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In this question, I asked whether a finite group $G$ could always be expressed as the semidirect product of its commutator subgroup $[G,G]$ and the abelian quotient group $G/[G,G]$. The answer is no.

So now we move on to the more general question, which is whether there is any nice characterization of the extension of $G/[G,G]$ by $[G,G]$ that equals $G$. As I am interested in $G$ nonabelian, I realize the general extension problem is very hard. But I am hoping that working with the commutator subgroup makes this problem easier. The research I have done so far is not encouraging, however, so I can only be pleasantly surprised at this point!

Motivation: It is tempting to think of the derived series of a finite group as decomposable into a "product" of abelian communtator quotient groups:

$G=G/G'\times G'/G''\times ... \times G^{(n-1)}/G^{(n)} \times G^{(n)}$

where $G^{(n)}$ is the perfect group where the derived series terminates. If $G^{(n)}$ is not trivial, then $G$ is not solvable (and $G^{(n)}$ is necessarily nonabelian), and $G$ is solvable if $G^{(n)}$ is trivial. In the latter solvable case, all the "factors" are abelian, but the "product" can be nonabelian. If we could characterize the "product" as a semidirect product, then this interpretation might have some merit, but we cannot. But if the extension I am asking about can be characterized, this idea might be salvagable: that solvable groups are not "too nonabelian" because their "product" has no nonabelian "factors".

Thanks!

More motivation: DonAntonio wonders what I am up to. I believe that non-mathematicians hate/fear math mostly because it is boring. But what is boring is all the preliminary stuff you need to get through to see the beautiful stuff. Unfortunately, the beautiful stuff can be hard going without the preliminaries. So I am trying to develop an accessible presentation of Galois Theory: not to teach it thoroughly, but to show that there is really cool math hidden from view. There are many steps to this. This question deals with one of the last steps: why does the quintic formula become "too complex" to exist? The answer is that it is "too nonabelian": it contains a nonabelian "factor" that can't be eliminated (the perfect group of the derived series). But there are groups that are solvable and nonabelian: they have abelian "factors" but nonabelian "products". This all sounds good, but I'd like it to actually be correct! I was hoping that semidirect products would fill the bill, but they don't. At best, the characterization of the "product" of commutator quotient groups is going to be more complicated that I hoped, and at worst there will be no such characterization. Even though I am trying to popularize, and am trying to simplify, I don't want to be wrong.

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For solvable groups you have two very different cases: nilpotent and non-nilpotent. If $G$ is nilpotent (and only if, assuming $G$ is finite) then $[G,G]$ is contained within the Frattini subgroup. Such extensions are called “Frattini extensions” and are diametrically opposed to semi-direct products (for $K$ a normal subgroup of $G$, not only is there no $H$ with $H \cap K=1$ and $HK=G$, there is not even a $H \neq G$ with $HK =G$).

Dichotomy

Not all extensions are either semi-direct or Frattini, but if we require $K$ to be a minimal normal (abelian) subgroup then we do get this dichotomy. The $K$ that give semi-direct products are called complemented and the $K$ that give Frattini extensions are called Frattini.

Every solvable group is built up by a series (refining the derived series) of normal subgroups $K_i$ such that $K_i/K_{i+1}$ is a minimal normal subgroup of $G/K_{i+1}$. Each $K_i/K_{i+1}$ is either complemented or Frattini. These factors $K_i/K_{i+1}$ are called chief factors and are an important method of studying solvable groups.

Failure

If $G$ is not solvable, then there can be factors that are neither Frattini nor complemented. If $G$ is infinite, it need not have chief factors at all.

Even if $G$ is finite solvable, $[G,G]/1$ need not be a chief factor, so the extension of $G/[G,G]$ by $[G,G]$ may be neither complemented nor Frattini (the dihedral group of order 24 gives a reasonably obvious example, being both a version of the complemented $S_3$ and the Frattini $D_8$).

Adjust the question

There are a couple of things that might be worth changing in the question (not on stackexchange, but in your quest).

In Galois theory, solvable groups are not so much the groups with terminating derived series, but rather with a subnormal series with finite cyclic quotients (polycyclic groups) corresponding to taking $n$th roots, for $n$ the size of the quotient. Unfortunately, in general such a series has very little structure (it is basically a composition series).

The products you describe are generally called extensions, and in some ways extensions are associative, but actually several of the special types of extensions are not “associative.” For example, consider $G/H \times H/K \times K/L$, you might have that $G/H \times H/K$, $H/K \times K/L$, and even $(G/H \times H/K) \times K/L$ are all central extensions, but $G/H \times (H/K \times K/L)$ is not a central extension. This happens in $D_8$ and $Q_8$ for example. You'll need to decide where the parentheses go, or you'll need to phrase the question so parentheses don't matter. The parentheses can make a huge difference. If you do $(((ab)c)d)e$ then prime factorization is called a chief series, but if you do $a(b(c(de)))$ then a prime factorization is called a composition series. I'm more fond of the chief series approach.

