mathbb{Z}/p\mathbb{Z}$ by itself, since we have the obvious exact sequence $$0 \to \mathbb{Z}/p\mathbb{Z} \to \mathbb{Z}/p^2\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}\to 0$$ but this sequence is not split! Otherwise, since $\mathbb{Z}/p^2\mathbb{Z}$ is abelian, we would have $\mathbb{Z}/p^2\mathbb{Z}\simeq \mathbb{Z}/p\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}$.

Now, given groups $K$ and $Q$, describing group extensions of $K$ by $Q$ is extremely difficult, which is why we cannot say that the classification of finite simple groups suddenly gave us a description of all finite groups. For instance, if we assume that $K$ is abelian, then those extensions are classified by: an action of $Q$ on $K$ by automorphisms, and a cohomology class in $H^2(Q,K)$ (the semi-direct product corresponds to the trivial class).

  • $\begingroup$ Btw, is $1 \mapsto W \mapsto F \mapsto G \mapsto 1$ also an exact sequence, where $W$ is the set of words in $G$ that gets canceled, and $F$ is the free words generated by generators of $G$? I am wondering that whether such manner of presentation (having $G$ in the 4th slot) is related to the extension which you describe here. $\endgroup$ – Violapterin Feb 5 '20 at 17:43
  • $\begingroup$ Yes, any quotient gives an exact sequence, more or less by definition. But I'm not sure I see the connection with the matter of extensions of finite groups. It was already a little out of,place in your question. $\endgroup$ – Captain Lama Feb 5 '20 at 17:57
  • $\begingroup$ I see. I think my comment above is indeed irrelevant. Why I am confused in the post itself, is that I suppose D&F sec. 6.3 on free groups, is part of the effort to construct finite groups, and I am mistaken that they may be some continuation of the Hölder program; I was surely overthinking. $\endgroup$ – Violapterin Feb 5 '20 at 18:01
  • $\begingroup$ I was wondering that how do we study those "non-tangible" groups, and we can only (in my limited knowledge) use group presentations, right? So I mentioned that in the post. And study of presentations surely sounds like a difficult discipline. $\endgroup$ – Violapterin Feb 5 '20 at 18:04

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