Talk:Solvable group

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Polya's dictum : "if there's a problem you can't figure out, there's a simpler problem you can't (?) figure out" seems wrong. Moreover, the opposite sentence "if there's a problem you can't figure out, there's a simpler problem you can figure out" is obviously a reformulation from the works of René Descartes.

"as every simple, abelian group must be cyclic of prime order" seems to be wrong; actually as every simple, abelian group must be products of cyclic groups (may not be of prime order).

Every simple abelian group is cyclic of prime order. For an abelian group to be simple it must not have any proper non-trivial subgroups, because all its subgroups are normal. --Zundark 07:49, 9 Apr 2004 (UTC)

Zundark's right

Every commutater is of course even. So the commutator subgroup is A3. Am I missing something?Rich 10:06, 6 January 2007 (UTC)Reply

The commutator subgroup of ${\displaystyle S_{3}}$ is indeed ${\displaystyle A_{3}}$. The next term of the lower central series of ${\displaystyle S_{3}}$ is also ${\displaystyle A_{3}}$. So ${\displaystyle S_{3}}$ is not nilpotent. --Zundark 11:30, 6 January 2007 (UTC)Reply

I see, thankyou.Rich 16:41, 6 January 2007 (UTC)Reply

I noticed I've been wanting articles that talk about pi-nilpotence, pi-separable, pi-soluble, pi-constrained, etc. This article already has supersoluble, but I wasn't sure if generalizations were also good. These terms are all basically defined in terms of normal series where the factors are restricted in some obvious way. For p-soluble for instance, it just means that the chief factors with order divisible by p are in fact p-groups (so elementary abelian p-groups). If this is not the right article for them, what would be? I don't like short stubs that just define a term. It is easier just to define the term within a larger article, but even better to group related definitions together and give context, relations, examples, etc. —Preceding unsigned comment added by JackSchmidt (talk • contribs) 06:38, 14 July 2007

For finite groups, do the notions solvable and supersolvable coincide? —Preceding unsigned comment added by 129.70.14.134 (talk) 22:24, 30 September 2007 (UTC)Reply

No. The alternating group A₄ is solvable, but not supersolvable. --Zundark 07:57, 1 October 2007 (UTC)Reply

I'm pretty sure that solvable groups are not a variety. Just take a 1-solvable group, a 2-solvable group, a 3-solvable group, etc... and take their direct product. It's pretty easy to show that the result isn't k-solvable for any k. Maybe "pseudovariety" was meant? 128.208.1.222 (talk) 21:35, 9 September 2008 (UTC)Reply

I agree. I added the fact that metabelian groups form a variety, as well as the N-solvables for any N, but that the direct product of a sequence of solvable groups of unbounded derived length is not solvable. I think this is in Neumann's book, probably page 1, but I don't have it at hand. The unbounded derived length direct product fact is stated and used in Holt's Perfect Groups if someone desperately wants a reference. The gist is merely that the derived subgroup of a direct product is the direct product of the derived subgroup, and iterate. If you only need to do it finitely many times, then all is good, and if not, then all is bad. JackSchmidt (talk) 21:55, 9 September 2008 (UTC)Reply

To provide useful info for beginners of subject matter it is desirable to provide a rather obvious and quite "simple" example among more complex examples. Introduction to math sub/structures also include some trivial examples, which are basically useless, but show a mathematical principle on the most elementary sub/structure (e.g. zero vector space). It would be great to improve this by adding those changes suggesteted. Mathnovice (talk) 15:51, 26 November 2021 (UTC)Reply