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  • $\begingroup$ Great answer, and this looks like it may be as close as I can get. I was hoping through the derived series that the extension problem was simpler, but it seems not in general. It looks like the chief series is actually what I was after, and since I am only interested in finite solvable groups, this always exists, correct? $\endgroup$ – user452 Dec 10 '13 at 0:59
  • $\begingroup$ Yes, chief series always exist for finite groups. For finite solvable groups, the factors are always abelian groups (of exponent $p$) and each one is either Frattini or complemented. They are similar to composition series, but keep much more information about the group (except that for nilpotent groups, every chief series is a composition series). $\endgroup$ – Jack Schmidt Dec 10 '13 at 2:31
  • $\begingroup$ Perfect, this is likely as close to the characterization I was looking for. The characterization of Frattini subgroups as consisting of "non-generators" is also very helpful. Thanks again! $\endgroup$ – user452 Dec 10 '13 at 2:35
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    $\begingroup$ Yup, and the complemented factors count how many generators the group needs. I think that is in Robinson's group theory textbook. I think the original paper was "Eulerian functions for finite solvable groups" by Hall (1936), and then the same basic title by Gaschütz (1959) ams.org/mathscinet-getitem?mr=107670 $\endgroup$ – Jack Schmidt Dec 10 '13 at 2:38
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I'm not sure what are you up to, but observe that a rather "nice" (who decides?) group fulfills your conditions: the symmetric group $\;S_n\;$, since

$$S_n=A_n\rtimes C_2\;,\;\;\text{and}\;\;S_n'=A_n\;,\;\;S_n/S_n'=C_2$$

So you can always say that at least, by Cayley's Theorem, any finite group is a subgroup of a group with the form you want.

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  • $\begingroup$ But $S_n$ is not solvable for $n>4$. I'm trying to characterize finite solvable groups as abelian "products" for a suitable "product", which may not exist. So this may be futile. $\endgroup$ – user452 Dec 4 '13 at 0:35
  • $\begingroup$ Where in your question do you mention "characterizing" solvable groups? You only mention them as some cases and the derived series and etc. $\endgroup$ – DonAntonio Dec 4 '13 at 4:45
  • $\begingroup$ Isn't a group with trivially terminating derived series solvable? I'm focusing on the derived series because of the way commutator subgroups measure how nonabelian is a group. $\endgroup$ – user452 Dec 4 '13 at 12:11
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Well, probably not that useful (or insightful), but it's guaranteed if $[G,G]$ and $G/[G,G]$ have coprime order.

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  • $\begingroup$ Just to be clear, because I think you mentioned it in my last question, what's guaranteed is we have a semidirect product for coprime order. Yes? $\endgroup$ – user452 Dec 4 '13 at 2:24
  • $\begingroup$ Yes, it follows directly from (Philip) Hall's Theorem, which is an extension of Sylow's Theorem. $\endgroup$ – Doc Dec 4 '13 at 2:30
  • $\begingroup$ Nicely argued. ${}{}$ $\endgroup$ – mrs Dec 4 '13 at 2:44
  • $\begingroup$ @Doc, thinking about this more overnight, this is very helpful. You have now reduced the problem to looking at the case where $[G,G]$ and $G/[G,G]$ have orders with a common prime factor. That might be a hard problem, but it's progress. $\endgroup$ – user452 Dec 4 '13 at 12:28
  • $\begingroup$ @trb456 I do have a small bit to add in the case in which the commutator subgroup and the abelianization are not of coprime order, but it's a very special case. I'm posting it as another answer. $\endgroup$ – Doc Dec 4 '13 at 14:22
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Suppose $N$ is a normal abelian $p$-group of $G$, and that $p\mid |G/N|$. Then $G$ splits over $N$ if and only if $P$ splits over $N$, where $P$ is a Sylow $p$-subgroup of $G$.

This is a theorem of Gashutz.

So let's imagine we're in the following situation:

For some Sylow $p$-subgroup $P$ of $G$ we have (i) $P$ normal in $G$, (ii) $[P,P]=[G,G]\cap P$, (iii) $[P,P]$ is abelian. (Note that we assume that $p$ divides both the orders of $[G,G]$ and $G/[G,G]$, as we've already handled the case where the orders are coprime.)

Suppose $G$ splits over $[G,G]$, say $G=[G,G]\rtimes K$. Then $P=[P,P]\rtimes (K\cap P)$. Thus, by Gashutz, we may eliminate all solvable $G$ that satisfy the above conditions for some $p$, provided the Sylow $p$-subgroup for that $p$ does not split over its commutator subgroup.

This is a work in progress. I know I can extend this argument, though I'm not sure by how much.

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